Q. 394 - Intermediate Algebra

Question

In the following exercises, factor completely using the difference of squares pattern, if possible.

$6p^2{q}^2-54p^2$

Step-by-Step Solution

Answer

The Factored form of given polynomial is, 

$6p^2(q+3)(q-3)$

Step 1. Given Information

We are given a polynomial,  

$6p^2{q}^2-54p^2$

The formula used for factoring using the difference of squares pattern is, 

$a^2-b^2=(a+b)(a-b)$

Step 2. Factorizing the polynomial

Factoring out greatest common factor $6p^2$, we get

$$\begin{gathered} 6p^2{q}^2-54p^2=6p^2(q^2-9) \\ 6p^2{q}^2-54p^2=6p^2(q^2-3^2) \end{gathered}$$

Using $a^2-b^2=(a+b)(a-b)$, we get 

$6p^2{q}^2-54p^2=6p^2(q+3)(q-3)$

Step 3. Checking the factorization by multiplying

Multiplying the factors, we get 

$$\begin{gathered} 6p^2(q+3)(q-3)=6p^2{q}^2-54p^2 \\ 6p^2(q^2-3q+3q-9)=6p^2{q}^2-54p^2 \\ 6p^2(q^2-9)=6p^2{q}^2-54p^2 \\ 6p^2{q}^2-54p^2=6p^2{q}^2-54p^2 \\ LHS=RHS \end{gathered}$$

Hence the factorization is correct.