Question

Factor completely.

6x²+ 3x - 63


A. 3(2x + 7)(x - 3)
B. (3x + 9)(2x - 7)
C. (6x + 21)(x - 3)
D. 3(2x²+ x - 21)

Answer 1

it would be A. work the equation backwards. 3(2x+7)(x-3) * represents multiply
3*2x=6x
3*7=21 leaves you with (6x+21)(x-3)
follow FOIL: first outer inner last
F:6x*x=6x²
O:6x*-3=-18x
I:21*x=21x
L:21*-3=-63
6x² -18x+21x=3x -63
you would get 6x²+3x-63 so A is the answer

Answer 2

If you wanted to work this question forwards instead of backwards, this is how you would do it:

First, factor out a 3, since all of the coefficients are divisible by it.


`6x^2+3x-63 \\ 3(2x^2+x-21)`

You want to split the middle into two numbers that multiply to equal the product of the value of x² and the constant and add to equal the value of x.
Now, look at the coefficients here. We're looking for two numbers that mutliply to get -42 (2 times -21) and add to make 1.

Think of factors of 42. 7 times 6 is 42. 7 minus 6 is 1...your numbers are 7 and -6. Now all we have to do is split the middle and factor.


`3(2x^2+7x-6x-21)\\3(x(2x+7)-3(2x+7))\\\boxed{3(2x+7)(x-3)}`