Chapter 6.2, Problem 3MRE

To determine

To calculate: The factors of the polynomial $16x^4-81$.

Answer to Problem 3MRE

The polynomial $16x^4-81$ is factored as $\left(2x-3\right)\left(2x+3\right)\left(4x^2+9\right)$.

Explanation of Solution

Given information:

Thepolynomial $16x^4-81$.

Formula used:

The difference of square of two numbers a and bis $a^2-b^2=\left(a-b\right)\left(a+b\right)$.

Calculation:

Consider the polynomial $16x^4-81$.

Recall that difference of square of two numbers a and bis $a^2-b^2=\left(a-b\right)\left(a+b\right)$.

The term $16x^4$ is the square of $4x^2$ and 81 is the square of 9.

Therefore, the polynomial is rewritten as,

  $\begin{array}{c}16x^4-81={\left(4x^2\right)}^2-9^2\\ =\left(4x^2-9\right)\left(4x^2+9\right)\end{array}$

Again term $4x^2$ is the square of $2x$ and 9 is the square of 3. Apply the concept that difference of square of two numbers a and bis $a^2-b^2=\left(a-b\right)\left(a+b\right)$.

  $\begin{array}{c}16x^4-81={\left(4x^2\right)}^2-9^2\\ =\left(4x^2-9\right)\left(4x^2+9\right)\\ =\left[{\left(2x\right)}^2-3^2\right]\left(4x^2+9\right)\\ =\left(2x-3\right)\left(2x+3\right)\left(4x^2+9\right)\end{array}$

Thus, thepolynomial $16x^4-81$ is factored as $\left(2x-3\right)\left(2x+3\right)\left(4x^2+9\right)$.