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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:
The given quadratic polynomials are:
(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 $(a-17)(a+3)$.
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