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 $(2x-5)(2x-1)$.

Akhileshwar Nani

Updated on: 12-Apr-2023

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