Chapter 8.7, Problem 13CYU

To determine

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

Answer to Problem 13CYU

The factored polynomial is $20r^4-45n^4=5\left(\sqrt{2}r-\sqrt{3}n\right)\left(\sqrt{2}r+\sqrt{3}n\right)\left(2r^2+3n^2\right)$ .

Explanation of Solution

Given information:

The given polynomial is $20r^4-45n^4$ .

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 $E=20r^4-45n^4$ that can be written as

  $E=5\left(4r^4-9n^4\right)$

  $E=5\left\{{\left(2r^2\right)}^2-{\left(3n^2\right)}^2\right\}$    Difference of two squares

  $\Rightarrow E=5\left(2r^2-3n^2\right)\left(2r^2+3n^2\right)$ .

  $\Rightarrow E=5\left({(\sqrt{2}r)}^2-{(\sqrt{3}n)}^2\right)\left(2r^2+3n^2\right)$    Again difference of two squares

  $\Rightarrow E==5\left(\sqrt{2}r-\sqrt{3}n\right)\left(\sqrt{2}r+\sqrt{3}n\right)\left(2r^2+3n^2\right)$

Thus, the factored polynomial is $20r^4-45n^4=5\left(\sqrt{2}r-\sqrt{3}n\right)\left(\sqrt{2}r+\sqrt{3}n\right)\left(2r^2+3n^2\right)$ .

(The polynomial $20r^4-45n^4$ is not a prime polynomial.)

Conclusion:

The factored polynomial is $20r^4-45n^4=5\left(\sqrt{2}r-\sqrt{3}n\right)\left(\sqrt{2}r+\sqrt{3}n\right)\left(2r^2+3n^2\right)$ .