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Factorize each of the following quadratic polynomials by using the method of completing the square:
(i) $a^2+2a-3$
(ii) $4x^2-12x+5$
Given:
The given quadratic polynomials are:
(i) $a^2+2a-3$
(ii) $4x^2-12x+5$
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 $a^2+2a-3$.
Here,
The coefficient of $a^2$ is $1$
The coefficient of $a$ is $2$
The constant term is $-3$
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,
$a^2+2a-3=a^2+2a-3+1^2-1^2$ [Since $\frac{1}{2}\times2=1$]
$a^2+2a-3=a^2+2a+1^2-3-1$
$a^2+2a-3=a^2+2(a)(1)+1^2-4$
$a^2+2a-3=(a+1)^2-4$ (Completing the square)
Now,
$(a+1)^2-4$ can be written as,
$(a+1)^2-4=(a+1)^2-2^2$ [Since $4=2^2$]
Using the formula $a^2-b^2=(a+b)(a-b)$, we can factorize the given expression as,
$(a+1)^2-4=(a+1)^2-2^2$
$(a+1)^2-4=(a+1+2)(a+1-2)$
$(a+1)^2-4=(a+3)(a-1)$
Hence, the given expression can be factorized as $(a-1)(a+3)$.
(ii) The given expression is $4x^2-12x+5$.
We can write $4x^2-12x+5$ as,
$4x^2-12x+5=4(x^2-3x+\frac{5}{4})$
Here,
The coefficient of $x^2$ is $1$
The coefficient of $x$ is $-3$
The constant term is $\frac{5}{4}$
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,
$4x^2-12x+5=4(x^2-3x+\frac{5}{4})$
$4x^2-12x+5=4[x^2-3x+\frac{5}{4}+(\frac{3}{2})^2-(\frac{3}{2})^2]$ [Since $\frac{1}{2}\times3=\frac{3}{2}$]
$4x^2-12x+5=4[x^2-3x+(\frac{3}{2})^2+\frac{5}{4}-\frac{9}{4}]$
$4x^2-12x+5=4[x^2-2(x)(\frac{3}{2})+(\frac{3}{2})^2+\frac{5-9}{4}]$
$4x^2-12x+5=4[(x-\frac{3}{2})^2+\frac{-4}{4}]$ (Completing the square)
$4x^2-12x+5=4[(x-\frac{3}{2})^2-1]$
Now,
$(x-\frac{3}{2})^2-1$ can be written as,
$(x-\frac{3}{2})^2-1=(x-\frac{3}{2})^2-1^2$
Using the formula $a^2-b^2=(a+b)(a-b)$, we can factorize the given expression as,
$4x^2-12x+5=4[(x-\frac{3}{2})^2-1^2]$
$4x^2+12y+5=4[(x-\frac{3}{2}+1)(x-\frac{3}{2}-1)]$
$4x^2-12x+5=4[(x+\frac{-3+2}{2})(x+\frac{-3-2}{2})]$
$4x^2-12x+5=4[(x+\frac{-1}{2})(x+\frac{-5}{2})]$
$4x^2-12x+5=4[(x-\frac{1}{2})(x-\frac{5}{2})]$
$4x^2-12x+5=2(x-\frac{1}{2})2(x-\frac{5}{2})]$
$4x^2-12x+5=(2x-2\times\frac{1}{2})(2x-2\times\frac{5}{2})]$
$4x^2-12x+5=(2x-1)(2x-5)]$
Hence, the given expression can be factorized as $(2x-5)(2x-1)$.
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