Question

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

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

Short answer

The Factored form of given polynomial is,

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

Step-by-step solution

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

localid="1644736388246" $\begin{array}{l}48m^4{n}^2-243n^2=3n^2\left(16m^2-81n^2\right)\\ 48m^4{n}^2-243n^2=3n^2\left((4m{)}^2-(9n{)}^2\right)\end{array}$

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.