Algebraic Expressions and Identities PDF

ALGEBRAIC EXPRESSIONS AND I DENTITIES 93 CHAPTER Algebraic Expressions and Identities...

ALGEBRAIC EXPRESSIONS AND I DENTITIES 93 CHAPTER Algebraic Expressions and Identities 8 8.1 Addition and Subtraction of Algebraic Expressions In earlier classes, we have already become familiar with what algebraic expressions (or simply expressions) are. Examples of expressions are: x + 3, 2y – 5, 3x2, 4xy + 7 etc. In the earlier classes, we have also learnt how to add and subtract algebraic expressions. For example, to add 7x2 – 4x + 5 and 9x – 10, we do 7x2 – 4x + 5 + 9x – 10 7x2 + 5x – 5 Observe how we do the addition. We write each expression to be added in a separate row. While doing so we write like terms one below the other, and add them, as shown. Thus 5 + (–10) = 5 –10 = –5. Similarly, – 4x + 9x = (– 4 + 9)x = 5x. Let us take some more examples. Example 1: Add: 7xy + 5yz – 3zx, 4yz + 9zx – 4y , –3xz + 5x – 2xy. Solution: Writing the three expressions in separate rows, with like terms one below the other, we have 7xy + 5yz –3zx + 4yz + 9zx – 4y + –2xy – 3zx + 5x (Note xz is same as zx) 5xy + 9yz + 3zx + 5x – 4y Thus, the sum of the expressions is 5xy + 9yz + 3zx + 5x – 4y. Note how the terms, – 4y in the second expression and 5x in the third expression, are carried over as they are, since they have no like terms in the other expressions. Reprint 2024-25 94 MATHEMATICS Example 2: Subtract 5x2 – 4y2 + 6y – 3 from 7x2 – 4xy + 8y2 + 5x – 3y. Solution: 7x2 – 4xy + 8y2 + 5x – 3y 5x 2 – 4y2 + 6y – 3 (–) (+) (–) (+) 2x2 – 4xy + 12y2 + 5x – 9y + 3 Note that subtraction of a number is the same as addition of its additive inverse. Thus subtracting –3 is the same as adding +3. Similarly, subtracting 6y is the same as adding – 6y; subtracting – 4y2 is the same as adding 4y2 and so on. The signs in the third row written below each term in the second row help us in knowing which operation has to be performed. EXERCISE 8.1 1. Add the following. (i) ab – bc, bc – ca, ca – ab (ii) a – b + ab, b – c + bc, c – a + ac 2 2 2 2 (iii) 2p q – 3pq + 4, 5 + 7pq – 3p q (iv) l2 + m2, m2 + n2, n2 + l2, 2lm + 2mn + 2nl 2. (a) Subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3 (b) Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz (c) Subtract 4p2q – 3pq + 5pq2 – 8p + 7q – 10 from 18 – 3p – 11q + 5pq – 2pq2 + 5p2q 8.2 Multiplication of Algebraic Expressions: Introduction (i) Look at the following patterns of dots. Pattern of dots Total number of dots 4×9 5×7 Reprint 2024-25  ALGEBRAIC EXPRESSIONS AND I DENTITIES 95 To find the number of dots we have to multiply the expression for the number of rows by the m×n expression for the number of columns. Here the number of rows is increased by 2, i.e., m + 2 and number (m + 2) × (n + 3) of columns increased by 3, i.e., n + 3. (ii) Can you now think of similar other situations in which two algebraic expressions have to be multiplied? Ameena gets up. She says, “We can think of area of a rectangle.” The area of a rectangle is l × b, where l is the length, and b is breadth. If the length of the rectangle is increased by 5 units, i.e., (l + 5) and breadth is decreased by 3 units , i.e., (b – 3) units, To find the area of a rectangle, we the area of the new rectangle will be (l + 5) × (b – 3). have to multiply algebraic expressions like l × b or (iii) Can you think about volume? (The volume of a (l + 5) × (b – 3). rectangular box is given by the product of its length, breadth and height). (iv) Sarita points out that when we buy things, we have to carry out multiplication. For example, if price of bananas per dozen = ` p and for the school picnic bananas needed = z dozens, then we have to pay = ` p × z Suppose, the price per dozen was less by ` 2 and the bananas needed were less by 4 dozens. Then, price of bananas per dozen = ` (p – 2) and bananas needed = (z – 4) dozens, Therefore, we would have to pay = ` (p – 2) × (z – 4) Reprint 2024-25  96 MATHEMATICS TRY THESE Can you think of two more such situations, where we may need to multiply algebraic expressions? [Hint: Think of speed and time; Think of interest to be paid, the principal and the rate of simple interest; etc.] In all the above examples, we had to carry out multiplication of two or more quantities. If the quantities are given by algebraic expressions, we need to find their product. This means that we should know how to obtain this product. Let us do this systematically. To begin with we shall look at the multiplication of two monomials. 8.3 Multiplying a Monomial by a Monomial Expression that contains only one term is called a monomial. 8.3.1 Multiplying two monomials We begin with 4 × x = x + x + x + x = 4x as seen earlier. Notice that all the three Similarly, 4 × (3x) = 3x + 3x + 3x + 3x = 12x products of monomials, 3xy, Now, observe the following products. 15xy, –15xy, are also monomials. (i) x × 3y = x × 3 × y = 3 × x × y = 3xy (ii) 5x × 3y = 5 × x × 3 × y = 5 × 3 × x × y = 15xy (iii) 5x × (–3y) = 5 × x × (–3) × y = 5 × (–3) × x × y = –15xy Some more useful examples follow. Note that 5 × 4 = 20 (iv) 5x × 4x2 = (5 × 4) × (x × x2) i.e., coefficient of product = coefficient of = 20 × x3 = 20x3 first monomial × coefficient of second monomial; (v) 5x × (– 4xyz) = (5 × – 4) × (x × xyz) and x × x2 = x 3 = –20 × (x × x × yz) = –20x2yz i.e., algebraic factor of product Observe how we collect the powers of different variables = algebraic factor of first monomial in the algebraic parts of the two monomials. While doing × algebraic factor of second monomial. so, we use the rules of exponents and powers. 8.3.2 Multiplying three or more monomials Observe the following examples. (i) 2x × 5y × 7z = (2x × 5y) × 7z = 10xy × 7z = 70xyz (ii) 4xy × 5x2y2 × 6x3y3 = (4xy × 5x2y2) × 6x3y3 = 20x3y3 × 6x3y3 = 120x3y3 × x3y3 = 120 (x3 × x3) × (y3 × y3) = 120x6 × y6 = 120x6y6 It is clear that we first multiply the first two monomials and then multiply the resulting monomial by the third monomial. This method can be extended to the product of any number of monomials. Reprint 2024-25  ALGEBRAIC EXPRESSIONS AND I DENTITIES 97 TRY THESE We can find the product in other way also. 4xy × 5x2y2 × 6x3 y3 Find 4x × 5y × 7z = (4 × 5 × 6) × (x × x2 × x3) × (y × y2 × y3) First find 4x × 5y and multiply it by 7z; = 120 x6y6 or first find 5y × 7z and multiply it by 4x. Is the result the same? What do you observe? Does the order in which you carry out the multiplication matter? Example 3: Complete the table for area of a rectangle with given length and breadth. Solution: length breadth area 3x 5y 3x × 5y = 15xy 9y 4y2.............. 4ab 5bc.............. 2l2m 3lm2.............. Example 4: Find the volume of each rectangular box with given length, breadth and height. length breadth height (i) 2ax 3by 5cz (ii) m2 n n 2p p2 m (iii) 2q 4q2 8q3 Solution: Volume = length × breadth × height Hence, for (i) volume = (2ax) × (3by) × (5cz) = 2 × 3 × 5 × (ax) × (by) × (cz) = 30abcxyz for (ii) volume = m2n × n2p × p2m = (m2 × m) × (n × n2) × (p × p2) = m3n3p3 for (iii) volume = 2q × 4q2 × 8q3 = 2 × 4 × 8 × q × q2 × q3 = 64q6 EXERCISE 8.2 1. Find the product of the following pairs of monomials. (i) 4, 7p (ii) – 4p, 7p (iii) – 4p, 7pq (iv) 4p3, – 3p (v) 4p, 0 2. Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively. (p, q); (10m, 5n); (20x2, 5y2); (4x, 3x2); (3mn, 4np) Reprint 2024-25 98 MATHEMATICS 3. Complete the table of products. First monomial → 2x –5y 3x 2 – 4xy 7x 2y –9x2y2 Second monomial ↓ 2x 4x 2..................... –5y –15x2y......... 3x 2.................................... – 4xy 7x 2y.................. –9x2y2.................. 4. Obtain the volume of rectangular boxes with the following length, breadth and height respectively. (i) 5a, 3a2, 7a4 (ii) 2p, 4q, 8r (iii) xy, 2x2y, 2xy2 (iv) a, 2b, 3c 5. Obtain the product of (i) xy, yz, zx (ii) a, – a2, a3 (iii) 2, 4y, 8y2, 16y3 (iv) a, 2b, 3c, 6abc (v) m, – mn, mnp 8.4 Multiplying a Monomial by a Polynomial Expression that contains two terms is called a binomial. An expression containing three terms is a trinomial and so on. In general, an expression containing, one or more terms with non-zero coefficient (with variables having non negative integers as exponents) is called a polynomial. 8.4.1 Multiplying a monomial by a binomial Let us multiply the monomial 3x by the binomial 5y + 2, i.e., find 3x × (5y + 2) = ? Recall that 3x and (5y + 2) represent numbers. Therefore, using the distributive law, 3x × (5y + 2) = (3x × 5y) + (3x × 2) = 15xy + 6x We commonly use distributive law in our calculations. For example: 7 × 106 = 7 × (100 + 6) = 7 × 100 + 7 × 6 (Here, we used distributive law) = 700 + 42 = 742 7 × 38 = 7 × (40 – 2) = 7 × 40 – 7 × 2 (Here, we used distributive law) = 280 – 14 = 266 Similarly, (–3x) × (–5y + 2) = (–3x) × (–5y) + (–3x) × (2) = 15xy – 6x and 5xy × (y2 + 3) = (5xy × y2) + (5xy × 3) = 5xy3 + 15xy. What about a binomial × monomial? For example, (5y + 2) × 3x = ? We may use commutative law as : 7 × 3 = 3 × 7; or in general a × b = b × a Similarly, (5y + 2) × 3x = 3x × (5y + 2) = 15xy + 6x as before. TRY THESE Find the product (i) 2x (3x + 5xy) (ii) a2 (2ab – 5c) Reprint 2024-25  ALGEBRAIC EXPRESSIONS AND I DENTITIES 99 8.4.2 Multiplying a monomial by a trinomial Consider 3p × (4p2 + 5p + 7). As in the earlier case, we use distributive law; 3p × (4p2 + 5p + 7) = (3p × 4p2) + (3p × 5p) + (3p × 7) = 12p3 + 15p2 + 21p Multiply each term of the trinomial by the monomial and add products. TRY THESE Observe, by using the distributive law, we are able to carry out the Find the product: multiplication term by term. (4p2 + 5p + 7) × 3p Example 5: Simplify the expressions and evaluate them as directed: (i) x (x – 3) + 2 for x = 1, (ii) 3y (2y – 7) – 3 (y – 4) – 63 for y = –2 Solution: (i) x (x – 3) + 2 = x2 – 3x + 2 For x = 1, x2 – 3x + 2 = (1)2 – 3 (1) + 2 =1–3+2=3–3=0 (ii) 3y (2y – 7) – 3 (y – 4) – 63 = 6y2 – 21y – 3y + 12 – 63 = 6y2 – 24y – 51 For y = –2, 6y2 – 24y – 51 = 6 (–2)2 – 24(–2) – 51 = 6 × 4 + 24 × 2 – 51 = 24 + 48 – 51 = 72 – 51 = 21 Example 6: Add (i) 5m (3 – m) and 6m2 – 13m (ii) 4y (3y2 + 5y – 7) and 2 (y3 – 4y2 + 5) Solution: (i) First expression = 5m (3 – m) = (5m × 3) – (5m × m) = 15m – 5m2 Now adding the second expression to it,15m – 5m2 + 6m2 – 13m = m2 + 2m (ii) The first expression = 4y (3y2 + 5y – 7) = (4y × 3y2) + (4y × 5y) + (4y × (–7)) = 12y3 + 20y2 – 28y The second expression = 2 (y – 4y + 5) = 2y3 + 2 × (– 4y2) + 2 × 5 3 2 = 2y3 – 8y2 + 10 Adding the two expressions, 12y3 + 20y2 – 28y + 2y3 – 8y2 + 10 14y3 + 12y2 – 28y + 10 Example 7: Subtract 3pq (p – q) from 2pq (p + q). Solution: We have 3pq (p – q) = 3p2q – 3pq2 and 2pq (p + q) = 2p2q + 2pq2 Subtracting, 2p2q + 2pq2 3p2q – 3pq2 – + 2 –pq + 5pq2 Reprint 2024-25 100 MATHEMATICS EXERCISE 8.3 1. Carry out the multiplication of the expressions in each of the following pairs. (i) 4p, q + r (ii) ab, a – b (iii) a + b, 7a2b2 (iv) a2 – 9, 4a (v) pq + qr + rp, 0 2. Complete the table. First expression Second expression Product... (i) a b+c+d... (ii) x+y–5 5xy (iii) p 6p2 – 7p + 5... (iv) 4p2q2 p2 – q2...... (v) a+b+c abc 3. Find the product.  2   −9 2 2  (i) (a2) × (2a22) × (4a26) (ii)  xy ×  x y   3   10   10 3   6 3  (iii)  − pq  ×  p q  5  (iv) x × x2 × x3 × x4 3 1 4. (a) Simplify 3x (4x – 5) + 3 and find its values for (i) x = 3 (ii) x =. 2 (b) Simplify a (a2 + a + 1) + 5 and find its value for (i) a = 0, (ii) a = 1 (iii) a = – 1. 5. (a) Add: p ( p – q), q ( q – r) and r ( r – p) (b) Add: 2x (z – x – y) and 2y (z – y – x) (c) Subtract: 3l (l – 4 m + 5 n) from 4l ( 10 n – 3 m + 2 l ) (d) Subtract: 3a (a + b + c ) – 2 b (a – b + c) from 4c ( – a + b + c ) 8.5 Multiplying a Polynomial by a Polynomial 8.5.1 Multiplying a binomial by a binomial Let us multiply one binomial (2a + 3b) by another binomial, say (3a + 4b). We do this step-by-step, as we did in earlier cases, following the distributive law of multiplication, (3a + 4b) × (2a + 3b) = 3a × (2a + 3b) + 4b × (2a + 3b) Observe, every term in one = (3a × 2a) + (3a × 3b) + (4b × 2a) + (4b × 3b) binomial multiplies every = 6a2 + 9ab + 8ba + 12b2 term in the other binomial. = 6a2 + 17ab + 12b2 (Since ba = ab) When we carry out term by term multiplication, we expect 2 × 2 = 4 terms to be present. But two of these are like terms, which are combined, and hence we get 3 terms. In multiplication of polynomials with polynomials, we should always look for like terms, if any, and combine them. Reprint 2024-25  ALGEBRAIC EXPRESSIONS AND I DENTITIES 101 Example 8: Multiply (i) (x – 4) and (2x + 3) (ii) (x – y) and (3x + 5y) Solution: (i) (x – 4) × (2x + 3) = x × (2x + 3) – 4 × (2x + 3) = (x × 2x) + (x × 3) – (4 × 2x) – (4 × 3) = 2x2 + 3x – 8x – 12 = 2x2 – 5x – 12 (Adding like terms) (ii) (x – y) × (3x + 5y) = x × (3x + 5y) – y × (3x + 5y) = (x × 3x) + (x × 5y) – (y × 3x) – ( y × 5y) = 3x2 + 5xy – 3yx – 5y2 = 3x2 + 2xy – 5y2 (Adding like terms) Example 9: Multiply (i) (a + 7) and (b – 5) (ii) (a2 + 2b2) and (5a – 3b) Solution: (i) (a + 7) × (b – 5) = a × (b – 5) + 7 × (b – 5) = ab – 5a + 7b – 35 Note that there are no like terms involved in this multiplication. (ii) (a2 + 2b2) × (5a – 3b) = a2 (5a – 3b) + 2b2 × (5a – 3b) = 5a3 – 3a2b + 10ab2 – 6b3 8.5.2 Multiplying a binomial by a trinomial In this multiplication, we shall have to multiply each of the three terms in the trinomial by each of the two terms in the binomial. We shall get in all 3 × 2 = 6 terms, which may reduce to 5 or less, if the term by term multiplication results in like terms. Consider (a + 7) × (a 2 + 3a + 5) = a × (a2 + 3a + 5) + 7 × (a2 + 3a + 5)

   binomial trinomial [using the distributive law] 3 2 2 = a + 3a + 5a + 7a + 21a + 35 = a3 + (3a2 + 7a2) + (5a + 21a) + 35 = a3 + 10a2 + 26a + 35 (Why are there only 4 terms in the final result?) Example 10: Simplify (a + b) (2a – 3b + c) – (2a – 3b) c. Solution: We have (a + b) (2a – 3b + c) = a (2a – 3b + c) + b (2a – 3b + c) = 2a2 – 3ab + ac + 2ab – 3b2 + bc = 2a2 – ab – 3b2 + bc + ac (Note, –3ab and 2ab are like terms) and (2a – 3b) c = 2ac – 3bc Therefore, (a + b) (2a – 3b + c) – (2a – 3b) c = 2a2 – ab – 3b2 + bc + ac – (2ac – 3bc) = 2a2 – ab – 3b2 + bc + ac – 2ac + 3bc = 2a2 – ab – 3b2 + (bc + 3bc) + (ac – 2ac) = 2a2 – 3b2 – ab + 4bc – ac Reprint 2024-25 102 MATHEMATICS EXERCISE 8.4 1. Multiply the binomials. (i) (2x + 5) and (4x – 3) (ii) (y – 8) and (3y – 4) (iii) (2.5l – 0.5m) and (2.5l + 0.5m) (iv) (a + 3b) and (x + 5) (v) (2pq + 3q2) and (3pq – 2q2) (vi) 2. Find the product. (i) (5 – 2x) (3 + x) (ii) (x + 7y) (7x – y) 2 2 (iii) (a + b) (a + b ) (iv) (p2 – q2) (2p + q) 3. Simplify. (i) (x2 – 5) (x + 5) + 25 (ii) (a2 + 5) (b3 + 3) + 5 (iii) (t + s2) (t2 – s) (iv) (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd) (v) (x + y)(2x + y) + (x + 2y)(x – y) (vi) (x + y)(x2 – xy + y2) (vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y (viii) (a + b + c)(a + b – c) WHAT HAVE WE DISCUSSED? 1. Expressions are formed from variables and constants. 2. Terms are added to form expressions. Terms themselves are formed as product of factors. 3. Expressions that contain exactly one, two and three terms are called monomials, binomials and trinomials respectively. In general, any expression containing one or more terms with non-zero coefficients (and with variables having non- negative integers as exponents) is called a polynomial. 4. Like terms are formed from the same variables and the powers of these variables are the same, too. Coefficients of like terms need not be the same. 5. While adding (or subtracting) polynomials, first look for like terms and add (or subtract) them; then handle the unlike terms. 6. There are number of situations in which we need to multiply algebraic expressions: for example, in finding area of a rectangle, the sides of which are given as expressions. 7. A monomial multiplied by a monomial always gives a monomial. 8. While multiplying a polynomial by a monomial, we multiply every term in the polynomial by the monomial. 9. In carrying out the multiplication of a polynomial by a binomial (or trinomial), we multiply term by term, i.e., every term of the polynomial is multiplied by every term in the binomial (or trinomial). Note that in such multiplication, we may get terms in the product which are like and have to be combined. Reprint 2024-25