MHF4UI Textbook - Chapter 2 Polynomials (pg. 37-70) PDF

I ProbIemSo1ving: Using the Strategies 1. A school’s debating team consists of four cross, and it takes Larry 8 mill. If they cross in females and four males. The team captain is a pairs, they cross at the speed of the slower female. The team is entered in a regional...

I ProbIemSo1ving: Using the Strategies 1. A school’s debating team consists of four cross, and it takes Larry 8 mill. If they cross in females and four males. The team captain is a pairs, they cross at the speed of the slower female. The team is entered in a regional person. How can they all cross and all catch competition. Howevei each school is allowed to the train? send only six team members, three males and 4. The sum of the numbers of sides of two three females, to the competition. The team convex polygons is 11, and the sum of the captain must be among the members who numbers of their diagonals is 14. Name the two compete. Other members are equally qualified types of polygons. to compete. How many possible combinations of members could make up the team that 5. A cube is 20 cm on an edge. competes? a) How can you cut the cube into six identical pyramids? 2. You have a 5-L container and a 9-L b) What are the dimensions of each pyramid? container and plenty of water. You want exactly 6 L of water in the 9-L container. How can you 6. Must each calendar year have at least one use the two containers to measure exactly 6 L Friday the 13th? Explain. of water? 7. A circle is inscribed in an equilateral 3. Four people need to cross a footbridge at triangle, and a square is inscribed in the circle. night to catch a train, which leaves in more than 15 mm but less than 16 mm. The bridge can only hold two people at a time. The group has one flashlight, which must be used for each crossing of the bridge. The flashlight must be carried by hand and cannot be thrown back. Because of their different degrees of nervousness, the people take different times to cross the bridge. Juan can cross in 1 mm. Sue Determine the exact value of the ratio of the can cross in 2 mill. It takes Alicia 5 mill to area of the triangle to the area of the square. Problem Solving: Using the Strategies MHR 37 II  I Chapter 2 Polynomials Specific Expectations Section Determine, through investigation, using graphing calculators or graphing software, various properties of the graphs of polynomial functions. 21 2 6 Compare the nature of change observed in polynomial functions of higher degree with that observed in linear and quadratic functions. 2 1 2, 2.3, Solve problems involving the abstract extensions of algorithms. Demonstrate an understanding of the remainder and factor theorems. 2.3, 2.4 Factor polynomial expressions of degree greater than two, using the factor theorem. 2.4 Determine, by factoring, the real or complex roots of polynomial equations of degree greater than two. 25 Determine the real roots of non-factorable polynomial equations by interpreting the graphs of the corresponding functions, using graphing calculators or graphing software. 25 Write the equation of a family of polynomial functions, given the real or complex zeros. 2.5 Sketch the graph of a polynomial function whose equation is given in factored form. 2.6 Solve factorable polynomial inequalities. 2.6 Solve non-factorable polynomial inequalities by graphing the corresponding functions, using graphing calculators or graphing software and identifying intervals above and below 2.6 the x-axis. Describe the nature of change in polynomial functions of degree greater than two, using 27 finite differences in tables of values. Determine an equation to represent a given graph of a polynomial function, using methods apptopriate to the situation. 2.8 ‘S. M0 -4 F- l -m  Review of Prerequisite Skills 1. Simplifying expressions 6. Solving quadratic equations by graphing Expand and simplify. Solve by graphing. Round solutions to the a) 5(2x + 3) + 4(x 7) — nearest tenth, if necessary. b)2(y-3)-(y+4)-3(3y+5) a)2y2=y+3 c) 5(x2—x—2)—2(x2+3) b) 4t2+9=llt U) 2(w—3)2—(w+5)(2w—4) c) 2m2—Sm=O U) 6w —4=5w 2. Dividing a polynomial by a monomial Simplify each of the following. 7. Solving quadratic equations by factoring a’ 6x2 + 9x Solve by factoring. Check your solutions. 3x a) w2=8w 4x 3 —8x 2 —12x b) y2+7y+1O=O b c) x—x—21=O LX U) t2+20=9t c) 8m+16mn 4ii 8. Solving quadratic equations by factoring 9x4 —6x3 +3x2 —12x Solve by factoring. Check your solutions. U) a) 2x2—Sx+2=O 3x 2 b) 2y +7y+3=O 2 3. Factoring ax + bX + c, a 2 = I c) 3w2 2 = w — Factor. d) 6t2 + 13t = 5 a) t+9t b) w2 196 — 9. The quadratic formula c) x2 + 8x + 15 Solve using the quadratic formula. Express d) y2 — 18 — solutions both in exact form and as e) x2 + 15x + 56 approximations to the nearest tenth. f) w2—38w+361 a) X—X—5O g) t2—98t-99 b) y2+3x+1=O_— h) s2+90s+89 c) 2w2+w=4 U) O=4t2+2t—1 4. Factoring ax 2 + bx + c, a 1 Factor. JO. Complex numbers a) 2x2 7x + 3 — Simplify. b) 3y2 — — a) J4 b) J—ioo c) 4w2 +9w +2 c) U) U) 10a2-a-3 e) f) — e) 6t +7t+2 1) 25x 2 g) 1 h) 6ixS; — i) 6i2 j) (j’J)2 5. Simplifying rational expressions k) (—2i)(3i) I) (3iJ)(—21J) Simplify. State any restrictions on the variables. + x —2 11. Complex numbers a) Simplify. x—1 2 a) s+P b) x2_6x+9 b) _3_ c) —t—6 c) t2—2t—8 2 d) a29a+20 f4+%J U) a2 +a 30 7 40 MHR Chapter 2 12. Operations with complex numbers 18. Solving first-degree inequalities Simplify. Solve. Graph the solution on a number line. a) (3+2i)+(7—3i) a) 2x—4—4(2x—1)+6 e) 1.2x—0.1x—2 23 b) 52t x—2 g) ——1x+—1 c)L U) —5—41 2+3x 3x—2 2 1—i 1—21 2 > 6 3 14. The quadratic formula 19. Finite differences Find the exact solution using the quadratic Use finite differences to determine whether each formula. function is linear, quadratic, or neither. a) x2+3=2x b) y2+90 a) b) c) d) c) t2—t+3=0 x y x y x y xIIy U) 2m2+3m=—3 0 —5 0 —3 0 4 0 19 15. Evaluating functions 1 —3 1 —1 1 5 1 8 If f(x)= 7x—4, find 2 —1 2 5 2 12 2 5 a) f(2) b) f(—7) c) ff0.5) 3 1 3 15 3 31 10 3 U) f(—1.5) e) 1(200) f) f(—150) 4 3 4 29 4 68 4 23 16. Evaluating functions 2x—1 e) f) g) h) If h(x)= 2 x y x y x y x y a) h(7) b) h(0) c) h(0.5) 0 —4 —2 —27 —4 33 —3 —5.5 U) hf—6) e) h(—1.5) f) h(0.01) 2 —14 —1 —13 —2 17 —1 —2.5 17. Evaluating functions 4 —24 0 —11 0 9 1 0.5 Ifg(x)=x2 +2x—5, find 6 —34 1 —9 2 9 3 3.5 a) g(0) b) g(3) c) gf—2) U) g(0.6) e) g(—2.5) f) g(100) 8 —44 2 5 4 17 5 6.5 Historical Bite: Coordinate Systems Pierre de Fermat (1601—1665) was probably the first person to use the idea of a coordinate system. However, René Descartes (1596—1650) is generally given the credit for this invention. It is said that Descartes came across the idea while he was in military service. When he was lying on his cot one day, he stared at a fly that was hovering above him. It occurred to him that the fly’s position at any moment could be described by its distance from three mutually perpendicular intersecting lines, which we now call axes. Review of Prerequisite Skills MHR 41 Investigating Math: Polynomial Functions on a Graphing Calculator You have previously studied constant, linear, and quadratic functions. Equations for these functions have the form 1(x) = c, 1(x) = mx + b, and f(x) = ax2 + bx + c. These functions are special cases of a class of functions called polynomial functions. Other examples of polynomial functions are f(x) = 2x2 + 3x 1, f(x) = x3 + 7, and f(x) = 2x4 3x3 + 14. — — Examples of functions that are not polynomials are 1(x) = f(x) = 2x, and f(x) = x3. A polynomial function of degree ii, where n is a positive integer, has the form ii Il—I f(x) = a,,x + a1x +... + ax + a0 where a,2 0. The numbers a,,, a,,1,..., a1, a0 are called the coefficients of the polynomial. The number a0 is the constant coefficient. The number a,,, the coefficient of the highest power, is the leading coefficient. In this section, we explore the features of the graphs of polynomial functions. rMLK7J , Quadratic functions are polynomial functions of degree 2. They are also called second-degree functions. Quadratic functions have two zeros, that is, two values of the ETEI variable that make the value of the function zero. The corresponding quadratic equations can have two real or two complex roots. If the roots are real, they can be distinct or equal. - II Fi rn ‘—I —. I; Two complex roots Two distinct real roots Two equal real roots When working with functions of degree greater than 2, it is helpful to know that a function of degree n has n zeros. For example, a cubic function, such as y = x3 + 3x2 3x 4, — — has three zeros. Cubic functions are also called third-degree functions. Window variables: x e [—4.7, 4.7], y a [—9.3, 9.3] Cubic Functions 1. Graph each cubic (third-degree) function in the standard viewing window, using the ZStandard instruction. Sketch the graphs in your notebook. a) y=x 3 +2x 2 —3x—4 b) y=x 3 +2x 2 —4x—8 c) y=x 3 —2x 2 —2x—3 U) y=x3—3x2+3x—1 e) y=—x3—3x2+x+3 f) y=—x3+3x2—Sx+6 g) y=—x3+x2+Sx+3 h) y=—x3+3x2—3x+1 42 MHR Chapter 2 2. Using the graphs from step 1, describe the general shape of the graphs of cubic functions. 3. How is the shape of the graph of a cubic function different from the shape of the graph of a quadratic function? 4. The general form of the equation of a cubic function is y = ax3 + bx2 + cx + d, where a 0. Describe how the graphs of cubic functions for which a is positive differ from those for which a is negative. 5. What does the value of d represent on the graph of a cubic function? 6. a) Graph the cubic function y = x3 2x2 Sx + 6. Find the x-intercepts. Sketch the — — graph and label the x-intercepts. b) Do the x-intercepts from part a) satisfy the related cubic equation x3 2x2 Sx + 6 = 0? — — Explain. 7. The real roots of a cubic equation are the x-intercepts of the related cubic function. How many real roots does the cubic equation corresponding to each cubic function graphed in step 1 have? 8. State the possible numbers of each of the following types of roots for a cubic equation. Sketch a graph to illustrate each possibility. a) distinct roots b) equal real roots c) complex roots 9. Can a cubic equation have three complex roots? Explain. 10. A cylinder is inscribed in a sphere, as shown. The radius of the spheie is 2 units a) Write a formula for the volume, V, of the cylinder in terms i1 of x. What kind of function is V? — ‘ x — — b) Argue that negative values of x are reasonable, provided that 1 0 we think of x as a coordinate and not as a distance. What is the j. domain of the function in part a)? c) Determine the roots of the function in part a). Explain the 1 - meaning of the roots. Ii. Summarize what you have learned about cubic functions and their graphs. Quartic Functions A quartic function, such as y = x4 — 6x2 — x + 3, has four zeros. Quartic functions are also called fourth-degree functions. P1.’i P1.12 P1’ I1 X”4—6X’2—X+3II I Window variables: x a [-4.7, 4.7], y a [—9.3, 9.3] 1. Graph each quartic function in the standard viewing window. Sketch the graphs. a) y=x4—5x2+2x+2 b) y=x4+3x3—x—3 c) y=x4+2x3+2x+6 U) y=x4—4x3+6x2—4x+1 e) y=x4—2x2+1 f) y=—x4+5x2+4 2.1 Investigating Math: Polynomial Functions on a Graphing Calculator MHR 43 ZZJI g) y—x4+x3±3x2—2x—5 h) y=—x4—5x3—5x2+5x+6 i) y=—x +3x +3x —7x—6 j) y—x —4x —5x —4x—4 2. Using your graphs from step 1, describe the general shape of the graphs of quartic functions. 3. How is the shape of the graph of a quartic function different from the shape of the graph of a cubic function? a quadratic function? 4. The general form of the equation of a quartic function is y = ax4 + bx3 + cx2 + dx + e, where a 0. Describe how the graphs of quartic functions for which a is positive differ from those for which a is negative. 5. What does the value of e represent on the graph of a quartic function? 6. a) Graph the quartic function y = x4 + x3 7x2 x + 6. Find the x-intercepts. Sketch — — the graph and label the x-intercepts. b) Do the x-intercepts from part a) satisfy the related quartic equation x 4 +x 3 —7x 2 —x+6=0?Expiain. 7. The real roots of a quartic equation are the x-intercepts of the related quartic function. How many real roots does the quartic equation corresponding to each quartic function graphed in step 1 have? 8. State the possible numbers of each of the following types of roots for a quartic equation. Sketch a graph to illustrate each possibility. a) distinct real roots b) equal real roots c) complex roots 9. An archway for the entrance to the county fair is designed to look somewhat like a rounded letter M. A scale model of the arch passes through the points (0, 45), (1, 4$), (—1, 4$), (2, 45), and (—2, 45), where all distances are in centimetres. The x-coordinates of the points represent locations across the entrance and the y-coordinates represent heights above the ground. a) What kind of polynomial should be used to model the arch? b) Use an appropriate regression to determine the function that models the arch. c) Determine the roots of the function in part b). U) Explain what the roots in part c) represent. 10. Summarize what you have learned about quartic functions and their graphs. Quintic Functions 1. The function y = x(x2 - 1)(x2 - 4) is a quintic (fifth-degree) function. Explain why. 2. What are the zeros of y = x(x — 1)(x — 4)? Explain without graphing. 3. Predict the shape of the graph of y = x(x2 — 1)(x2 — 4). Then, sketch the graph. 4. a) Graph y x(x2 1)(x2 4) in the standard viewing window. Sketch the graph and — — label the x-intercepts. b) Describe the similarities and differences between your prediction and the actual graph. 5. Graph each function in the standard viewing window. Describe and explain the numbers of distinct real roots, equal roots, and complex roots in each case. 22 2 2 a)y=x(x+1)(x—2) b) y=(x—1)(x—4)(3—x) c) y=(x2-x-2)(x2-1)(x+1) U) y=(x2+x—2)(x2—1)(x+1) 44 MHR Chapter 2 6. Is the graph of a quintic function more like the graph of a quadratic function or more like the graph of a cubic function? Explain. 7. Part of a roller coaster track is being designed. For this part of the track, the coaster will come down a long way, reach a low point, rise up to a high point, go down again to another low point, rise up again to another high point, and then plunge down. In this exercise, you will model this part of the track with a polynomial function. a) What should the degree of the polynomial be? b) A coordinate grid is placed on a diagram of the track. The track crosses the x-axis five times, and adjacent crossing points are separated by 2 units. Conjecture a formula for the function that models the track. c) Is there just one formula that satisfies the conditions in part b)? Explain. U) Use technology to determine the heights of the high and low points of the track. Are these reasonable values for a realistic roller coaster track? If not, explain how to modify your conjectured formula to make the model reasonable. 8. Summarize what you have learned about quintic functions and their graphs. Features of Polynomial Graphs The graph of a polynomial function is continuous, that is, there are no breaks in the graph. You can trace the graph without lifting your pencil. The graph of the cubic function y = 2x3 — 6x 1 is shown. The EE — graph has a peak at (—1, 3) and a valley at (1, —5). The point (—1, 3) is called a local maximum point of the function. This point does not have the greatest y-coordinate of any point of the function, but no nearby points have as great a y-coordinate. Similarly, the point (1, —5) is called a local minimum point of the function. It is the lowest point on the graph among nearby points. Local maximum points and local minimum points are also known as turning points. The end behaviour of the graph of a function describes the sign of the values of the function for the left-most and right-most parts of the graph. Polynomial functions can have four types of end behaviour...ftfl R [Li The right-most y-values are positive. 3) — co as x —> 00..[N4\j The right-most y-values are negative. y — —co as x —> 00 y —* a means that the values for y get larger and larger, without bound, in the positive direction. The left-most y-values are The left-most y-values are y —* —co means that the values positive. negative. for yget larger and larger, without 3) —+ co as x —* —00 y —* —co as x —> —00 bound, in the negative direction. 2.1 Investigating Math: Polynomial Functions on a Graphing Calculator MHR 45 1. III Z The right-most y-values are The right-most y-values are positive, negative. y-JDasx-+co y—>—coasx—*cID The left-most y-values are The left-most y-values are negative, positive. y—*—c’asx—>—c1D y—+coasx--+— 1. Graph the following functions in the standard viewing window. Copy and complete the table. End Observed Number of -y a) yx+2 Function Degree Behaviour Turning Points Ib) y-3x + 1 c) y=x2—4 d)y=—2x2+3x+2 e) y = x3 — 3x f) y=—x3+2x—1 4 2 g) y=x —4x +5 h) y=—x4+4x2+x—2 4 2 3 i) y=x 5 —2x —3x +Sx j) y=x5—2 k) y = —x5 + 4x3 + 2 2. For two functions with the same degree, how does the sign of the leading coefficient affect the end behaviour of the graph? 3. A function whose degree is an even number is called an even-degree function. a) How do the end behaviours of even-degree functions compare? b) Test your conjecture by graphing functions of even degree higher than 4. 4. A function whose degree is an odd number is called an odd-degree function. a) How do the end behaviours of odd-degree functions compare? b) Test your conjecture by graphing functions of odd degree higher than 5. 5. Sketch the graph of a polynomial function that satisfies each of the following sets of conditions. a) an even-degree function with two local maximum points and one local minimum point b) an even-degree function with two local minimum points and one local maximum point 46 MHR Chapter 2 c) an odd-degree function with two local maximum points and two local minimum points, and the left-most y-values are negative d) an odd-degree function with three local maximum points and three local minimum points, and the right-most y-values are negative e) an even-degree function with four local maximum points and three local minimum points 6. a) What appears to be the maximum possible number of turning points for a linear function? a quadratic function? a cubic function? a quartic function? b) What appears to be the relationship between the degree of a function and the maximum number of turning points? c) If the graph of a function has five turning points, what is the minimum possible degree of the function? 7. The graph of f(x) x(x 2)(x + 3) is shown. — I7Ji The x-intercepts are 2, 0, and —3. The three x-intercepts divide the x-axis into four intervals. Window (—cc, —3) (—3, 0) (0, 2) (2, cc) variables: Determine the sign of f(x) in each interval. x e [—4.7, 4.7], 8. For each of the following functions, sketch y a [—9.3, 9.3] the graph and determine the sign of f(x) in the intervals determined by the x-intercepts. a) 1(x) = (x + 1)(x —3) b) f(x) = x(x + 2)(x —4) c) 1(x)=—(x—2)(x+3) d) f(x)=x(x—l)(x+4) e) f(x)=(x+3)(x-3)(x+1) f) f(x)=-(x-5)(x+3)(x+3) g) 1(x)=x(x—3)(x—5)(x+4) h) f(x)=—(x+4)(x+ 1)(x—2)(x—5) 9. Summarize what you have learned about the graphs of polynomial functions by listing the key concepts from this section. 2.1 Investigating Math: Polynomial Functions on a Graphing Calculator MHR 47 : Dividing a Polynomial by a Polynomial We have previously multiplied polynomials. For example, (x + 2)(x 5) = x2 3x — — — 10. We have also factored polynomials, which is the reverse process. For example, 7 7 2x 3 —2x+3x—3=2x(x—1)+3(x--1) = (x 1)(2x2 + 3) — But polynomials are not always simple to factor. In Sections 2.3 and 2.4, we learn to identify a factor of a polynomial such as x3 6x2 + 7x + 6. In order to find the other — factors, we must be able to divide a polynomial by a binomial, such as x 1. In this — section, we illustrate this process. In Example 1, division of a polynomial by a binomial is compared to division of a number by a number. Example 1 Dividing a Polynomial by a Binomial of the Form x — b Divide x2 + Sx + 6 by x + 3. State the restriction on the variable. Solution Division of a polynomial by a binomial Division of a number by a number The first steps are: X Thjnkx2÷xx. 2 Think 7÷3=2(remainder 1). x+3)x2+Sx+6 3)72 x2 + 3x Multiply x(. + 3). 6 Multiply 2(3). 2x +6 Subtract. Bring down the 6. 12 Subtract. Bring down the 2. The final steps are: x +2 Think 2x÷x = 2. 24 Think 12÷3 = 4. x+3)x2+5x+6 3Jñ x2+3x 6 2x+6 12 2x +6 Multiply 2(x + 3). 12 Multiply 4(3). 0 Subtract to get the remainder. 0 Subtract to get the remainder. Because division by zero is not defined, x + 3 0, that is, x —3. So, (x2 + 5x + 6) ÷ (x + 3) = x + 2, where x —3. The polynomial that is being divided is called the numerator or dividend, the polynomial that the numerator is divided by is called the denominator or divisor, and the result of the division is called the quotient. quotient denominator) numerator Note that just as we write numbers in order of place value, we must write polynomials in order of degree before we divide. Consider Example 2. 48 MHR Chapter 2  Example 2 Dividing a Polynomial by a Binomial of the Form ax — b i 2 3 Drviae —19x + 6x + 18x — 20 by 2x 5. State the restriction on the variable. — Solution First, rewrite the polynomial with the terms in order of degree, 6x3 19x2 + 1$x — — 20. 3x2— 2x+ 4 2x_5)6x3 —19x2 +18x —20 6x3 —15x2 -4x2+ 18x —4x2+lOx 8x—20 8x—20 Because division by zero is not defined, 2x — 5 0, so x. Thus, (6x3—19x2+18x-20)÷(2x-5)=3x2-2x+4,x. In Example 2, since 2x 5 divides into 6x3 19x2 + 18x 20 evenly (that is, with a remain — — — der of zero), 2x 5 is a factor of 6x3 19x2 + 18x 20. Another way of writing the result is — — — 6x3— 19x2+18x—20=(2x—5)(3x2—2x+4). Similarly, in Example 1, x + 3 is a factor of x2 + 5x + 6. When the divisor is not a factor of the dividend, the remainder is not zero, as shown in Example 3. Example 3 Division With a Non-Zero Remainder a) Divide —9x 3 + 6x3 4x2 by 2x2 3. — — — b) State the restrictions on the variable. Solution a) First, rewrite the polynomial as 6x3 — 4x2 — 9x — 3. 3x—2 2x2 —3) 6x3 4x2 9x —3 — — 6x3 +0x2 —9x Note the use of Ox2 and Ox as placeholders. —4x2 +Ox-3 — 4x2 + Ox +6 —9 Thus, the remainder is —9. That is, (-9x -3 + 6x3 - 4x2) ÷ (2x2 —3) = 3x —2 b) Since 2x 2x3 3 0, — 2x2 3 23 x x ± 2.2 Dividing a Polynomial by a Polynomial MHR 49 I Polynomials can be named, using function notation, as F(x), Q(x), and so on. The result of the division of a polynomial can be represented as F(x) R(x) = which is equivalent to P(x) = D(x)Q(x) + R(x) Here, D(x) is the divisor, Q(x) is the quotient, and R(x) is the remainder. Written in these forms, the result of Example 3 can be expressed as follows. F(x) R(x) 6x3 4x2 9x —3 9 =3x—2— — — — D(x) D(x) 2x2 —3 2x2 —3 P(x)= D(x)Q(x)+R(x) 6x3 —4x2 —9x —3 =(2x2 —3)(3x—2)—9 Division of polynomials can help us understand the formula for annuities. Example 4 Accumulated Value of an Annuity An annuity is an investment where the same amount of money F is deposited at the end of each month. Interest is applied to the investment at the end of each month starting with the second month. The accumulated amount A of the annuity after 11 months is (X —1 A F x—1 where x = I + r, and r is the interest rate per month, expressed as a decimal. Use long division to verify the formula for A when n 3. Solution For n = 3, the formula for A is x—1 Using long division, we determine the quotient in parentheses. x2+ x+1 x i) x3 — + Ox2 + Ox —1 x3— x2 x2 +Ox x2— x x—1 x—1 0 Thus, A=P(x2+x+ 1) 50 MHR Chapter 2  There are three deposits altogether, made at the end of the first, second, and third months. The deposit made at the end of the third month earns no interest, so its contribution to the final amount is P. The deposit made at the end of the second month earns interest for one month, so its contribution to the final amount is P(1 + i). Since x = 1 + i, P(1 + i) = Px. Finally, the deposit made at the end of the first month earns interest for two months, so its contribution to the final amount is P(1 + i)2 = Px2. Thus, the total accumulated amount is P + Px + Px2 P(x2 + x + 1) which verifies the formula. Key Concepts Polynomials can be divided using long division. The result of the division of a polynomial P(x) by the polynomial D(x) can be represented as P(x) R(x) or P(x)=D(x)Q(x)+R(x) Communicate Your Understanding 1. Describe how you would rewrite the expression (x + 4x3 3) ÷ (2x 1) before — — dividing. 2. Explain how to determine the restriction on the variable for the division (2x3—x2+7x-4)÷(2x+S). 3. Explain the steps for dividing 4x3 2x2 + 7 by 2x + 3 using long division. — Practise c) (Sr2 31r + 6) ÷ (Sr 1) — - In each of the following, state any restrictions d) (4t2—3—4t)÷(2t+1) on the variables. e) (14 + 6r2 25r) ÷ (3r 2) — — f) (21x+10x2+9)÷(Sx+3) 1. Divide. g) (8x2+14x+15)÷(4x—3) a) (x2+8x+15)÷(x+3) b) (a2—7a+10)÷(a—S) 4. Divide. c) (y y 12) (y 4)— — a) (x3+2x2—x—2)÷(x2—1) — — d) (t2-4)÷(t+2) b) (x3 3x2 + 4x 12) ÷ (x2 + 4) — — c) (y3+3y+ 15 +5y2)÷(y2+3) 2. Divide. d) (4—4a—a+a÷(a—4) a) (x3+2x2+3x+2)÷(x+1) 5. Divide. b) (t3+3t2—St-4)÷(t+4) (2x 3 —2x+3x—3)÷(x—1) 7 a) c) (a3—3a2—a+3)÷ (a—3) 7 b) (3z3+6z2+Sz+ 10) ÷ (z+2) d) (4x—13x+x 3 +8)÷(x—1) c) (4m3+2m2—6m—3)÷(2m+1) e) (m3—4+3m2)÷(m+2) U) (6n3—9n2—8n+12)÷(2n—3) f) (y3+8—2y—4y2)÷(y-4) e) (lSd2 + 4d + 6 + 10d3) ÷ (Sd2 +2) 3. Divide. f) (4x + 6x2 3) ÷ (2x2 + 1) - — a) (2x2+llx+1S)÷(2x+5) g) (352+1253_s_2o5)÷(352_s) b) (3y2+8y-3)÷(3y—1) h) (8 + 21t3 — 28t2 — 6t) ÷ (7t2 — 2) 2.2 Dividing a Polynomial by a Polynomial MHR 51 I 6. Identify the numerator, denominator, 10. Divide. quotient, and remainder. a) (x5—1)÷(x-1) b) (x5—32)÷(x—2) a, 125io 11. Divide. 12 12 a) (6x3 + 4x2 + 4x + 3) ÷ (2x2 + 1) b) b) (3y3-9y2—3)÷(3y2+1) c) (2t3 + 4t2 4t 19) ÷ (2t2 4) — — — c) a2 Sa +6 d) (6d4 13d2 + d + 4) ÷ (2d2 3) =a—2 — — — a—3 12. In Exercise 11, does the value of the 2x+5x—2 U, =x+3+ remainder depend on the value of the variable? 2x—1 2x—1 Explain. t3+t2+t—3 4 e) =t+1+— t2+1 t2+1 13. Find the whole number value of k such that 3x+2 a) x + 3 divides x2 x k evenly 6x3+3x2—x — — f) =2x+1+ b) 2y 1 is a factor of 6y2 + y k — — 3x2—2 3x2—2 c) the remainder is 3 when 4x2 + 9x + k is 7. Identify the dividend, divisor, quotient, and divided by x 1 — remainder. Can you be sure about the quotient d) the remainder is —5 when 2x3 + 7x2 + Sx — k and the divisor? Explain. is divided by 2x + 1 a) 255=(11)(23)+2 b) 8y3+6y2—4y—5=(4y+3)(2y2—1)—2 14. The area of a triangle is represented by the c) (x)(x+1)+3x2+x+3 expression 6x2 5x 4. — — 8. Divide. Write each result in the forms = Q(x) + and P(x) = D(x)Q(x) + R(x). a) (x2+4x+2)÷(x+4) =6x2—Sx—4 b) (x2—3x—8)÷(x-3) c) (x2-7x+1O)÷(x-4) If the height is 3x — 4, what is the base? U) (2x2+x—3)÷(x+2) 15. The area of a trapezoid is represented by the e) (4x —7x—7)÷(x—2) expression 12y2 fly + 2. f) (x2-8)±(x—3) — g) (5x2+14x+11)÷(x+4) 3y+1 h) (6x2—Sx—5)÷(x—4) Apply, Solve, Communicate A=12Y2_11Y+\ 9. Communication a) Divide each of the following. I) x3 —1 ii) x3 —8 x—l x—2 5y—3 ii) x3—27 x—3 The bases are 3y + 1 and Sy 3. What is the b) Describe the pattern in the coefficients of the — height? quotients. c) Predict the coefficients for the quotients of 16. Inquiry/Problem Solving Dividing 5x2 + 14x — 3 x3—64 and x3—125 by x + 3 gives a quotient of Sx 1. Explain — x—4 x—5 why 5x2 +14x—3 is not the same as Sx 1. — Then, divide to check your prediction. x+3 52 MHR Chapter 2  17. On February 1, Diasetz invests $500 in a constant multiple is called the common ratio of guaranteed investment certificate (GIC). At the the series. For example, 3 + 6 + 12 + 24 + 48 is end of each month, interest is added to the a geometric series with common ratio 2. account at the rate of 1 % per month, a) Is the series 1 + x + x2 + x3 + + xk. compounded monthly. What is the value of geometric? If so, state the common ratio. the GIC after If not, explain why. a) 10 months? b) Use the results of question 19 to obtain b) 25 months? a formula for the sum of the series c) 10 years? 1+x + x2 + x3 + + xk. 18. Inquiry/Problem Solving After graduating from c) Suppose $1 is invested at the end of each university, Selena decides to live with her par month at an interest rate of i% per month, ents and devote her time to unpaid work as a compounded monthly. If r is the interest rate volunteer for a charity. She has saved $10 000 written as a decimal, determine an expression and placed it in a bank account that pays inter for the total value of the investment after est of 1% per month, compounded monthly. She i) 1 month ii) 2 months withdraws $500 from the account each month iii) 3 months iv) k months to pay for her living expenses. d) Using the results of parts b) and c), a) How long will she be able to continue before determine the total value of an investment of she exhausts her money and has to obtain paid $1 per month after 10 years, if the interest rate work? (Hint: A spreadsheet may be helpful.) is 1% per month, compounded monthly. b) Construct a scatter plot using the data in e) Repeat part d) if the monthly investment is part a). Plot time on the horizontal axis and the $1000. current value of the GIG on the vertical axis. 21. Communication a) When dividing a c) Draw a smooth curve through the points on polynomial by a linear polynomial, can the the scatter plot of part b). What is the remainder be a quadratic polynomial? A linear x-intercept? Explain the meaning of the polynomial? Explain. x-intercept. b) When dividing a polynomial by a quadratic 19. Inquiry/Problem Solving a) Determine each polynomial, can the remainder be a quadratic quotient. polynomial? A linear polynomial? Explain. x—1 c) Generalize parts a) and b) by making a x—1 x—1 general statement about the possible form of x—1 the remainders when polynomials are divided x—1 x—1 iii) iv) by polynomials. x—1 x—1 22. Mario manages to save $150 every month. b) Using the results of part a), conjecture a At the end of each month, he deposits this formula for the quotient , where n is a money into a savings account paying 0.2% positive integer. interest per month, compounded monthly. Every c) Test the conjecture of part b) by judicious three months, he transfers the balance of his choices of specific numbers for x and 71. Are you savings account into a GIG that pays 5.2% confident that your conjecture is correct? Explain. interest per year, compounded quarterly. a) How much money does Mario transfer every 20. Application A series of numbers is said to be a three months? geometric series if each number is a constant b) What is the total value of Mario’s GIG after multiple of the one before it in the series. The seven years? 2.2 Dividing a Polynomial by a Polynomial MHR 53 The Remainder Theorem In Section 2.4 we will study the factor theorem, which is used to help determine factors of polynomial expressions. Studying the remainder theorem in this section will help in understanding the factor theorem in the following section. When a polynomial function is divided by a binomial that is not a factor, there will be a non-zero remainder. In the following investigation, you will explore a possible relationship between such remainders and certain values of the polynomials. Investigate & Inquire: The Remainder Theorem 1. Copy and complete the table by dividing the polynomial P(x) = 2x3 — 2x2 — 3x + 3 by the given divisor. Compare your completed table with a classmate’s. Divisor, x — b b Quotient and Remainder, !i)_ P(b) a)x—3 b)x-2 J c)x+2 d) 1 x — e)x+3 f)x+1 2. Use the results of step 1 to make a conjecture about the relationship between the remainder and the value of F(b). 3. a) Divide the polynomial x3 + 2x2 7x 2 by each binomial to find the remainder. — — I) x—1 ii) x+1 iii) x—2 iv) x+3 b) Use your conjecture from step 2 to find the remainder when x3 + 2x2 7x 2 is divided — — by each binomial in part a). c) Compare your results from parts a) and b). 4. When a divisor is a factor of a polynomial, what is the remainder? Explain. The division of polynomials by divisors of the form x b, as in the investigation, can be — generalized as follows. If a polynomial P(x), of degree at least 1, is divided by (x b), the remainder is a constant, — R. Furthermore, the value of the constant is equal to the value of the polynomial at x = which can be shown as follows. From Section 2.2, we have the relation (for constant R) P(x)=Q(x)(x—b)+R Substituting b for x, we get P(b)=(b—b)Q(b)+R =R 54 MHR Chapter 2 This relationship between P(b) and 1? is known as the remainder theorem. Remainder theorem: When a polynomial P(x) is divided by x — b, the remainder is P(b). Example I Determining a Remainder Determine the remainder when P(x) = 2x3 — 4x2 + 3x — 6 is divided by x +2. Solution Write x + 2 as x (—2). The remainder theorem states that when P(x) — = 2x3 — 4x2 + 3x — 6 is divided by x (—2), the remainder is F(—2). — F(—2) = 2(—2) 4(_2)2 + 3(—2) 6 — — = —44 The remainder is —44. Example 2 Determining a Coefficient When x3 — kx2 + 17x + 6 is divided by x — 3, the remainder is 12. find the value of k. Solution LetP(x)=x3-kx2+ 17x+6 From the remainder theorem, P(3) is the remainder of the division by x — 3, so P(3) = 12. F(3) = (3)3 k(3)2 + 17(3) + 6 - = —9k + $4 —9k + $4 = 12 —9k = —72 k=$ The value of k is 8. Example 3 Using a System of Linear Equations When the polynomial F(x) = 3x3 + cx2 + dx 7 is divided by x 2, the remainder is —3. — — When P(x) is divided by x + 1, the remainder is —18. What are the values of c and ci? Solution Use the given information to write a system of linear equations. P(x)=3x3+cx2+dx—7 P(2) = 3(2) + c(2)2 + d(2) 7 - =4c+2d+ 17 From the remainder theorem, we know that P(2) = —3. So, 4c+2d+ 17=—3 4c+2d=—20 2c+d=—1O (1) 2.3 The Remainder Theorem MHR 55  = 3(_f)3 c(-1)2 d(-1) 7 F(—1) + + — =c-d— 10 From the remainder theorem, we know that F(—1) = —18. So, c—d— 10=—18 c—d=—8 (2) Solve the system of linear equations. 2c+d=—10 (1) c—d=--8 (2) Add (1) and (2): 3c=—18 Solve for c: c = —6 Substitute —6 for c in (2). c—d=—8 —6 —d=—8 d=2 Check in (1). Check in (2). L.S. =2c + d R.S. -10 L.S. = c d— R.S. = —8 =2(—6)+2 =—6—2 =—10 L.S. = R.S. L.S. = R.S. The value of c is —6, and the value of d is 2. The remainder theorem can be extended to include divisors in which the coefficient of x is not 1. If P(x) is divided by ax b and the division is continued until the quotient is Q(x) — and the remainder is a constant, R, then the following division statement is true for all values of x. P(x)=(ax—b)Q(x)+R Substitute for x: a Thus, a more general form of the remainder theorem is as follows. When a polynomial P(x) is divided by (ax — b), the remainder is 56 MHR Chapter 2  Example 4 Dividing by ax — b for a 1 Determine the remainder when 2x3 + 3x2 — 7x — 3 is divided by 2x + S. Solution I Solution 2 Graphing Calculator Method Paper and Pencil Method Input the function y = 2x3 + 3x2 7x 3 in the — — LetP(x)=2x 3 +3x—7x—3. , Y= editor of a graphing calculator. 53 (5)2 (5) = 2 =2 f 125 +3 f25” 35 —--j 1— Press pj 1 1 1 to call Yl to evaluate the 5 — 125 75 70 12 function at x = — —. — 4+4+4 4 =2 The remainder is 2. The remainder is 2. Canadian Catherine Bond-Mills won the bronze medal in the heptathlon at the Commonwealth Games in Victoria, British Columbia, in 1994. In the heptathion, athletes compete in seven events. One of them is the shot-put. For a certain “put,” the function h(t) = —St2 + 9t + 2 approximates the height of the shot, in metres, as a function of the time, t, in seconds, since it was released. The formula is only valid until the shot hits the ground. To find the height at various times, we substitute values of t into the equation for h(t). Doing this can also help us learn more about the equation. Example 5 Shot-Put Height Formula Consider the shot-put formula h(t) = —St2 + 9t + 2. a) In the division relationship = Q(t) + where b is a constant, explain what Q(t) and R mean, both geometrically and physically. b) What is the remainder when h(t) = —St2 + 9t + 2 is divided by t 2? What does the — remainder represent physically? 2.3 The Remainder Theorem MHR 57 Solution a) Recall that Q(t) represents the quotient and R represents the remainder when h(t) is divided by t b. Recall from the remainder theorem that — R h(b). Thus, R is the y-value on the graph of h(t) at the point where x = b. The physical mean ing of R, therefore, is the height of the shot after b seconds. To understand the meaning of Q(t), it is helpful to solve the division relationship R —s h(t) =Q(t)+— for Q(t) to obtain h(t) R z—‘t—b t—b —h(t)—R t—b Thinking in terms of rise over run, we can see that the previous equation for Q(t) represents the slope of the line segment joining the points (b, R) and (t, h(t)). Physically, Q(t) represents the change in height of the shot divided by the time for the time interval that begins b seconds after the shot is launched and ends t seconds after the shot is launched. b) Using the remainder theorem, the remainder when h(t) is divided by t 2 is h(2). — h(Z) _5(2)2 + 9(2)’+ 2 0 The remainder is 0 when h(t) = —St2 + 9t + 2 is divided by t 2. — Since the remainder represents the height of the shot after 2 s, and this height is 0, we conclude that the shot hits the ground after 2 s. In Chapters 3 and 4, we will see that the quantity Q(t) from Example S can also be interpreted as the average velocity of the shot over the time interval from b to t. In general, such quotients can be interpreted as average rates of change. Key Concepts When a polynomial P(x) is divided by x — b, the remainder is P(b). When a polynomial P(x) is divided by ax — b, the remainder is In the division relationship P(x) R = Q(x) and R can be interpreted as quantities on the graph of y = P(x). The quantity R is the height of the graph at x = b; that is, R = P(b). The quantity Q(x) is the slope of the line segment joining the points (b, F(b)) and (x, P(x)). 58 MHR Chapter 2  __ L - Communicate Your Understanding 1. Describe how you would determine the remainder when the polynomial x3 3x2 + 4x 7 is divioed ny x 2. — — — 2. Describe how you would determine the remainder when the polynomial 2x’ 5x + 4 is divided by 2x 3. — — 3. When a divisor is a factor of a polynomial, what is the remainder? Explain. Practise I) (2—x+x 2 —x 3 —x)÷(x+2) 4 m) (3x2_x+3)±(x+) 1. For P(x) = 2x2 — 3x 2, determine the 42 JI) — n) (2r4 9) — (r following. — — a) P(1) b) P(0) 6. Use the remainder theorem to determine the c) F(2) U) F(—2) remainder for each division. a) (2x2 + 5x + 7) ÷ (2x 3) 2. For g(y) 2y3 3y2 5, determine the — = — + b) (6a2+5a-4)÷(3a+4) following. c) (x3+2x2—4x+1)÷(2x—1) a) g(—2) b) g(4) U) (2y3+y2—6y+3)÷(2y+1) c) g(0) d) g() e) (9m3—6m2+3rn+2)÷(3m—1) f) (8t3+4t2+17)÷(2t+3) 3. Use the remainder theorem to. determine the remainder when each polynomial is divided 7. For each polynomial, determine the value of byx—2. k if the remainder is 3. 7 a) x—5x—3 2 b) 2x +x—10 a) (kx2+3x+1)÷(x+2) c) x3+2x2—8x+1 U) x3—3x2+5x—2 b) (x3+x2+kx-17)÷(x-2) e) 2x3+3x2—9x—10 f) 3x3—12x+2 c) (x3+4x2—x+k)÷(x—1) U) (x3+kx2+x+2)÷(x+1) 4. Use the remainder theorem to determine the 8. When the polynomial 4x 3 + mx + ;ix + his 7 remainder when 2x3 — x2 — 4x 4 is divided by divided by x + 2, the remainder is —7. When the — each binomial. a) x—1 polynomial is divided by x 1, the remainder is b) x+1 — 14. What are the values of m and n? c) x—2 U) x+3 9. The polynomial px3 x2 + qx 2 has no Use the remainder theorem to determine the — — 5. remainder when divided by x 1 and a remainder for each division. — remainder of —18 when divided by x + 2. a) (x2+2x+4)÷(x—2) Determine the values of p and q. b) (4n2+7n—5)÷(n+3) c) (x3+2x2—3x+1)÷(x—1) 10. The polynomial 3x3 + vx2 5x + w has a — U) (x 3 +6x—3x 2 —x+8)÷(x+1) 7 remainder of —f when divided by x + 2 and a remainder of 109 when divided by x 3. What e) (2w3+3w2—5w+2)÷(w+3) — are the values of v and w? f) (y3-$)÷(y+2) g) (x3+x2+3)(x+4) h) (1-x3)÷(x—1) Apply, Solve, Communicate 4 3 2 i) (m —2m +m +12m—6)÷(m—2) 11. The divisions (2x3 + 4x2 kx + 5) ÷ (x + 3) — j) (2—3i+t2+t3)÷(t—4) and (6y3 3y2 + 2y + 7) ÷ (2y 1) have the — — k) (2y 4 —3y 2 +1)÷(y—) -, same remaindei Determine the value of k. 2.3 The Rmainder Theorem MHR 59 12. When kx3 — 3x2 + 5x 8 is divided by x 2, — — 17. Communication, Inquiry/Problem Solving A park the remainder is 22. What is the remainder ing officer keeps, a funning total of the tickets when kx3 3x2 + Sx $ is divided by x + 1? — — that she gives during an 8-hour shift. Analysing her results later, she uses regression to approxi 13. The area, A(h), of a triangle is represented by mate the total number of tickets, F, given up to the expression h2 + 0.5h, where h is the height. time t by the formula P(t) = 0.5t2 + 2t, where t a) Determine the remainder when the is measured in hours and t a [0, 8]. expression is divided by 2h 7. a) Divide P(t) by t 2. State the quotient Q(t) — b) Interpret the remainder. — and the remainder R. Explain the meaning of 14. The product of two numbers is represented Q(t) and R. by the expression 6n2 Sn + 8, where n is a — b) Does the formula for P(t) seem realistic? positive integer greater than 4. Explain. a) Determine the remainder when the c) On another day, the parking officer’s formula expression is divided by 2n + 1. for the running total of tickets given is b) Interpret the remainder. P(t) = —t2 + lOt for t a [0, 8]. Is this formula 15. Application The main span of the Tsing Ma realistic, or has a mistake been made? Explain. Bridge in Hong Kong is the longest span of any 18. Inquiry/Problem Solving A tree farm operator suspension bridge in the world. If the origin models the cumulative total number of Christmas is placed on the roadway of the main span, trees sold, C(t), during December by the formula below the lowest point on the support cable, the C(t) = 0.1t2+ lOt+25, where ta[1, 24], t= 1 shape of the cable can be modelled by the represents December 1, t = 2 represents function h(d) = 0.0003d2 + 2. December 2, and so on. where h(d) metres is the height of the cable above a) Explain the strengths and weaknesses of the the roadway, and d metres is the horizontal formula for C(t). Do you think that it is an distance from the lowest point on the cable. exact formula or an approximation? Explain. a) Determine the remainder when 0.0003d2 + 2 b) Divide C(t) by t 5. State the quotient Q(t) — is divided by d 500. — and the remainder R. Explain the meaning of b) Determine the remainder when 0.0003d2 + 2 Q(t) and R. is divided byd+500. c) Do you think that the same formula for C(t) c) Compare the results from parts a) and b). is still valid for t 25? Explain. U) Use the graph of the function 19. For what values of k does the function h(d) = 0.0003d2 + 2 to explain your findings. 1(x) x3 + 6x2 + kx 4 give the same remainder — 16. Communication The hammer throw is an when divided by either x 1 or x + 2? — Olympic throwing event. The path of the 20. When x2 + Sx + 7 is divided by x + k, the hammer in one throw can be modelled by the remainder is 3. Determine k. function 21. When the polynomial bx2 + cx + a is divided h(d) = —0.017d2 + 1.3d + 2.5 by x a, the remainder is zero. where h(d) metres is the height of the hammer, — a) What can you conclude from this result? and d metres is the horizontal distance of the b) Write an equation that expresses a in terms hammer from the point where it was released. of b, c, and a. a) Divide the polynomial —0.01 7d2 + 1.3d + 2.5 byd—50. 22. A polynomial, P(x), is divided by x b to — b) Interpret the remainder from part a). give a quotient, Q(x), and a remainder, R. c) Divide the polynomial —0.017d2 + 1.3d + 2.5 a) Predict the quotient and the remainder when byd—80. F(x)isdividedbyb—x. U) Does the remainder from part c) have any b) Use an example with a non-zero remainder meaning for the hammer throw? Explain. to test your prediction. 60 MHR Chapter 2 c) Use the division statements for the division 24. Divide using long division. by x b and the division by b x to explain x4 —1 x3+2x2+Sx+1 — — your findings. a) b) x- +1 xl + I 23. Write a polynomial that satisfies each set of conditions. Check that the polynomial satisfies x8—1 2x3—5x2+6x—7 the conditions, using a different method than the c) d) x2 —1 x_ —x one you used to create the polynomial. a) a quadratic polynomial that gives a 25. Application, Inquiry/Problem Solving In light of remainder of —4 when it is divided by x 3— question 24 and its results, is it possible to b) a cubic polynomial that gives a remainder of extend the remainder theorem to cover cases 3 when it is divided by x + 2 such as these? If so, conjecture a generalization c) a quartic polynomial that gives a remainder of the remainder theorem, and test your of 1 when it is divided by 2x 1 — conjecture. If not, explain why. ICareer Connection Computer Science — Like mathematics, a career in computer science is potentially a good “fit” for anyone who enjoys creating and comprehending abstract thought structures. Many computer applications draw heavily on mathematical theory. For example, algorithms are among the most fundamental structures in all computer software, and they are often best expressed in mathematical terms. A computer algebra system is able to replicate many of the processes studied in this chapter, because the software has been developed out of an excellent mathematical understanding of algorithms such as polynomial division to find a quotient and remainder. Software applications from telecommunications networks to video game graphics demand similar skills and aptitudes. 2.3 The Remainder Theorem MHR 61 The Factor Theorem In this section we study the factor theorem, which is helpful in factoring polynomial expressions. The integer factors of the integer 6 are 1, 2, 3, 6, —1, —2, —3, and —6. The factors all divide 6 evenly, that is, they give a remainder of zero. This concept also applies to polynomials. For example, the quadratic x2 Sx + 6 can be factored as follows. — x2—Sx+6=(x-2)(x-3) The remainder theorem can be used to find the remainder when x2 — Sx + 6 is divided by each of its factors. P(x) = x2 — 5x + 6 P(x) x2 — 5x + 6 = 32 F(2) = 22 — 5(2) + 6 F(3) 5(3) + 6 =0 =0 Thus, as with integers, dividing a polynomial by one of its factors gives a remainder of zero. Note also that 2 and 3 are factors of the constant term, 6, of the quadratic. In the following investigation, you will test these ideas for other polynomials. Investigate & Inquire: The Factor Theorem 1. Copy and complete the table for the polynomial P(x) = x3 + 2x2 — Sx — 6 using the remainder theorem. Divisor, x — b b Remainder, Pfb) Is x b a factor of P(x) -- j_J - a)x+1 —1 0 b)x—2 ji c)x—3 d)x+3 e)x±4 1) x g)x±6 2. When a binomial is a factor of a polynomial, is the constant term of the binomial a factor of the constant term of the polynomial? Explain. 3. If the constant term of a binomial is a factor of the constant term of a polynomial, is the binomial always a factor? Explain. 4. a) What are the possible values of b for the binomial factors, x — of x3 + 2x2 13x + 10? - b) Test the possible values of b to find the three binomial factors of x3+ 2x2 13x + 10. — 62 MHR Chapter 2 The factor theorem, which is a special case of the remainder theorem, can be stated as follows. Factor theorem: A polynomial F(x) has x — b as a factor if and only if P(b) = 0. The factor theorem can be established as follows. If a polynomial P(x) has x b as a factor, then — P(x) = (x b)Q(x) — Substitute b for x: F(b) = (b b)Q(b) — Conversely, if F(b) = 0, then, by the remainder theorem, the remainder is zero when P(x) is divided by x b, which means that x b is a factor of P(x). — — The factor theorem can be used to find or verify a factor of a polynomial. Example 1 Verifying a Factor x — b Show that x + 2 is a factor of x3 + 5x2 + 2x — 8 Solution I Paper and Pencil Method From the factor theorem, if P(—2) = 0, then x + 2 is a factor of P(x) = x3 + 5x2 + 2x — 8. P(-2) = (-2) + 5(_2)2 + 2(-2) 8 — =0 Since P(—2) = 0, x + 2 is a factor of x3 + 5x2 + 2x 8. — Solution 2 Graphing Calculator Method From the factor theorem, if P(—2) = 0, then x + 2 is a factor of P(x) = x3 + 5x2 + 2x 8.— Enter the function y = x3 + 5x2 + 2x 8 as Yi. Then, find Yi(—2) (press VARS — 1 1 W12W). FJ..11 Fic.t P1’ Since Yi(—2) = 0, that is, P(—2) = 0, x + 2 is a factor of x3 + 5x2 + 2x — 8. When attempting to factor a polynomial, it is helpful to know which values of b to try. The integral zero theorem, which can be expressed as follows, can help in making a decision. A zero of a polynomial Integral zero theorem: If x = b is an integral zero of a P(x) is a number bsuch polynomial F(x) with integral coefficients, then b is a that P(b) = 0. An integral factor of the constant term of the polynomial. zero is a zero b that is an integer. 2.4 The Factor Theorem MHR 63 I For example, the polynomial P(x) = x2 5x + 6 is expressed in factored form as — P(x) = (x 2)(x 3). The zeros of F(x) are therefore 2 and 3, both of which are factors — — of 6, the constant term of F(x). If x = b is an integral zero of a polynomial, then x b is a factor of the polynomial. — Therefore, we can use the factors of the constant term of the polynomial to look for possible factors of the polynomial. Example 2 Factoring Using the Integral Zero Theorem Factor x3 + 3x2 — 13x — 15. Solution I Paper and Pencil Method Find a factor by evaluating P(x) for values of x equal to possible values of b. The possible values of b are the factors of 15. The factors of 15 are ±1, ±3, ±5, and ±15. P(x) = x3 + 3x2 — 13x — 15 Start by testing the simplest possible values of b until a zero is found. = (;)3 + 3(f)2 P(1) - 13(1) - 15 = —24 F(—1) =(—l) + 3(_1)2 — 13(—1) — 15 =0 3 2 Since P(—1) = 0, x + 1 is a factor of x + x — 13x — 15. Use division to find another factor. 2 x+2x—1S From the long division, another factor is x + 2x — 15. x+1)x3+3x2_13x_ls So,x3+3x2—13x—15=(x+1)(x2+2x—15). Factoring x2 + 2x 15 gives (x 3)(x + 5). — — x3 + x2 So,x3+3x2— 13x- 15=(x+ 1)(x—3)(x+5). 2x2—13x 2x2+ 2x — lSx—15 — 15x— 15 0 Solution 2 Graphing Calculator Method Alternatively, the factors of the polynomial x3 + 3x2 13x — — 15 can be found by graphing the corresponding function P(x) = x3 + 3x 13x 15. — — PicU 7c12 1cI J Window variables: x e [—10, 10], y E [—25, 25] “ The graph has x-intercepts of 3, —1, and —5. These are the real zeros of the function. Using these zeros, the polynomial can be written in factored form as P(x)=(x-3)(x+1)(x+5) 64 MHR Chapter 2 Example 3 Factoring a Polynomial Factor x3 — 1. Solution The factors of 1 are ±1. x2 + x + 1 F(x) = X 1 — x — i) x3 + Ox2 + Ox—i P(1) = 1 - =0 Therefore, x 1 is a factor. Use long division to find another — x2 + Ox factor 2 From the long division, another factor of x3 1 is x2 + x + 1. x — x — The expression x2 + x + 1 cannot be factored. x —1 So,x3—i=(x—1)(x2+x+1). x—1 0 The factor theorem can be extended to include polynomials for which the leading coefficient is not 1. If a polynomial P(x) has ax b as a factor, then — P(x) = (ax b)Q(x)— Substitute - for x: = [a(-) — bjQ(-) Conversely, if = 0, then, by the remainder theorem, the remainder is 0 when F(x) is divided by ax — b, which means that ax — b is a factor of P(x). Thus, a polynomial P(x) has ax — b as a factor if and only if = 0. Example 4 Verifying a Factor ax — b, Where a 1 Verify that 2x — 3 is a factor of 2x3 — 5x2 — x + 6. Solution I Paper and Pencil Method If = 0, then 2x — 3 is a factor of P(x) = 2x3 - 5x2 - x + 6. 3 =2—33 32 3 +6 P-. —5— — — 2 2 2 2 27 45 6 24 =0 Since = 0, 2x - 3 is a factor of 2x3 - 5x2 — x + 6. 2.4 The Factor Theorem MHR 65 Solution 2 Graphing Calculator Method If = 0, then 2x — 3 is a factor of F(x) 2x3 — 5x2 — x + 6. Enter the function y = 2x3 — 5x2 — x + 6 into the function editor. Then, find Yi( ). Since Y1() = 0, that is, = 0, 2x — 3 is a factor of 2x3 — 5x2 — x + 6. When attempting to factor a polynomial, it is helpful to know which values of a and b to try. The rational zero theorem, as follows, can help in making a decision. Rational zero theorem: If x = is a rational zero of a polynomial A rational zero P(x) with integral coefficients, then b is a factor of the constant is a zero that is a term of the polynomial, and a is a factor of the leading coefficient rational number. of the polynomial. If x = is a rational zero of a polynomial, then ax — b is a factor of the polynomial. Therefore, we can use the factors of the constant term of the polynomial and the factors of the leading coefficient of the polynomial to look for factors of the polynomial. Example 5 Factoring Using the Rational Zero Theorem Factor 8x3 — 4x2 — 2x + 1 completely. Solution If 8x3 4x2 2x + 1 has a factor ax b, then b is a factor of the constant term, which is 1, — — — and a is a factor of the leading coefficient, which is 8. The factors of 1 are ±1. The factors of 8 are ±1, ±2, ±4, and ±8. Find a factor by evaluating P(x) for x-values that equal the possible values of b 1 1 —1 —1 1 1 —1 —1 1 1 —1 —1 1 1 —1 The possible values of — are , z’ y’ T’ —, ‘ ‘ ‘ ‘ ‘ ‘ ‘ ‘ ‘ and. Because some of these values are the same, the possible values of are ±1, ± ± , and ±. Now, test the values of until a zero is found. 66 MHR Chapter 2  P(x) — 4x2 — 2x + I FJ.’i P1.12 P1.,I P(i) = 3 P(—1) —9 p() 0 E21 Since = 0, 2x — 1 is a factor of 8x3 — 4x2 — 2x + 1. Use long division to find another factor. 4x +Ox—1 From the long division, 4x2 1 is another factor. — 2x — i) 8x3 — 4x2 2x + I — Since 4x2 1 can be factored further, we have — 8x3 —4x2 $x3—4x2—2x+1=(2x-1)(4x2—1) = (2x 1)(2x 1)(2x + 1) — — 0x—2x = (2x l)2(2x + 1) - 0x2—Ox —2x+1 —2x+1 0 The factoring tools developed in this section are used in Section 2.5 to help solve polynomial equations. Key Concepts A polynomial P(x) has x — b as a factor if and only if P(b) = 0. A polynomial F(x) has ax - b as a factor if and only if = 0. If x = is a rational zero of a polynomial P(x) with integral coefficients, then b is a factor of the constant term of the polynomial, and a is a factor of the leading coefficient of the polynomial. Communicate Your Understanding 1. Describe how to verify that x 3 is a factor of x3 — + 5x2 — 12x — 36. 2. Describe how to factor x3 2x2 Sx + 6. — — 3. Describe how to factor 2x3 3x2 lix + 6. — — 2.4 The Factor Theorem MHR 67  Practise f) 2x3—3x2+3x—10 g) 6x3 11x2 26x + 15 — — 1. Use the factor theorem to determine whether h) 4y3+8y2—y—2 W each polynomial has a factor of x 1. — I) 4x3+3x2—4x—3 a) x 3 —3x 2 +4x—2 3 2 b) 2x —x —3x—2 j) 6w3+16w2—21w+5 c) 3x3—x—3 U) 2x3+4x2—Sx—1 2. State whether each polynomial has a factor Apply, Solve, Communicate ofx+2. Application The area, A, in millions of a) 5x2+2x+4 b) x3+2x2—3x—6 U 8.hectares, of forest cut down in a certain c) 3x3+2x2—7x+2 U) x4—2x3+3x—4 part of the world in year t can be modelled 3. Verify that the binomial is a factor ofthe by A(t) = 3t2 + 2t, where t 0 represents the polynomial. year 2000, t = 1 represents the year 2001, a) x3+2x2+2x+1;x+1 and so on. 2 b) x 3 —3x +4x—4;x—2 a) Calculate the value of A(2) and explain its c) m3—3m2+m—3;m—3 meaning. d) x3+7x2+17x+15;x+3 b) Calculate the value of A(S) A(2) and — e) 4x2—9x—9;x—3 explain its meaning. f) 6x2—llx—17;x+1 c) Calculate the value of A(S) A(2) and — g) 2y3—5y2+2y+1;y—1 explain its meaning. 5—2 h) x3—6x—4;x+2 d) Explain the meaning of the quantity 4. State whether each polynomial has a factor A(t) A(2) — of 2x 1. Explain your reasoning. t—2 e) Do you think that t e [0, co) is a reasonable — a) 6x2 + 5x,— 4 b) 4x2 + 8x 7 domain for the function A(t)? Explain. — c) 2x3—x2—6x+3 U) 2x3+9x2+3x—4 e) 2x4—x3+3x—1 f) —4x3+4x2+x—1 9. For each polynomial function f(x), and each value b, 5. Show that the binomial is a factor of the i) factorf(x)—f(b) ii) simplify polynomial. 2 a) 2x3+x2+2x+1;2x+1 a) f(x) = x; b = 3 2 b) 2x3—3x2—2x+3;2x—3 b) f(x)=x +2x—1;b=2 c) 3y3+8y2+3y—2;3y—1 c) f(x)=2x3+4x2—Sx—7; b=—1 d) 6n3—7n2+1;3n+1 10. Consider question 9. e) 8x2—2x—1;2x—1 a) Rewrite the division statement f) 3x —x —3x+1;3x—1 P(x) R(x) = Q(x)+—, substituting f(x) for P(x), 6. Factor completely. 3 2 and substituting x b for D(x). a) x —6x +llx—6 b) x3+8x2+19x+12 — 3 2 b) Since x b has degree 1, it follows that R(x) — c) x —2x —9x+18 U) x3+4x2+2x—3 3 2 is actually a constant in part a). Determine the e) z3 + z2 22z 40 f) x +x —16x—16 value of the constant in terms of the function f. — — g) x3 2x2 6x 8 h) k3+6k2—7k—60 (Hint: Begin by multiplying each term by x b.) — — — x3 27x + 10 — j) x3+4x2—15x—18 Rewrite the division statement of part a) by — 7. Factor. substituting the constant that you have just a) 2x 3 9x 2 + lOx 3 - - determined. b) 4y33— 7y-3 c) Solve the division statement from part b) c) 3x —4x —17x+6 forQ(x). 2 d) 3x 3 2x — 12x + 8 — U) Do you think that f(x) f(b) can be — e) 2x3 + 13x2 + 23x + 12 factored for every polynomial function f? If so, 6$ MHR Chapter 2  explain why. If not, provide an example of a 15. For the polynomial polynomial function f for which f(x) f(b) — P(x) = 3x3 + 7x2 — 22x 8,— cannot be factored. (Hint: See part c).) P(2) = 0, P(-4) = 0, and P(L ) = 0. y LJ.qE. a) Find three factors of the polynomial. b) Are there any other factors? Explain. 16. The polynomial 6x3 + mx2 + nx S has a — factor of x + 1. When divided by x 1, the E EEZ ZH — remainder is —4. What are the values of m and ii? 17. Show that (x a) is a factor of the — polynomial x3 ax2 + bx2 abx + cx ac. — — — 10 18. Inquiry/Problem Solving a) Factor the e) Referring to the graph of the function polynomials in parts i) to vi). y = f(x),