How do you factor #3x^2 + 2x - 8# by grouping?

2 Answers
Apr 2, 2017

Given -

#3x^2+2x-8#

Find two numbers

The addition of the two numbers must be equal to the coefficient of #x#

The coefficient of #x=2#
The two numbers are #6# and #-4#
If you add you get #2#

The Product of the same two numbers must be equal to the product of the coefficient of #x^2# and the constant term in the expression.

The coefficient of #x^2# is #3#
The constant term is #-8#
The product of these two numbers are #3 xx (-8) = -24#

The product of the two numbers #6 xx (-4) = -24#

Then rewrite the original expression as -

#3x^2+6x-4x-8#

[You have replaced #2x# with #6x-4x#

#3x(x+2)-4(x+2)#]

#(3x-4)(x+2)#

Apr 2, 2017

Answer:

#3x^2+2x-8 = (3x-4)(x-2)#

Explanation:

This is a trinomial quadratic, which means it has only three terms instead of four, but we will end up with four terms in the answer.

The trinomial also has a leading value greater than #1# so we need to use a different procedure from that used with a single first term.

Given: #3x^2+2x-8#

Because the first term is greater than #1# we will need to use it as well as factors of the third term in our search for factors we can use in the solution.

To start with, multiply the first value by the third: #3*8=24#

Then we have to find factors of #24# that will subtract to leave us with #2#, the second term. We know that one of the factors has to be negative because the #8# is negative.

It looks like #6 and -4# will work: #6-4=2#

Then we can equate: #3x^2+2x-8 = 3x^2 +6x-4x-8#

Now we can factor: #3x^2 +6x-4x-8 = 3x(x+2)-4(x+2)#

Then removing the common factor #(x+2) and grouping terms:

#3x(x+2)-4(x+2) = (3x-4)(x+2)#

So: #3x^2+2x-8 = (3x-4)(x-2)#