Question

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

Answer 1

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`-6m^5+34m^3-40m`

`= -2m(sqrt(3)m-sqrt(5))(sqrt(3)m+sqrt(5))(m-2)(m+2)`

Explanation:

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

`-6m^5+34m^3-40m = -2m(3m^4-17m^2+20)`

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 `AC = 3*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:

`3m^4-17m^2+20 = (3m^4-12m^2)-(5m^2-20)`

`=3m^2(m^2-4)-5(m^2-4)`

`=(3m^2-5)(m^2-4)`

`=(3m^2-5)(m-2)(m+2)`

`=(sqrt(3)m-sqrt(5))(sqrt(3)m+sqrt(5))(m-2)(m+2)`

Putting it all together:

`-6m^5+34m^3-40m`

`= -2m(sqrt(3)m-sqrt(5))(sqrt(3)m+sqrt(5))(m-2)(m+2)`