How do you factor #9z^2 + 6z - 8#?

2 Answers
Dec 24, 2016

Answer:

#9z^2+6z-8 = (3z-2)(3z+4)#

Explanation:

Given:

#9z^2+6z-8#

This example can be factored using an AC method:

Find a pair of factors of #AC = 9*8=72# which differ by #B=6#

The pair #12, 6# works.

Use this pair to split the middle term, then factor by grouping:

#9z^2+6z-8 = 9z^2+12z-6z-8#

#color(white)(9z^2+6z-8) = (9z^2+12z)-(6z+8)#

#color(white)(9z^2+6z-8) = 3z(3z+4)-2(3z+4)#

#color(white)(9z^2+6z-8) = (3z-2)(3z+4)#

Dec 24, 2016

Answer:

#9z^2+6z-8 = (3z-2)(3z+4)#

Explanation:

Given:

#9z^2+6z-8#

We can factor this quadratic by completing the square.

We will also use the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

with #a=(3z+1)# and #b=3#

Note that:

#31^2 = 961#

So:

#(3z+1)^2 = 9z^2+6z+1#

[Think what happens when #z=10# to see why]

Hence:

#9z^2+6z-8 = 9z^2+6z+1-9#

#color(white)(9z^2+6z-8) = (3z+1)^2-3^2#

#color(white)(9z^2+6z-8) = ((3z+1)-3)((3z+1)+3)#

#color(white)(9z^2+6z-8) = (3z-2)(3z+4)#