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to factor means to under multiplication
you have to understand the mechanics of how the quadratic there was formed in order to know how to take it apart
I know I have to find the GCF first then the same with the exponent
err... thats a cubic lol; math notation helps
2x cubed-8x squared-9x+ 36 notates to: 2^3 -8x^2 -9x +36
lol...i didnt know how to type it in here
first group them; and see if it wroks... \[(2x^3-8x^2)+(-9x+36)\]
what can we factor out of the first part?
2x^2 (x -4) ; is what i get do you see why?
2 is part of it; but they also have some 'x's in common right?
yes and it would b 2x^2 right?
(2xxx - 8xx) have what in common: 2xx (x-4)
then you mult.?
they both have at least XX
then we see what the other part factors to.. (-9x +36) factor to what?
do u divide? woiuld it b 4x?
you do the same process as you did in the firs part ..... nothing new is happening here.
lets say its 4x and test that: 4x* ? = -9x ? 4x * ? = 36?
you cant factor bc there is no x with 36
your right; no 'x' factors out; but what do they have in common? not 'what do they have thats not in common'
9 goes into 36 4 times??
yes; very good :)
what we have done so far looks like this right? \[2x^3 -8x^2 -9x +36\] \[(2x^3-8x^2)+(-9x+36)\] \[2x^2(x-4)+9(-x+4)\]
but we want those parenthesis to be the same .... and right now they are close, but slightly different right?
right..the 2nd x needs to be positive??
how do we get: (x-4) to equal (-x+4) ? what can we factor out of ...lets say the one on the right?
close, real close; but lets try a (-1) \[(-x+4)\iff -1(x-4)\]
ohh ok....i always get confused with the negatives...but that would make it a positive and it would be (x-4)?
\[2x^2(x−4)+9(-x+4)\] \[2x^2(x−4)+9(-1)(x-4)\] \[2x^2(x−4)-9(x-4)\]
thank you so much for helping me. :)
youre welcome; but we got one last step
we gotta factor this last part again to get our final answer
\[2x^2(x−4)−9(x−4)\] what do these have in common that we can factor out? think of it as: \[2x^2(A)-9(A)\]
good :) so lets pull that out and see whats left
thats our answer
ok, got it :)
i knew you would ;)
again, thanx a lot!