How do you factor the quadratics #6x^2-x-2# and #x^2-7x+12# ?

1 Answer
Dec 8, 2017

Answer:

#6x^2-x-2 = (2x+1)(3x-2)#

#x^2-7x+12 = (x-3)(x-4)#

Explanation:

In order to find the appropriate split of the middle terms in each of these quadratics, we can use an AC method:

Example 1

Given:

#6x^2-x-2#

look for a pair of factors of #AC=6*2 = 12# which differ by #B=1#. Note that we look for a suitable difference because the coefficient of the constant term is negative.

The pair #4, 3# works in that #4 * 3 = 12# and #4 - 3 = 1#.

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

#6x^2-x-2 = (6x^2-4x)+(3x-2)#

#color(white)(6x^2-x-2) = 2x(3x-2)+1(3x-2)#

#color(white)(6x^2-x-2) = (2x+1)(3x-2)#

So this quadratic has zeros #x=-1/2# and #x=2/3#

Example 2

Given:

#x^2-7x+12#

look for a pair of factors of #AC=1*12 = 12# with sum #B=7#. Note that we look for a suitable sum because the coefficient of the constant term is positive.

The pair #4, 3# works in that #4 * 3 = 12# and #4 + 3 = 7#.

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

#x^2-7x+12 = (x^2-4x)-(3x-12)#

#color(white)(x^2-7x+12) = x(x-4)-3(x-4)#

#color(white)(x^2-7x+12) = (x-3)(x-4)#

So this quadratic has zeros #x=3# and #x=4#