How do you factor 6x^2 + 3x – 63?

1 Answer
May 15, 2015

6x^2+3x-63 = (3*2x^2)+(3*x)-(3*21)

= 3(2x^2+x-21)

To factor 2x^2+x-21 I will look for roots of the equation:

2x^2+x-21 = 0

This is of the form ax^2+bx+c = 0, which has solutions:

x=(-b+-sqrt(b^2-4ac))/(2a)

In our case a=2, b=1 and c=-21, so

x=(-b+-sqrt(b^2-4ac))/(2a)

=(-1+-sqrt(1^2-4*2*(-21)))/(2*2)

=(-1+-sqrt(1+168))/4

=(-1+-sqrt(169))/4

=(-1+-13)/4

So one root is 12/4 = 3 and the other is -14/4 = -7/2.

From this we can deduce that (x-3) is a factor and (2x+7) is the other factor.

In summary, we have 6x^2+3x-63 = 3(x-3)(2x+7).