Factoring question, please help!

Can someone please help me with a factoring question?

2c^2 + 6cd - 8d^2

answer: 2(c + 4d)(c - d)

Your help will be most appreciated!

Original:
2c^(2)+6cd-8d^(2)

Factor out the GCF of 2 from each term in the polynomial.
2(c^(2))+2(3cd)+2(-4d^(2))

Factor out the GCF of 2 from 2c^(2)+6cd-8d^(2).
2(c^(2)+3cd-4d^(2))

In this problem 4*-1=-4 and 4-1=3, so insert 4 as the right hand term of one factor and -1 as the right-hand term of the other factor.
2(c+4d)(c-d)

Since c and d are the variables and a term with a square of each and term withe the product of the two you know that you are going to have factors that look like F(Wc+Xd)(Yc+Zd)

We can factor 2 out of the whole equation
2(c^2 + 3cd - 4d^2) so F=2
Factor of the multiplier for the c^2 term is just 1 so both W and Y equal 1
Factors of the multiplier for the d^2 term are -1, 2 and 4
Now we know that W*Z+X*Y=6 becomes Z+X=3 therefore X=3-Z
We also know that X*Z=-4
Now you can substitute the factors manually in both equations until you find a pair that works
for 2 and -2 we can't find a combination where X=3-Z so we know X and Z have to be either 1 and -4 or 4 and -1
we find that 4=3-(-1) so we know that X=4 and Z= -1
Substituting W, X, Y and Z into the original equation gives
2(1c+4d)(1c-1d)=2(c+4d)(c-d)