Question

$\frac{(x-2)(x-4)(x-7)}{(x+2)(x+4)(x+7)} > 1.$

Answer 1

$(-\infty, -7) \cup (-4, -2)$

Answer 2

To solve the inequality $\frac{(x-2)(x-4)(x-7)}{(x+2)(x+4)(x+7)} > 1$, we first rewrite it as:

$\frac{(x-2)(x-4)(x-7)}{(x+2)(x+4)(x+7)} - 1 > 0$

Combining the terms on the left side gives:

$\frac{(x-2)(x-4)(x-7) - (x+2)(x+4)(x+7)}{(x+2)(x+4)(x+7)} > 0$

Expanding the numerator:

$\frac{(x^3 - 13x^2 + 50x - 56) - (x^3 + 13x^2 + 50x + 56)}{(x+2)(x+4)(x+7)} > 0$

Simplifying the numerator:

$\frac{-26x^2 - 112}{(x+2)(x+4)(x+7)} > 0$

Since $-26x^2 - 112$ is always negative, we need the denominator to be negative as well:

$(x+2)(x+4)(x+7) < 0$

The roots of the denominator are $x = -7, -4, -2$. Testing intervals:

• $x < -7$: $(x+2)$, $(x+4)$, and $(x+7)$ are all negative, so the product is negative.

• $-7 < x < -4$: $(x+2)$ and $(x+4)$ are negative, $(x+7)$ is positive, so the product is positive.

• $-4 < x < -2$: $(x+2)$ is negative, $(x+4)$ and $(x+7)$ are positive, so the product is negative.

• $x > -2$: All factors are positive, so the product is positive.

Thus, the solution is $x \in (-\infty, -7) \cup (-4, -2)$.