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

#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)#

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)#