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 $(q-7)(q-3)$.

Updated on: 12-Apr-2023

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