Question

I'm starting to forget this factoring thing, where the coefficient is not 1. can some one please go in details and explain?

Answer 1

I will show you a trick that works

Answer 2

suppose
\[4x^2-16x+15\]

Answer 3

Take the 4 and multiply it to the 15 and get 60
rewrite as
\[x^2-16x+60\]
Factor
(x-6)(x-10)
since you took out the 4 by multiplying it, you must return it by dividing it
(x-(6/4))(x-(10/4)) Whatever number you remove, you must return in
(x-(3/2))(x-(5/2)) I just reduced the fractions
since 3/2 is 1.5 I need to move the 2 to the front of the x
since 5/2 is 2.5 I need to move the 2 to the front of the x
(2x-3)(2x-5)
You can check it by multiplying it out

Answer 4

precal thank you very much.. you sir are a life sever

Answer 5

I love to use this trick, it always works. Don't forget to divide the number back in...

Answer 6

what about having big numbers please

Answer 7

still works but if you have big numbers, you can always factor a number out
example 4x^2+2x+2 or act like I actually used big numbers
2(2x^2+1x+1) then do the trick inside