Algebraic Expressions PDF

### Factors of Algebraic Expressions So, we can express any number as a product of two or more prime factors. Similarly, to the factors of an algebraic expression, we shall write it as the product of two or more algebraic expressions. Each expression is known as the factor of the given algebraic ex...

### Factors of Algebraic Expressions So, we can express any number as a product of two or more prime factors. Similarly, to the factors of an algebraic expression, we shall write it as the product of two or more algebraic expressions. Each expression is known as the factor of the given algebraic expression. So, we can conclude that the process of writing a given algebraic expression as a product of two or more algebraic factors is known as factorisation. ### Factors of Algebraic Expressions To understand the concept of factorisation of algebraic expression, we begin with the simplest algebraic expression which is a monomial and write it as the product of two more factors. Let us illustrate it with the help of an example. **Example 18:** Write 8a²b in the form of product of some factors. **Solution:** 8a²b = 1 x 8a²b 8a²b = 2a² x 4b 8a²b = 8 x a²b 8a²b = 2b x 4a² 8a²b = 8a x ab 8a²b = 4ab x 2a 8a²b = 8ab xa 8a²b = 2ab x 4a 8a²b = 8a² x b 8a²b = 2 x 2ab x 2a 8a²b = a² x 8b So, 1, 8a²b; 8, a²b; 8a, ab; 8ab, a; 8a², b; a², 8b; 2a², 4b; 2b, 4a²; 4ab, 2a; 2ab, 4a; 2, 2ab x 2a are some possible factors of 8a²b. ### Common Factors of Monomials To understand the term common factors, let us write some factors of 4xy and 9x. Factors of 4xy are 1, 4xy, 4, x, 4x, y, 4y, xy, 2x, 2y, 2xy, 2. Factors of 9x are 1, x, 9, 9x, 3, 3x. Factors common to both 9x and 4xy are 1 and x. We, therefore, infer that the factors which are common to both monomials are common factors. It means that these factors will occur in both the monomials. **Example 19:** Find the common factors of 3x² and 6xy. **Solution:** Factors of 3x² are 1, 3x², 3, x², 3x, x. Factors of 6xy are 1, 6xy, 6, xy, 6x, y, 6y, x, 2, 3xy, 2x, 3y, 2xy, 3x, 6. The .... ... ### Highest Common Factor of Monomials Highest Common Factor (H.C.F.) of given monomials is a common factor having greatest coefficient and highest power of the variable. To understand the concept more clearly, let us take an example. **Example 20:** Find the H.C.F. of 4a³b³ and 12abc. **Solution:** Common factors of 4a³b³ and 12abc are 1, 2, 4, a, b, 4a, 4b, ab, 4ab, b², 4b², ab², 4ab². Of all these, let us consider the factor 4ab². 1. Here, 4 represents the H.C.F. of numerical coefficients of two given monomials 2. a is the highest common power of variable a and b² is the highest common power of the variable b, in the two monomials. So 4ab² is the H.C.F. of 4a³b³ and 12abc. **Remember** To find the H.C.F. of two or more monomials, we follow the given steps: - Find the numerical coefficients and calculate their H.C.F. - Find the common variables appearing in the given monomials. - Find the highest common power of each variable in the given monomials. - The product of H.C.F. of numerical coefficients and the highest common powers of the variables gives the H.C.F. of the monomials. **Example 21:** Find the H.C.F. of 21x³y² and 35x⁵y. **Solution:** H.C.F. of 21 and 35 is 7. Common variables here are x and y. Highest common power of x in two monomials = x³ Highest common power of y in two monomials = y So H.C.F. = 7x³y ### Factorisation by Taking Out a Common Factor We know that an algebraic expression is a sum or difference of monomials. Let us take up an example to find the factorisation of algebraic expression consisting of a common monomial. **Example 22:** Factorise the binomial 3x²y - 6xy². **Solution:** We know H.C.F. shall be the product of the highest common coefficient and highest common powers of the variables respectively. H.C.F. of 3x²y and 6xy² is (3) (xy) or 3xy. So, 3x²y-6xy² = 3xy (x-2y). 3xy and (x-2y) are the two factors of the given expression. **Remember** For factorisation by taking out common factor- - Find the H.C.F. of all the terms in an expression. - Express each term of the given expression as a product of H.C.F. and the quotient when it is divided by H.C.F. - Use the distributive property of multiplication over addition, to express the given expression as a product of H.C.F. and the quotient of the given expression divided by H.C.F. **Example 23:** Find the H.C.F. of the terms and factorise the expression, 18x³y² + 36xy - 24x²y³ **Solution:** H.C.F. of 18x³y², 36xy and - 24x²y³ is 6xy² 18x³y² + 36xy - 24x²y³ = 6xy² (3x² + 6² - 4xy) ### Worksheet 5 1. Express the following as a product of its any two factors (in four different ways). - (i) 12x²y - (ii) 18ab² - (iii) 24cb 2. Find the H.C.F. of the following monomials. - (i) 2a⁵ and 12a² - (ii) 9x³y and 18x²y³ - (iii) a²b³ and a³b² - (iv) 15a³, - 45a², 150a - (v) 2x³y², 10x²y³, 14x²y² 3. Find the H.C.F. of the terms and factorise. - (i) 5y-15y² - (ii) 16m-4m² - (iii) 8x³y² + 8x³ - (iv) 20x³- 40x² + 80x - (v) x² - 3x²y² - 6xy² - (vi) 8x²y² - 16xy² + 24xy ### Factorisation of Algebraic Expression When a Binomial is a Common Factor To factorise an algebraic expression containing a binomial as a common factor, we write the expression as a product of the binomial and quotient obtained by dividing the given expression by its binomial. The following examples will illustrate the procedure: **Example 24:** Factorise (x + 2) y + (x + 2) x **Solution:** Here, (x + 2) is a binomial common to both the terms of a given expression. So, we have, *(x + 2) y + (x + 2) x = (x + 2) (y + x)* = (x + 2) (y + x) (x + 2) and (y + x) are two factors of the given algebraic expression. **Example 25:** Factorise (5x² - 10xy) - 4x + 8y **Solution:** Taking 5x common from (5x² - 10xy), we get *5x (x-2y) - 4x + 8y* Now, taking - 4 common from - 4x + 8y, we get *5x (x-2y) - 4 (x-2y)* = *(x-2y) (5x-4)* [Taking (x - 2y) common from both products] **Example 26:** Factorise (2x - 3y) (a + b) + (3x - 2y) (a + b) **Solution:** Here, in the given expression, (a + b) is common to both the terms, so = *(a + b) (2x-3y + 3x - 2y)* = *(a + b) (5x – 5y)* Here, 5 is common to the term in (5x – 5y), so we have = *(a + b) 5 (x - y) = 5 (a + b) (x - y).* ### Factorisation by Regrouping the Terms In the previous section, we have been able to factorise the given expression by taking a factor common from all the terms. But it is not always possible to have a common factor. So we try to form groups in such a way that a factor can be taken out from each group and the expression can be factorised. Let us take an example. Now,... ### You Must Know 1. An algebraic expression is a combination of constants and variables connected by +, - , x and ÷ . 2. Algebraic expression with one term is called a monomial, with two terms, a binomial and with three terms, a trinomial. 3. Product of variables follows the rule of exponents: x^m x x^n = x ^m+n. 4. Product of monomials is the product of constants of the given monomials and their variables. 5. To multiply a monomial and a binomial, the monomial is multiplied with each term of the binomial and the products are added. 6. To multiply two binomials, each term of one binomial is multiplied with each term of the second binomial and the products are then added. 7. To multiply a binomial and a trinomial, each term of the binomial is multiplied with each term of the trinomial and the products are then added. ### Multiplication of a Binomial and a Trinomial Do you remember what a trinomial is? It is an algebraic expression containing three terms, for instance, 9x²-5y² + y. You can look upon a trinomial as sum of three monomials with unlike terms (x² -5² and y in the above example) To multiply a binomial of the type (A + B) and a trinomial of the type (C + D + E), we follow the following steps. Step 1: Multiply A by the trinomial (C + D + E). Step II: We get sum of the products of the form AC + AD + AE Step III: Now, multiply B by the trinomial (C + D + E). ### Multiplication of Binomials You are now familiar with multiplication of a monomial and a binomial. Do you remember how to evaluate (a + b) (c + d) where, a, b, c, d are numbers? Recall that you do it by using distributive property twice as shown in the following: *(a + b) (c + d) = a (c + d) + b (c + d)* = ac + ad + bc + bd *(a + b) (c + a) = (a + b) c + (a + b) d* or, = ac + bc + ad + bd which is same as (1) You will use the same property for multiplication of binomials. Let us take an example. **Example 12:** Find the product of (x + 3y) and (2x + y). **Solution:** (x+3y) (2x + y) = x(2x + y) + 3y(2x + y) ### Multiplication of a Monomial and a Binomial Recall that a binomial is an algebraic expression having two terms. You may look upon a binomial as sum or difference of two monomials having unlike terms, e.g. binomial 3xy is sum of monomials 3xy and 2y². Now, to illustrate multiplication of a monomial and a binomial, we consider the following examples. **Example 6:** Multiply 3xy and 8x² + 7xy. **Solution:** *3xy(8x² + 7x³) = 3xy x 8x² + 3xy x 7xy* = 24x²y + 21x²y² Thus, to multiply a monomial and a binomial we take the following steps: Step 1: Multiply the monomial with first term of the binomial. Step II: Multiply the monomial with second term of the binomial. Step III: Add the products obtained in Step I and Step II if the terms of the binomial are separated by '+' sign. Subtract the terms if separated by '-' sign. **Example 7:** Find the product 2a² (9b² + 5ab). **Solution:** Step 1: 2a²x9b² = 18a²b² Step II: 2a² x 5ab = 10a²b Step III: 18a²b²+10a²b So, 2a² (9b² + 5ab) = 18a²b²+10a²b **Solution:** Step 1: *axa² = a³* Step II: *(-21) x a x bx c = -21a²bc* Thus... *(-3ab) x (7a²c) = - 21a²bc* **Note:** For finding the product of more than two monomials, you will follow the same steps. **Example 3:** Find the product (6x²y) x (2/3 xy²)x(-5yz²). **Solution:** Step 1: *6x²/3 x -5 = -20* Step II: *xxx = x³* *yxy²xy = y⁴* Step III: *(-20) x x³ x y⁴ x z² = -20x³y⁴z²* So, *(6x²y) x (2/3 xy²) x (-5yz²) = -20x³y⁴z²)* **Note:** Since the order in which numbers are added or multiplied does not affect the sum or product of numbers, monomials can be arranged in any order while multiplying. **Example 4:** Find the product (5x²y)x(-3/5 y²z)x(2xz²). Also verify the result for y = -1 and z = 2. **Solution:** *(5x²y)x(-3/5 y²z)x(2xz²)= (5x^-3/5)x(xxx)x(yxy²)x(zzx²)* = -6x⁴y³z² Therefore, *(5x²y)x(-3/5 y²z)x(2xz²) = - 6x⁴y³z²* When x = 1, y = - 1, z= 2, let us verify the equation (1). ... ### Chapter 6: Algebraic Expressions **Introduction** You have studied algebraic expressions in Class -VI. Do you remember what an algebraic expression is? An algebraic expression is a combination of constants and variables connected by means of four fundamental operations (+, -, x and ÷). For instance, 2x + 3, 8a²b+ a²b, 10y–x– 1/2 + 3/4 are algebraic expressions. You may also recall that: 1. Parts of an algebraic expression separated by the symbols + or - are called terms of the algebraic expression. 2. An algebraic expression having only one term is called a monomial 3. An algebraic expression having two terms is called a binomial. 4. An algebraic expression having three terms is called a trinomial. You know addition and subtraction of algebraic expressions. Let us look at the following examples to brush up our memory. **Example 1:** Add - 8x² + 6y² + 5 and 2x² - 3y². **Solution:** Adding by column method: - 8x² + 6y² + 5 +2x² - 3y² - 6x² + 3y² + 5 (Observe that the like terms are written below each other and they are added) Now, adding by horizontal method: (-8x² + 6y² + 5) + (2x² - 3y²) = (-8x² + 2x²) + (6y² - 3y²) + 5 = - 6x² + 3y² +5 (Like terms are combined together)