How do you factor completely -6m^5+34m^3-40m−6m5+34m3−40m?

1 Answer
George C.
Jul 2, 2016

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-6m^5+34m^3-40m−6m5+34m3−40m

= -2m(sqrt(3)m-sqrt(5))(sqrt(3)m+sqrt(5))(m-2)(m+2)=−2m(√3m−√5)(√3m+√5)(m−2)(m+2)

Explanation:

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

-6m^5+34m^3-40m = -2m(3m^4-17m^2+20)−6m5+34m3−40m=−2m(3m4−17m2+20)

The remaining quartic expression can be treated as a quadratic in m^2m2.

We can use an AC method to find quadratic factors:

Find a pair of factors of AC = 3*20 = 60AC=3⋅20=60 with sum B=17B=17.

The pair 12, 512,5 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)3m4−17m2+20=(3m4−12m2)−(5m2−20)

=3m^2(m^2-4)-5(m^2-4)=3m2(m2−4)−5(m2−4)

=(3m^2-5)(m^2-4)=(3m2−5)(m2−4)

=(3m^2-5)(m-2)(m+2)=(3m2−5)(m−2)(m+2)

=(sqrt(3)m-sqrt(5))(sqrt(3)m+sqrt(5))(m-2)(m+2)=(√3m−√5)(√3m+√5)(m−2)(m+2)

Putting it all together:

-6m^5+34m^3-40m−6m5+34m3−40m

= -2m(sqrt(3)m-sqrt(5))(sqrt(3)m+sqrt(5))(m-2)(m+2)=−2m(√3m−√5)(√3m+√5)(m−2)(m+2)

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