Question

How do you factor completely: `6x^4-9x^3-36x^2+54x`?

Answer 1

Factor by grouping to get:

`6x^4-9x^3-36x^2+54x`

`=3x(x^2-6)(2x-3)`

`=3x(x-sqrt(6))(x+sqrt(6))(2x-3)`

Explanation:

First, note that every term is divisible by `3x`, so

`6x^4-9x^3-36x^2+54x = 3x(2x^3-3x^2-12x+18)`

Next, to factor `2x^3-3x^2-12x+18`, notice that the ratio of the 1st and 2nd terms is the same as the ratio of the 3rd and 4th terms. So factoring by grouping will work:

`2x^3-3x^2-12x+18`

`=(2x^3-3x^2)-(12x-18)`

`=x^2(2x-3)-6(2x-3)`

`=(x^2-6)(2x-3)`

Next, use the difference of squares identity:

`a^2-b^2 = (a-b)(a+b)`,

with `a=x` and `b=sqrt(6)` to get:

`(x^2-6)(2x-3)`

`=(x^2-sqrt(6)^2)(2x-3)`

`=(x-sqrt(6))(x+sqrt(6))(2x-3)`

Answer 2

The answer is `3x(2x-3)(x+sqrt6)(x-sqrt6)`.

Explanation:

Factor by grouping.

`(6x^4-9x^3)-(36x^2-54x)`

Factor out the GCF for each group.

`3x^3(2x-3)-18x(2x-3)`

Factor out `(2x-3)`.

`(3x^3-18x)(2x-3)`

`(3x^3-18x)` can be factored further.

`3x(x^2-6)`

Substitute it back into the previous factorization.

`3x(x^2-6)(2x-3)`

`(x^2-6)` can be factored further.

`(x+sqrt6)(x-sqrt6)`

Put all of the factors together.

`3x(2x-3)(x+sqrt6)(x-sqrt6)`