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

2 Answers
May 16, 2015

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)

May 26, 2015

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)