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Simplify to the lowest terms 36x^2-64/12x^2-8x-32
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the numerator is the difference of two square it factors as \((6x+8)(6x-8)\) no doubt the denominator factors as well then cancel
The choice I have are A.) 9x-8/3x+4 B.) 3x-4/2x-4 C.) 3x-4/x-2 D.) 3x+4/x+2
k then factor the numerator as \(4(3x+4)(3x-4)\) then factor the denominator as well something will cancel
I have done it but something isn't adding up somewhere. I have had to retake my exam 3 times now...
\[12x^2-8x-32 \] each term has a common factor of \(4\) so you can start with \[4(3x^2-2x-4)\]
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then factor as \(4(3x+4)(x-2)\) then cancel common factors top and bottom
typo above, should have been \(4(3x^2-2x-8)\) answer is right though
Thanks. I got it.
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