How do you factor $4m^4 – 12m^2 + 8m$ ?

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Answer 1 · 2015-07-18T15:53:41

Separate out the common $4m$ factor, then identify roots in the remaining polynomial to find:

$4m^4-12m^2+8m = 4m(m-1)(m-1)(m+2)$

$4m^4-12m^2+8m = 4m(m^3-3m+2)$

Notice that the sum of the coefficients of $m^3-3m+2$ is $0$ . So $m=1$ is a root of $m^3-3m+2 = 0$ and $(m-1)$ is a factor.

$m^3-3m+2 = (m-1)(m^2+m-2)$

Now the sum of the coefficients of $m^2+m-2$ is also $0$ . So $m=1$ is a root of $m^2+m-2 = 0$ and $(m-1)$ is a factor.

$m^2+m-2 = (m-1)(m+2)$

Put these all together to get:

$4m^4-12m^2+8m = 4m(m-1)(m-1)(m+2)$

How do you factor 4m^4 – 12m^2 + 8m4m^4 – 12m^2 + 8m?