How do you factor #y= 4t^5-12t^3+8t^2# ?

1 Answer
Dec 27, 2015

Answer:

Separate out the common factor #4t^2#, then use the coefficient zero sum check to find:

#4t^5-12t^3+8t^2 = 4t^2(t-1)(t-1)(t+2)#

Explanation:

First notice that all of the terms are divisible by #4t^2#, so separate out that factor first:

#y = 4t^5-12t^3+8t^2#

#=4t^2(t^3-3t+2)#

Next notice that the sum of the coefficients of the terms of #t^3-3t+2# is zero. That is, #1-3+2 = 0#. So #t=1# is a zero of this cubic and #(t-1)# is a factor:

#(t^3-3t+2) = (t-1)(t^2+t-2)#

Notice that the sum of the coefficients of the terms of #(t^2+t-2)# is also zero, so there is another factor #(t-1)#:

#(t^2+t-2) = (t-1)(t+2)#

Putting this all together:

#4t^5-12t^3+8t^2 = 4t^2(t-1)(t-1)(t+2)#