# How do you factor completely 5x^4 +10x^2 -15?

May 10, 2016

$5 {x}^{4} + 10 {x}^{2} - 15$

$= 5 \left({x}^{2} + 3\right) \left(x - 1\right) \left(x + 1\right)$

$= 5 \left(x - \sqrt{3} i\right) \left(x + \sqrt{3} i\right) \left(x - 1\right) \left(x + 1\right)$

#### Explanation:

• Separate out the common scalar factor $5$.

• Factor as a quadratic in ${x}^{2}$.

• Use the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

as follows:

$5 {x}^{4} + 10 {x}^{2} - 15$

$= 5 \left({x}^{4} + 2 {x}^{2} - 3\right)$

$= 5 \left({\left({x}^{2}\right)}^{2} + 2 \left({x}^{2}\right) - 3\right)$

$= 5 \left({x}^{2} + 3\right) \left({x}^{2} - 1\right)$

$= 5 \left({x}^{2} + 3\right) \left(x - 1\right) \left(x + 1\right)$

Then if we allow Complex coefficients...

$= 5 \left({x}^{2} - {\left(\sqrt{3} i\right)}^{2}\right) \left(x - 1\right) \left(x + 1\right)$

$= 5 \left(x - \sqrt{3} i\right) \left(x + \sqrt{3} i\right) \left(x - 1\right) \left(x + 1\right)$

May 10, 2016

$5 \left({x}^{2} + 3\right) \left(x + 1\right) \left(x - 1\right)$

#### Explanation:

Divide the common factor of 5 out first.

This is a disguised quadratic: $5 \left({\left({x}^{2}\right)}^{2} + 2 {\left(x\right)}^{2} - 3\right)$

Find factors of 3 which subtract to give 2.

=$5 \left({x}^{2} + 3\right) \left({x}^{2} - 1\right)$

=$5 \left({x}^{2} + 3\right) \left(x + 1\right) \left(x - 1\right)$