Question

How do you solve `x^2+7x-18=0` by factoring?

Answer 1

While it will not always be the case, start by assuming we are only dealing with integers and the factors of
`x^2+7x-18`
have the form
`(x+a)(x-b)` we know `a` and `-b` have different signs since their product is negative (`-18`)
We also know that `a` is greater than `b` since the coefficient of `x` is greater than zero.

There are only a limited number of possible integer values for `a` and `b` with `a>b` and `ab = 18`

`(a,b) = (18,1)` which would give `a-b = 17`; not what we want

`(a,b) = (9,2)` which would give `a-b = 7`; this matches the `x` coefficient of the given equation

`(a,b)= (6,3)` which would give `a-b=3`; not what we want

There are no other integer possibilities.

The factoring is
`(x+9)(x-2) = x^2 + 7x -18`

Since
`x^2 + 7x - 18 = 0`
either
`x+9 = 0` or `x-2=0`

So
either `x=-9` or `x=2`