Question

In the following exercises, factor completely.

$m^4-81$

Short answer

The factored form of the polynomial is$\left(m^2+9\right)\left(m^2-9\right).$

Step-by-step solution

Step 1. Given Information.

The given polynomial is $m^4-81.$

Step 2. Factoring by the difference of squares

There is no GCF in the given polynomial.

The polynomial is a binomial and it is the type of difference of squares.

To factor, the given polynomial completely writes it as a product of conjugates.

So,

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

The polynomial is factored completely.

Thus, the factored form is $\left(m^2-9\right)\left(m^2+9\right).$

Step 3. Check the answer 

To check the answer, expand the factored form we get,

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

L.H.S = R.H.S

Thus, the answer is right.