Question

How do you factor `6z^3-3z^2-30z`?

Answer 1

`3z(2z-5)(z+2)`

Explanation:

Factor the the GCF, which is `3z`

`3z(2z^2 -1z -10)`

To see if the quadratic can be factored any further, find the two roots `r_1` and `r_2`, if they exist the quadratic will be able to be written in the form of `(z-r_1)(z-r_2)`.

The roots are `z = (1+- sqrt(1-4*2*(-10)))/4`, from that we get

`z = (1+- sqrt(1-(-80)))/4 = (1+- sqrt(81))/4`
`z = (1 +- 9)/4`

So we have
`r_1 = (1+9)/4 = 10/4 = 5/2`
`r_2 = (1-9)/4 = -8/4 = -2`

Now, since the parabola there has a two as the leading coefficient and we have a root who's a fraction two, we can say that the factored form is

`3z(2z-5)(z+2)`