Question

How do you factor the trinomial `8z^2+20z-48`?

Answer 1

`4(z+4)(2z-3)`

Explanation:

you can divide all the terms by a common factor `4` to get

`2z^2 + 5z - 12`

this expression can be factorised by grouping:

the first and last coefficients are `2` and `-12`.
the product of this is `-24`.

then, you can find `2` numbers that add to give the second coefficient, and multiply to give the product of the first and last coefficient.

`8 + -3 = 8 - 3 = 5`
`8 * -3 = -24`

the two numbers that add to give the second coefficient are `8` and `-3`.
therefore the second term can be expressed as the sum of two terms, `8z` and `-3z`.

this gives the expression `2z^2 - 3z + 8z - 12`.

then you can factorise each adjacent pair:

`2z^2 - 3z = z(2z-3)`
`8z - 12 = 4(2z-3)`

therefore `2z^2 - 3z + 8z - 12` can be written as `z(2z-3) + 4(2z-3)`.
this sum is the same as `(z+4)(2z-3)`.

then you have the expression `2z^2+5z-12` grouped as `(z+4)(2z-3)`.

the expression given in the question is `8z^2+20z-48`.
that is `4` times `2z^2+5z-12`, or the grouped expression `(z+4)(2z-3)`.
therefore it can be written as `4(z+4)(2z-3)`.