Question

How do you factor quadratic `x^2-x-56`?

Answer 1

Solve `x^2 - x - 56 = 0`
using the formula
`x_0 = (-b+-sqrt(b^2 - 4ac))/(2a)`

to get solutions
`x_0=a` and `x_0=b`
(in this particular case `x_0=8` and `x_0=-7`)

Since the original quadratic is equal to zero when `x` is equal to either of the `x_0` values
`(x-a)` and `(x-b)` are factors of the original quadratic.

Using `a=8` and `b=-7`
`(x-8)(x+7) = x^2-x -56`
and the factorization is complete.

Answer 2

To factor `x^2-x-56` by trial and error, use the following.

(This takes longer to explain than it does to do.)

We're looking for
`(x+b)(x+d)=x^2+(d+b)x+(bd)=x^2-x-56`

the products used to get 56 (`bd`) using whole numbers are
`1*56`
`2*28`

`4*14` (`3, 5, 6` are not factors of `56`)

`7*8` (the next number, `8` has already appeared on the right, so we're done)

We want `-56`, so we need `b` and `d` to have opposite signs. (One is positive and the other negative) they need to add up to `-1` because we want a `-x=-1x` in the middle. `+(d+b)x=-1x`, so `d+b=-1`.

Here's the list again:
`1*56`
`2*28`
`4*14`
`7*8`
If one is positive and the other negative which pair do I want to get a sum of `-1`?

Looks like I want `7` and `-8`

Check to be sure

`(x-8)(x+7)=x^2+7x-8x-56=x^2-x-56` Good, that works.

Note
If there is a number (other than `1`) in front of the `x^2`, we need

`(ax+b)(cx+d)=(ac)x^2+(ad+bc)x+(bd)`