How do you factor # a^2+a-12#?

2 Answers
Jun 25, 2018

Answer:

#(a-3)(a+4)#

Explanation:

Solving the quadratic equation by the quadratic Formula we get

#a_(1,2)=-1/2pmsqrt(1/4+48/4)#
#a_(1,2)=-1/2pm7/2#

we get

#a_1=3#

#a_2=-4#

so our factorization is given by

#(a-3)(a+4)#

Jun 25, 2018

Answer:

#(a+4)(a-3)#

Explanation:

If we factor it into linear brackets #(a-b)(a-c)#, then #a^2+a-12=a^2-(b+c)a+bc#. So #bc=-12# and #-(b+c)=1#. So we're looking for values that are of opposite sign, multiply to -12, and whose absolute values are 1 apart. 4 and 3 is where we are looking, and specifically, -4 and +3 are the answers, which we find by a small amount of trial and error.

So #a^2+a-12=(a+4)(a-3)#.