Chapter 8.7, Problem 14CYU

To determine

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

Answer to Problem 14CYU

The factored polynomial is $256n^4-c^4=\left(4n-c\right)\left(4n+c\right)\left(16n^2+c^2\right)$ .

Explanation of Solution

Given information:

The given polynomial is $256n^4-c^4$ .

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 $E=256n^4-c^4$ that can be written as

  $E={\left(16n^2\right)}^4-{\left(c^2\right)}^4$    Difference of two squares

  $\Rightarrow E=\left(16n^2-c^2\right)\left(16n^2+c^2\right)$ .

  $\Rightarrow E=\left({(4n)}^2-{(c)}^2\right)\left(16n^2+c^2\right)$    Again difference of two squares

  $\Rightarrow E=\left(4n-c\right)\left(4n+c\right)\left(16n^2+c^2\right)$

Thus, the factored polynomial is $256n^4-c^4=\left(4n-c\right)\left(4n+c\right)\left(16n^2+c^2\right)$ .

(The polynomial $256n^4-c^4$ is not a prime polynomial.)

Conclusion:

The factored polynomial is $256n^4-c^4=\left(4n-c\right)\left(4n+c\right)\left(16n^2+c^2\right)$ .