Chapter 8.7, Problem 11CYU

To determine

To factor: The given polynomial if possible. If polynomial cannot be factored write it a prime.

Answer to Problem 11CYU

The factored polynomial is $u^4-81=\left(u-3\right)\left(u+3\right)\left(u^2+9\right)$ .

Explanation of Solution

Given information:

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

Concept used:

The difference of two squares can be factored as $a^2-b^2=\left(a-b\right)\left(a+b\right)$ , and the difference $x^4-y^4=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)$ .

Calculations:

The given polynomial E (say) is $E=u^4-81$ that can be written as

  $E={\left(u^2\right)}^2-{\left(9\right)}^2$    Difference of two squares

  $\Rightarrow E=\left(u^2-9\right)\left(u^2+9\right)$ .

  $\Rightarrow E=\left({(u)}^2-{(3)}^2\right)\left(u^2+9\right)$    Again difference of two squares

  $\Rightarrow E=\left(u-3\right)\left(u+3\right)\left(u^2+9\right)$

Thus, the factored polynomial is $u^4-81=\left(u-3\right)\left(u+3\right)\left(u^2+9\right)$ .

(The polynomial $u^4-81$ is not a prime.)

Conclusion:

The factored polynomial is $u^4-81=\left(u-3\right)\left(u+3\right)\left(u^2+9\right)$ .