Q. 398 - Intermediate Algebra

Question

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

$48m^4{n}^2-243n^2$

Step-by-Step Solution

Answer

The Factored form of given polynomial is,

$3n^2(4m-9n)(4m+9n)$

Step 1. Given Information

 We are given a polynomial,

$48m^4{n}^2-243n^2$

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 $3n^2$, we get

$\begin{array}{l}48m^4{n}^2-243n^2=3n^2\left(16m^2-81n^2\right)\\ 48m^4{n}^2-243n^2=3n^2\left((4m{)}^2-(9n{)}^2\right)\end{array}$

Using $a^2-b^2=(a+b)(a-b)$, we get

$48m^4{n}^2-243n^2=3n^2(4m-9n)(4m+9n)$

Step 3. Checking the factorization by multiplying

$$\begin{gathered} 3n^2(4m-9n)(4m+9n)=48m^4{n}^2-243n^2 \\ 3n^2(16m^2+36mn-36mx-81n^2)=48m^4{n}^2-243n^2 \\ 3n^2(16m^2-81n^2)=48m^4{n}^2-243n^2 \\ 48m^4{n}^2-243n^2=48m^4{n}^2-243n^2 \\ LHS=RHS \end{gathered}$$

Hence the factorization is correct.