Question

How do you factor `4z^2+4z-15`?

Answer 1

`4z^2+4z-15` is of the form `az^2+bz+c` with `a=4`, `b=4` and `c=-15`.

The discriminant of this quadratic is

`Delta = b^2 - 4ac = 4^2 - (4xx4xx-15)`

`=16+240=256=16^2`.

Being a positive perfect square, we know that `4z^2+4z-15=0` has distinct rational roots.

The roots of `4z^2+4z-15=0` are

`z = (-b +-sqrt(Delta))/(2a) = (-4 +- 16)/(2xx4) = (-4 +-16)/8`

That is `z = -20/8 = -5/2` or `z = 12/8 = 3/2`.

We can deduce that `(2z+5)` and `(2z-3)` are both factors.

So `4z^2+4z-15 = (2z+5)(2z-3)`

Answer 2

Another solution is called the AC Method.

First multiply the coefficient (`A=4`) of the `z^2` term by the coefficient (`C=15`) of the constant term - ignoring the sign.

`AC = 4*15 = 60`

Notice that the sign of the constant (`C`) term is negative.

According to the AC Method, we need to look for a factorisation of our `AC` value into a pair of factors, whose difference is the middle coefficient `B=4`. Well `10 xx 6 = 60` and `10 - 6 = 4`. So the pair we want is `10` and `6`.

Now use this pair to split the middle term then factor by grouping:

`4z^2 + 4z - 15`

`= 4z^2 + 10z - 6z - 15`

`= (4z^2 + 10z) - (6z + 15)`

`=2z(2z + 5) - 3(2z + 5)`

`=(2z-3)(2z+5)`