Question

How do you factor `20w ^ { 2} + 33w - 27`?

Answer 1

Notice that `(aw+b)(cw+d)=(ac)w^2+(ad+bc)w+(bd)`.

For the polynomial `20w^2+3w-27`, we have
`ac=20`, `ad+bc=33`, and `bd=-27`.

To find `a` and `c`, factorise 20 into two positive factors: 1 and 20, 2 and 10, 4 and 5. Then to find `b` and `d`, factorise -27 into two factors with opposite signs: 1 and -27, 3 and -9, 9 and -3, 27 and -1.

Now choose one factorisation from each so that the sum of the cross multiplication (`ad+bc`) equals 33. In this case, `(5)(9)+(-3)(4)=33`, so `a=5`, `b=-3`, `c=4`, and `d=9`.

Substituting this into the first line, we obtain `(5w-3)(4w+9)`.

Answer 2

Use an AC method to find:

`20w^2+33w-27 = (5w-3)(4w+9)`

Explanation:

Given:

`20w^2+33w-27`

We can use an AC method:

Look for a pair of factors of `AC = 20*27 = 540` which differ by `B=33`. (Note: we look for a pair with difference `B` rather than sum `B` since the sign of the constant term is negative).

Now:

`540 = 2^2*3^3*5`

and the difference we are looking for is divisible by `3`, but not by `2`.

So one of the pair of factors is odd and the other even, but both of the factors are divisible by `3`.

So try:

`2^2*3 = 12" "xx" "45 = 3^2*5`

Note that:

`45 - 12 = 33`

just as we require. So we were (slightly) lucky.

Use the pair `45, 12` to split the middle term and factor by grouping as follows:

`20w^2+33w-27 = 20w^2+45w-12w-27`

`color(white)(20w^2+33w-27) = (20w^2+45w)-(12w+27)`

`color(white)(20w^2+33w-27) = 5w(4w+9)-3(4w+9)`

`color(white)(20w^2+33w-27) = (5w-3)(4w+9)`