How do you factor completely -6m^5+34m^3-40m?

Jul 2, 2016

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$- 6 {m}^{5} + 34 {m}^{3} - 40 m$

$= - 2 m \left(\sqrt{3} m - \sqrt{5}\right) \left(\sqrt{3} m + \sqrt{5}\right) \left(m - 2\right) \left(m + 2\right)$

Explanation:

First note that all of the terms are divisible by $- 2 m$. So we can separate that out as a factor:

$- 6 {m}^{5} + 34 {m}^{3} - 40 m = - 2 m \left(3 {m}^{4} - 17 {m}^{2} + 20\right)$

The remaining quartic expression can be treated as a quadratic in ${m}^{2}$.

We can use an AC method to find quadratic factors:

Find a pair of factors of $A C = 3 \cdot 20 = 60$ with sum $B = 17$.

The pair $12 , 5$ works, so we can use that to split the middle term and factor by grouping:

$3 {m}^{4} - 17 {m}^{2} + 20 = \left(3 {m}^{4} - 12 {m}^{2}\right) - \left(5 {m}^{2} - 20\right)$

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

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

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

$= \left(\sqrt{3} m - \sqrt{5}\right) \left(\sqrt{3} m + \sqrt{5}\right) \left(m - 2\right) \left(m + 2\right)$

Putting it all together:

$- 6 {m}^{5} + 34 {m}^{3} - 40 m$

$= - 2 m \left(\sqrt{3} m - \sqrt{5}\right) \left(\sqrt{3} m + \sqrt{5}\right) \left(m - 2\right) \left(m + 2\right)$