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

2 Answers
Apr 3, 2017

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).

Apr 3, 2017

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)