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)#