Question

$\frac{(2x-3)(7x+2)(5-3x)}{(3-x)} > 0$

Answer 1

$(-\infty, -2/7) \cup (3/2, 5/3) \cup (3, \infty)$

Answer 2

The inequality to solve is $\frac{(2x-3)(7x+2)(5-3x)}{(3-x)} > 0$.

First, we find the critical points by setting the numerator and the denominator equal to zero.

Numerator roots:

$2x-3 = 0 \implies x = 3/2$

$7x+2 = 0 \implies x = -2/7$

$5-3x = 0 \implies 3x = 5 \implies x = 5/3$

Denominator roots:

$3-x = 0 \implies x = 3$

The critical points are $-2/7$, $3/2$, $5/3$, and $3$. Let's arrange these points in increasing order on the number line:

$-2/7 \approx -0.286$

$3/2 = 1.5$

$5/3 \approx 1.667$

$3 = 3$

So the ordered critical points are $-2/7 < 3/2 < 5/3 < 3$.

These critical points divide the number line into five intervals:

$(-\infty, -2/7)$, $(-2/7, 3/2)$, $(3/2, 5/3)$, $(5/3, 3)$, $(3, \infty)$.

Now, we determine the sign of the expression $\frac{(2x-3)(7x+2)(5-3x)}{(3-x)}$ in each interval. We can pick a test value from each interval or use the wavy curve method.

Let's use the wavy curve method. To apply it easily, we rewrite the expression so that the coefficient of $x$ in each factor is positive. The term $(5-3x)$ can be written as $-3(x - 5/3)$. The term $(3-x)$ can be written as $-(x - 3)$. The expression becomes:

$\frac{(2x-3)(7x+2)(-3(x - 5/3))}{-(x - 3)} = \frac{-3(2x-3)(7x+2)(x - 5/3)}{-1(x - 3)} = \frac{3(2x-3)(7x+2)(x - 5/3)}{(x - 3)}$

The roots of the factors in the modified expression are the same: $-2/7, 3/2, 5/3, 3$. The factors are $(2x-3), (7x+2), (x-5/3), (x-3)$. All have positive coefficients for $x$. The overall constant multiplier is $3$, which is positive.

We place the critical points on the number line: $-2/7, 3/2, 5/3, 3$.

Starting from the rightmost interval $(3, \infty)$, the expression $\frac{3(2x-3)(7x+2)(x - 5/3)}{(x - 3)}$ will be positive because all factors $(2x-3), (7x+2), (x-5/3), (x-3)$ are positive for $x > 3$.

Since all factors in the original expression $\frac{(2x-3)(7x+2)(5-3x)}{(3-x)}$ have odd powers (power 1), the sign of the expression will change at each critical point as we move from right to left.

Sign analysis on the intervals:

• Interval $(3, \infty)$: Pick $x=4$. $\frac{(+)(+)(-)}{(-)} = (+)$. The expression is positive.

• Interval $(5/3, 3)$: Pick $x=2$. $\frac{(+)(+)(-)}{(+)} = (-)$. The expression is negative.

• Interval $(3/2, 5/3)$: Pick $x=1.6$. $\frac{(+)(+)(+)}{(+)} = (+)$. The expression is positive.

• Interval $(-2/7, 3/2)$: Pick $x=0$. $\frac{(-)(+)(+)}{(+)} = (-)$. The expression is negative.

• Interval $(-\infty, -2/7)$: Pick $x=-1$. $\frac{(-)(-)(+)}{(+)} = (+)$. The expression is positive.

We are looking for the intervals where the expression is strictly greater than zero ($> 0$). These intervals are $(-\infty, -2/7)$, $(3/2, 5/3)$, and $(3, \infty)$.

The denominator cannot be zero, so $x \neq 3$. This is already handled as $3$ is an open endpoint. Since the inequality is strict ($>0$), the points where the numerator is zero are also excluded. These are $x = -2/7, 3/2, 5/3$. These are also open endpoints in the intervals.

Thus, the solution set is the union of the intervals where the expression is positive:

$(-\infty, -2/7) \cup (3/2, 5/3) \cup (3, \infty)$.