Chapter 8.7, Problem 5BGP

To determine

To factor: The given polynomial.

Answer to Problem 5BGP

The factored polynomial is $6x^4-96=6\left(x-2\right)\left(x+2\right)\left(x^2+4\right)$ .

Explanation of Solution

Given information:

The given polynomial is $6x^4-96$ .

Concept used:

The difference of two squares can be factored as $a^2-b^2=\left(a-b\right)\left(a+b\right)$ .

Calculations:

The given polynomial E (say) is $E=6x^4-96$ that can be written as

  $E=6\left(x^4-16\right)$

  $\Rightarrow E=6\left\{{(x^2)}^2-{(4)}^2\right\}$

  $\Rightarrow E=6\left(x^2-4\right)\left(x^2+4\right)$

  $\Rightarrow E=6\left\{{(x)}^2-{(2)}^2\right\}\left(x^2+4\right)$

  $\Rightarrow E=6\left(x-2\right)\left(x+2\right)\left(x^2+4\right)$

Thus, the factored polynomial is $E=6x^4-96=6\left(x-2\right)\left(x+2\right)\left(x^2+4\right)$ .

Conclusion:

The factored polynomial is $6x^4-96=6\left(x-2\right)\left(x+2\right)\left(x^2+4\right)$ .