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Factorize each of the following quadratic polynomials by using the method of completing the square:
(i) $p^2+6p+8$
(ii) $q^2-10q+21$
Given:
The given quadratic polynomials are:
(i) $p^2+6p+8$
(ii) $q^2-10q+21$
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 $p^2+6p+8$.
Here,
The coefficient of $p^2$ is $1$
The coefficient of $p$ is $6$
The constant term is $8$
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,
$p^2+6p+8=p^2+6p+8+3^2-3^2$ [Since $\frac{1}{2}\times6=3$]
$p^2+6p+8=p^2+6p+3^2+8-9$
$p^2+6p+8=p^2+2(p)(3)+3^2-1$
$p^2+6p+8=(p+3)^2-1$ (Completing the square)
Now,
$(p+3)^2-1$ can be written as,
$(p+3)^2-1=(p+3)^2-1^2$
Using the formula $a^2-b^2=(a+b)(a-b)$, we can factorize the given expression as,
$(p+3)^2-1=(p+3)^2-1^2$
$(p+3)^2-1=(p+3+1)(p+3-1)$
$(p+3)^2-1=(p+4)(p+2)$
Hence, the given expression can be factorized as $(p+2)(p+4)$.
(ii) The given expression is $q^2-10q+21$.
Here,
The coefficient of $q^2$ is $1$
The coefficient of $q$ is $-10$
The constant term is $21$
Coefficient of $q^2$ is $1$. So, we can factorize the given expression by adding and subtracting square of half of coefficient of $q$.
Therefore,
$q^2-10q+21=q^2-10q+21+5^2-5^2$ [Since $\frac{1}{2}\times10=5$]
$q^2-10q+21=q^2-10q+5^2+21-25$
$q^2-10q+21=q^2-2(q)(5)+5^2-4$
$q^2-10q+21=(q-5)^2-4$ (Completing the square)
Now,
$(q-5)^2-4$ can be written as,
$(q-5)^2-4=(q-5)^2-2^2$ [Since $4=2^2$]
Using the formula $a^2-b^2=(a+b)(a-b)$, we can factorize the given expression as,
$(q-5)^2-4=(q-5)^2-2^2$
$(q-5)^2-4=(q-5+2)(q-5-2)$
$(q-5)^2-4=(q-3)(q-7)$
Hence, the given expression can be factorized as $(q-7)(q-3)$.
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