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# help

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Simplify $$\dfrac{x^3 - x^2 - 4x + 4}{x^2 + 6x + 8}$$

Jul 7, 2020

#1
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We factor by grouping on both the numerator and the denominator.

$$\quad \dfrac{x^3 - x^2 - 4x + 4}{x^2 + 6x + 8}\\ = \dfrac{(x^3 - x^2) - (4x - 4)}{(x^2 + 4x) + (2x + 8)}\\ = \dfrac{x^2(x - 1) - 4(x - 1)}{x(x + 4) + 2(x + 4)}\\ = \dfrac{(x^2- 4)(x - 1)}{(x + 2)(x + 4)}\\ = \dfrac{\color{red}(x + 2)\color{black}(x - 2)(x - 1)}{\color{red}(x + 2)\color{black}(x + 4)}\\$$

Now if I do more than that, that would be spoonfeeding.

Jul 7, 2020