Factorization of algebraic expressions when a binomial is a common factor:
The expression is written as the product of binomial and the quotient obtained by dividing the given expression is by its binomial.
Solved examples when a binomial is a common factor:
1. Factorize the expression (3x + 1)^{2} – 5(3x + 1)
Solution:
(3x + 1)^{2} – 5(3x + 1)
The two terms in the above expression are (3x + 1)^{2} and 5(3x + 1)
= (3x + 1) (3x + 1) – 5(3x + 1)
Here, we observe that the binomial (3x + 1) is common to both the terms.
= (3x + 1) [(3x + 1) – 5]; [taking common (3x + 1)]
= (3x + 1) (3x - 4)
Therefore, (3x + 1) and (3x - 4) are two factors of the given algebraic expression.
2. Factorize the algebraic expression 2a(b - c) + 3(b – c)
Solution:
2a(b - c) + 3(b – c)
The two terms in the above expression are 2a(b - c), 3(b – c)
Here, we observe that the binomial (b – c) is common to both
the terms, then we get
= 2a(b – c) + 3(b – c)
= (b – c) [2a + 3]; [taking common (b – c)]
Therefore, (b – c) and (2a + 3) are two factors of the given algebraic expression.
3. Factorize the expression (2a – 3b) (x – y) + (3a – 2b) (x – y)
Solution:
(2a – 3b) (x – y) + (3a – 2b) (x – y)
The two terms in the above expression are (2a – 3b) (x – y)
and (3a – 2b) (x – y)
Here, we observe that the binomial (x – y) is common to both the terms, then we get
= (x – y) [(2a – 3b) + (3a – 2b)]
= (x – y) [(2a – 3b) + (3a – 2b)]
= (x – y) [2a – 3b + 3a – 2b]
= (x – y) [5a - 5b]
Taking common 5, we get
= (x – y) 5(a – b)
= 5(x – y) (a – b)
Therefore, 5, (x – y) and (a – b) are three factors of the given algebraic expression.
8th Grade Math Practice
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