# Factorize each of the following quadratic polynomials by using the method of completing the square:(i) $4y^2+12y+5$(ii) $p^2+6p-16$

Given:

(i) $4y^2+12y+5$

(ii) $p^2+6p-16$

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 $4y^2+12y+5$.

We can write $4y^2+12y+5$ as,

$4y^2+12y+5=4(y^2+3y+\frac{5}{4})$

Here,

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

The coefficient of $y$ is $3$

The constant term is $\frac{5}{4}$

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

Therefore,

$4y^2+12y+5=4(y^2+3y+\frac{5}{4})$

$4y^2+12y+5=4[y^2+3y+\frac{5}{4}+(\frac{3}{2})^2-(\frac{3}{2})^2]$              [Since $\frac{1}{2}\times3=\frac{3}{2}$]

$4y^2+12y+5=4[y^2+3y+(\frac{3}{2})^2+\frac{5}{4}-\frac{9}{4}]$

$4y^2+12y+5=4[y^2+2(y)(\frac{3}{2})+(\frac{3}{2})^2+\frac{5-9}{4}]$

$4y^2+12y+5=4[(y+\frac{3}{2})^2+\frac{-4}{4}]$                     (Completing the square)

$4y^2+12y+5=4[(y+\frac{3}{2})^2-1]$

Now,

$(y+\frac{3}{2})^2-1$ can be written as,

$(y+\frac{3}{2})^2-1=(y+\frac{3}{2})^2-1^2$

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

$4y^2+12y+5=4[(y+\frac{3}{2})^2-1^2]$

$4y^2+12y+5=4[(y+\frac{3}{2}+1)(y+\frac{3}{2}-1)]$

$4y^2+12y+5=4[(y+\frac{3+2}{2})(y+\frac{3-2}{2})]$

$4y^2+12y+5=4[(y+\frac{5}{2})(y+\frac{1}{2})]$

$4y^2+12y+5=2(y+\frac{5}{2})2(y+\frac{1}{2})]$

$4y^2+12y+5=(2y+2\times\frac{5}{2})(2y+2\times\frac{1}{2})]$

$4y^2+12y+5=(2y+5)(2y+1)]$

Hence, the given expression can be factorized as $(2y+1)(2y+5)$.

(ii) The given expression is $p^2+6p-16$.

Here,

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

The coefficient of $p$ is $6$

The constant term is $-16$

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

Therefore,

$p^2+6p-16=p^2+6p-16+3^2-3^2$              [Since $\frac{1}{2}\times6=3$]

$p^2+6p-16=p^2+6p+3^2-16-9$

$p^2+6p-16=p^2+2(p)(3)+3^2-25$

$p^2+6p-16=(p+3)^2-25$                     (Completing the square)

Now,

$(p+3)^2-25$ can be written as,

$(p+3)^2-25=(p+3)^2-5^2$                       [Since $25=5^2$]

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

$(p+3)^2-25=(p+3)^2-5^2$

$(p+3)^2-25=(p+3+5)(p+3-5)$

$(p+3)^2-25=(p+8)(p-2)$

Hence, the given expression can be factorized as $(p-2)(p+8)$.

Updated on: 12-Apr-2023

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