Question

How do you solve `3w^2 + 28w + 9 = 0` by factoring?

Answer 1

The solutions are: `-1/3`, `-9`

`3w^2+28w+9=0`

If it can be factored using whole numbers it must be

`(3w+"something"_1)(w+"something"_2)`
So that the `w^2` term has a coefficient of `3`

`"something"_1 xx "something"_2` must be `9`
(the product of the constants (the lasts) must be the constant `9`)
To get a product of `9` using whole numbers our choices are
`1*9` or `3*3`.

But the first terms are different so we need to consider the choices:
`(3w+1)(w+9)`
`(3w+9)(w+1)`

`(3w+3)(w+3)` (Notice this would make the whole thing divisible by `3` -- which it isn't. This won't work.)

A quick check will convince you that

`(3w+1)(w+9)=3w^2+28w+9`

So, solving `3w^2+28w+9=0` is the same as solving:
`(3w+1)(w+9)=0`

And the only way for a product of numbers to be `0` is for at least one of the numbers to be `0`. So we need:

`3w+1=0` or `w+9=0`

In the first case we get `w=-1/3`, in the second `w=-9`.

The solutions are: `-1/3`, `-9`