# Factorize each of the following quadratic polynomials by using the method of completing the square:(i) $x^2+12x+20$(ii) $a^2-14a-51$

Given:

(i) $x^2+12x+20$

(ii) $a^2-14a-51$

To do:

We have to factorize the given quadratic polynomials.

Solution:

Factorizing algebraic expressions:

Factorizing an algebraic expression implies writing the expression as a product of two or more factors. Factorization is the reverse of distribution.

An algebraic expression is factored completely when it is written as a product of prime factors.

Completing the square is a method that is used to write a quadratic expression in a way such that it contains the perfect square.

(i) The given expression is $x^2+12x+20$.

Here,

The coefficient of $x^2$ is $1$

The coefficient of $x$ is $12$

The constant term is $20$

Coefficient of $x^2$ is $1$. So, we can factorize the given expression by adding and subtracting square of half of coefficient of $x$.

Therefore,

$x^2+12x+20=x^2+12x+20+6^2-6^2$              [Since $\frac{1}{2}\times12=6$]

$x^2+12x+20=x^2+12x+6^2+20-36$

$x^2+12x+20=x^2+2(x)(6)+6^2-16$

$x^2+12x+20=(x+6)^2-16$                     (Completing the square)

Now,

$(x+6)^2-16$ can be written as,

$(x+6)^2-16=(x+6)^2-4^2$               [Since $16=4^2$]

Using the formula $a^2-b^2=(a+b)(a-b)$, we can factorize the given expression as,

$(x+6)^2-16=(x+6)^2-4^2$

$(x+6)^2-16=(x+6+4)(x+6-4)$

$(x+6)^2-16=(x+10)(x+2)$

Hence, the given expression can be factorized as $(x+2)(x+10)$.

(ii) The given expression is $a^2-14a-51$.

Here,

The coefficient of $a^2$ is $1$

The coefficient of $a$ is $-14$

The constant term is $-51$

Coefficient of $a^2$ is $1$. So, we can factorize the given expression by adding and subtracting square of half of coefficient of $a$.

Therefore,

$a^2-14a-51=a^2-14a-51+7^2-7^2$              [Since $\frac{1}{2}\times14=7$]

$a^2-14a-51=a^2-14a+7^2-51-49$

$a^2-14a-51=a^2-2(a)(7)+7^2-100$

$a^2-14a-51=(a-7)^2-100$                     (Completing the square)

Now,

$(a-7)^2-100$ can be written as,

$(a-7)^2-100=(a-7)^2-10^2$                       [Since $100=10^2$]

Using the formula $a^2-b^2=(a+b)(a-b)$, we can factorize the given expression as,

$(a-7)^2-100=(a-7)^2-10^2$

$(a-7)^2-100=(a-7+10)(a-7-10)$

$(a-7)^2-100=(a+3)(a-17)$

Hence, the given expression can be factorized as $(a-17)(a+3)$.

Updated on: 12-Apr-2023

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