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Exercise

$\frac{\frac{x^2-16}{x-1}}{\frac{x^2+6x+8}{x^2+2x-3}}$

Step-by-step Solution

1

Simplify the fraction $\frac{\frac{x^2-16}{x-1}}{\frac{x^2+6x+8}{x^2+2x-3}}$

$\frac{\left(x^2-16\right)\left(x^2+2x-3\right)}{\left(x-1\right)\left(x^2+6x+8\right)}$
2

Factor the trinomial $\left(x^2+2x-3\right)$ finding two numbers that multiply to form $-3$ and added form $2$

$\begin{matrix}\left(-1\right)\left(3\right)=-3\\ \left(-1\right)+\left(3\right)=2\end{matrix}$
3

Rewrite the polynomial as the product of two binomials consisting of the sum of the variable and the found values

$\frac{\left(x^2-16\right)\left(x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x^2+6x+8\right)}$
4

Simplifying

$\frac{\left(x^2-16\right)\left(x+3\right)}{x^2+6x+8}$
5

Factor the trinomial $x^2+6x+8$ finding two numbers that multiply to form $8$ and added form $6$

$\begin{matrix}\left(2\right)\left(4\right)=8\\ \left(2\right)+\left(4\right)=6\end{matrix}$
6

Rewrite the polynomial as the product of two binomials consisting of the sum of the variable and the found values

$\frac{\left(x^2-16\right)\left(x+3\right)}{\left(x+2\right)\left(x+4\right)}$
7

Factor the difference of squares $\left(x^2-16\right)$ as the product of two conjugated binomials

$\frac{\left(x+4\right)\left(x+3\right)\left(x-4\right)}{\left(x+2\right)\left(x+4\right)}$
8

Simplify the fraction $\frac{\left(x+4\right)\left(x+3\right)\left(x-4\right)}{\left(x+2\right)\left(x+4\right)}$ by $x+4$

$\frac{\left(x+3\right)\left(x-4\right)}{x+2}$

Final answer to the exercise

$\frac{\left(x+3\right)\left(x-4\right)}{x+2}$

Try other ways to solve this exercise

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