Question

X^3+2x^2-16x-32/(x+2)(x+4)

Answer 1

$\frac{x^3+2x^2-16x-32}{(x+2)(x+4)} = \frac{(x^3-16x)+(2x^2-32)}{(x+2)(x+4)} = \frac{x(x^2-16)+(2(x^2-16)}{(x+2)(x+4)} =$

$=\frac{(x^2-16)(x+2)}{(x+2)(x+4)} =\frac{(x-4)(x+4)(x+2)}{(x+2)(x+4)} =x-4$

Answer 2

To simplify the expression $\frac{x^3 + 2x^2 - 16x - 32}{(x + 2)(x + 4)}$, you can start by factoring the numerator and denominator and then canceling out any common factors:

1. Factor the numerator: $x^3 + 2x^2 - 16x - 32 = x^2(x + 2) - 16(x + 2) = (x^2 - 16)(x + 2)$

2. Factor the denominator: $(x + 2)(x + 4)$

Now, you can see that both the numerator and denominator have a common factor of (x + 2). Cancel out this common factor:

$\frac{(x^2 - 16)(x + 2)}{(x + 2)(x + 4)} = \frac{(x^2 - 16)}{(x + 4)}$

Now, you have simplified the expression to $\frac{x^2 - 16}{x + 4}$. Further simplifying the numerator, which is a difference of squares:

$x^2 - 16 = (x + 4)(x - 4)$

So, the final simplified expression is:

$\frac{(x + 4)(x - 4)}{x + 4}$

Now, you can see that there is a common factor of (x + 4) in the numerator and denominator, which can be canceled out:

$\frac{\cancel{(x + 4)}(x - 4)}{\cancel{(x + 4)}} = x - 4$

The simplified expression is $x - 4$.