Question

How do you factor `4x^2 - 11x - 20`?

Answer 1

If you assume the coefficients of the factoring are `a, b, c, d` so that
your answer is `(ax+b)(cx+d)`
with
`ac = 4`
`ad+bc = -11` and
`bd = -20`
you will discover there are only a very few possible values to be considered (for example `ac=4 rarr {a,c} = {1,4} or {2,2})` and you could arrive at the solution by trial and error.

A more certain method is to use the formula:
`(-b+-sqrt(b^2-4ac))/(2a)`
to determine the values of `x` for which the quadratic is equal to `0`.

which in this case becomes
`(11 +- sqrt((-11)^2 - (4)(4)(-20)))/(2(4))`

`=(11 +-sqrt(21))/8`

So the quadratic is equal to `0` when
`x=4` and `x=-5/4`

Therefore two of the factors are
`(x-4)` and `(x+5/4)`

If we multiply these two factors together we notice that the result is `1/4` of the original quadratic,
so a full factoring would be:
`(4)(x-4)(x+5/4)`

Normally we would multiply the first and last term together to clear the fraction, so a more typical factorization would be:
`(x-4)(4x+5)`