Q 190 - 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 complete factor of ploynomial is $3n^2\left(2m-3\right)\left(2m+3\right)\left(4m^2+9\right)$

Step 1. Given information

Consider the given ploynomial.

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

Step 2. Formula used

$a^2-b^2=\left(a+b\right)\left(a-b\right)$

Step 3. Find the factor using difference of square pattern.

$$\begin{gathered} 48m^4{n}^2-23n^2=3n^2\left(16m^4-81\right) \\ =3n^2\left({\left(4m^2\right)}^2-9^2\right) \\ =3n^2\left(4m^2-9\right)\left(4m^2+9\right) \end{gathered}$$

Step 4. Write them as product of conjugate.

$$\begin{gathered} 3n^2\left(4m^2-9\right)\left(4m^2+9\right)=3n^2\left({\left(2m\right)}^2-3^2\right)\left(4m^2+9\right) \\ =3n^2\left(2m-3\right)\left(2m+3\right)\left(4m^2+9\right) \end{gathered}$$