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this the answer that I got

Hev123  Mar 5, 2018
 #1
avatar+7324 
+3

\(\quad\frac{\frac{x^2+x-12}{x-2}}{\frac{3x^2+11x-4}{x^2-4}}\\ =\\ \quad\frac{x^2+x-12}{x-2}\cdot\frac{x^2-4}{3x^2+11x-4} \\ =\\ \quad\frac{(x+4)(x-3)}{(x-2)}\cdot\frac{(x+2)(x-2)}{3x^2+12x-x-4} \\ =\\ \quad\frac{(x+4)(x-3)}{(x-2)}\cdot\frac{(x+2)(x-2)}{3x(x+4)-1(x+4)} \\ =\\ \quad\frac{(x+4)(x-3)}{(x-2)}\cdot\frac{(x+2)(x-2)}{(x+4)(3x-1)} \\ =\\ \quad\frac{(x+4)(x-3)(x+2)(x-2)}{(x-2)(x+4)(3x-1)} \\ =\\ \quad\frac{(x-3)(x+2)}{(3x-1)} \qquad\text{and}\qquad x\neq-4\, \qquad x\neq2\\ =\\ \quad\frac{x^2-x-6}{3x-1} \)

hectictar  Mar 5, 2018
 #2
avatar+68 
+2

\(\frac{\frac{x^2+x-12}{x-2}}{\frac{3x^2+11x-4}{x^2-4}}\)

Simplify the polynomials.

\(\frac{\frac{(x+4)(x-3)}{x-2}}{\frac{(3x-1)(x+4)}{(x+2)(x-2)}}\)

Multiply the top and bottom fractions by (x+2)(x-2).

\(\frac{(x+4)(x-3)(x+2)}{(3x-1)(x+4)}\)

Cross out like terms.

\(\frac{(x-3)(x+2)}{(3x-1)}\)

Multiply the top polynomial out.

\(\frac{x^2-x+6}{3x-1}\)

 

Just a different way to look at it. :D

CoopTheDupe  Mar 5, 2018

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