Question

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

Answer 1

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)`