Page 125 - Start Up Mathematics_8 (Non CCE)

Factorization of Algebraic Expressions with a Common Monomial in Each Term
Step 1: In the algebraic expression, find the highest common factor (HCF) of its terms.

Step 2: Express each term of the algebraic expression as the product of the HCF and the quotient, when it
is divided by the HCF.
Step 3: Using distributive property of multiplication over addition, express the given algebraic expression
as the product of its HCF and the quotient obtained in step 2.
3 2
2
3 2
2 3
4 3
2
Example 2: Factorize: (a) 4x + 12 (b) 2x y – 5x y (c) 15x y + 5x y – 20x y
Solution:
3 2
2 3
2
(a) 4x + 12 (b) 2x y – 5x y
3 2
2
4x = 2 × 2 × x × x 2x y = 2 × x × x × x × y × y
2 3
12 = 2 × 2 × 3 5x y = 5 × x × x × y × y × y
3 2
2
2 2
2 3
HCF of 4x and 12 = 4 HCF of 2x y and 5x y = x y
2 2
So, we take 4 common from both the So, we take x y common from both
terms. the terms.
3 2
2
2
2 2
2 3
\ 4x + 12 = 4(x + 3) \ 2x y – 5x y = x y (2x – 5y)
2
4 3
3 2
(c) 15x y + 5x y – 20x y Extension
3 2
15x y = 3 × 5 × x × x × x × y × y If a polynomial is divisible by one
2
5x y = 5 × x × x × y and itself only, it is said to be a prime
4 3
20x y = 2 × 2 × 5 × x × x × x × x × y × y × y polynomial. 2 2
For example, 3x + 4, 2x + 3y , etc.
3 2
2
2
4 3
HCF of 15x y , 5x y and 20x y = 5x y
2
Taking 5x y common from all the terms, we get
2
2 2
3 2
2
4 3
15x y + 5x y – 20x y = 5x y(3xy + 1 – 4x y )
Factorization of Algebraic Expressions Having a Binomial as Common Factor
Step 1: Locate the common binomial.
Step 2: Write the given algebraic expression as a product of this common binomial and the quotient obtained
on dividing the given algebraic expression by this binomial.
Example 3: Factorize the following:
2
(a) 5(3x – 4) + 7(3x – 4) (b) 8(5x – 9y) – 12(5x – 9y)
2
2
(c) (2x – 3y) – 6x + 9y (d) 9(x – y) + 6(y– x)
Solution: (a) 5(3x – 4) + 7(3x – 4) = (5 + 7)(3x – 4) {Taking (3x – 4) common}
= 12(3x – 4)
2
(b) 8(5x – 9y) – 12(5x – 9y) = 4 × 2(5x – 9y) × (5x – 9y) – 4 × 3(5x – 9y)
= 4(5x – 9y) {2(5x – 9y) – 3} {Taking 4(5x – 9y) common}
= 4(5x – 9y)(10x – 18y – 3)
2
2
(c) (2x – 3y) – 6x + 9y = (2x – 3y) – 3 × 2x + 3 × 3y
2
= (2x – 3y) – 3(2x – 3y) {Taking (–3) common from –6x + 9y}
= (2x – 3y)(2x – 3y – 3) {Taking (2x – 3y) common}
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