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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:
The given quadratic polynomials are:
(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 $(p-2)(p+8)$.
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