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Please Help me.. I don't understand it. 2. (3x^2-7x-6)/(x-3)
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Factor the numerator.
I got (3x^2) (-9+x) (x+2)/(x-3) :: Do i now divide by 3
\[ {3x^2 - 7x -6}\over{x-3}\] \[={{3x^2 - 9x + 2x - 6}\over{x-3}}={{{3x(x-3)+2(x-3)}}\over{x-3}}\] \[={{(3x+2)(x-3)}\over{x-3}}\]
Thank You. SoMuch..
hey alice first term of numerator is 3x^2 and last term is 6 multiply the coefficient of first and last term 3*6 = 18 now factorize 18 = 2 * 9 the middle term should be further written as (-9x+2x) = -7x 3x^2 - 7x - 6 = 3x^2 -9x +2x-6 = 3x(x-3)+2(x-3) = (x-3)(3x+2)
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My process was right just messed up on the (3x^2) (-9+x) (x+2)/(x-3) it was suppose to be 3x^2 -9x +2x-6 . Thank you again.
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