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