I'm starting to forget this factoring thing, where the coefficient is not 1. can some one please go in details and explain?
I will show you a trick that works
suppose \[4x^2-16x+15\]
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
precal thank you very much.. you sir are a life sever
I love to use this trick, it always works. Don't forget to divide the number back in...
what about having big numbers please
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
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