# 4.6 - Polynomial and Rational Inequalities

How to find what $x$ makes a polynomial or rational function positive or negative?

### Example. Solve $(x+4)(x-3)<0$.

#### Root Behavior:

|Factor | Zero | Multiplicity | Bounce or cross |
|---|---|---|---|
|$(x+4)$ | $x=-4$ | $1$ (odd) | Cross over x-axis |
|$(x-3)$ | $x=3$ | $1$ (odd) | Cross over x-axis |

1. **The graph crosses the x-axis at $x=-4$.**
2. **The graph crosses the x-axis at $x=3$.**

#### End Behavior:

FOILing

\begin{align}
(x+4)(x-3)&=(x)(x)+(x)(-3)+(4)(x)+(4)(-3) \\
&=x^2-3x+4x-12 \\
&=x^2+x-12
\end{align}

- The leading term is $x^2$.
- The leading coefficient is $1$ (positive).
- The degree is 2 (even).

3. **The graph rises to the left.**
4. **The graph rises to the right.**

#### Graph:

Sketching a graph with the following properties:
1. **The graph crosses the x-axis at $x=-4$.**
2. **The graph crosses the x-axis at $x=3$.**
3. **The graph rises to the left.**
4. **The graph rises to the right.**

Gives:


Since we're solving $(x+4)(x-3)<0$ we want to find when the graph is below the x-axis.
The graph is below the x-axis on the interval $(-4,3)$.

Answer: $$(-4,3)$$

---

### Example. Solve $(x+7)(x-2)>0$.

#### Root Behavior

| Factor | Zero | Multiplicity | Behavior |
|----------|----------|--------------|------------------------|
| $x+7$ | $x=-7$ | $1$ (odd) | crosses the x-axis |
| $x-2$ | $x=2$ | $1$ (odd) | crosses the x-axis |

1. **The graph crosses the x-axis at $x=-7$.**
2. **The graph crosses the x-axis at $x=2$.**

#### End Behavior

$$
(x+7)(x-2)=x^2 -2x +7x -14 = x^2 +5x -14.
$$

- Leading term is $x^2$
- Leading coefficient is $1$ (positive).
- Degree is 2 (even)

3. **The graph rises on the left.**
4. **The graph rises on the right.**

#### Sketch

Draw an upward‐opening parabola that crosses the x-axis at $-7$ and $2$:

1. **The graph crosses the x-axis at $x=-7$.**
2. **The graph crosses the x-axis at $x=2$.**
3. **The graph rises on the left.**
4. **The graph rises on the right.**


Since we're solving $(x+7)(x-2)>0$ we want to find when the graph is above the x-axis.
The graph is above the x-axis on the intervals $(-\infty,-7) \cup (2,\infty)$

Answer: $$(-\infty,-7) \cup (2,\infty)$$

---

### Example. Solve $x^2+5x+8\geq 6x+10$.

Subtract $6x$ and $10 from both sides to get:

$$x^2-x-2 \geq 0$$

Factoring the quadratic:

$$(x-2)(x+1) \geq 0$$

#### Root Behavior

| Factor | Zero | Multiplicity | Behavior |
|----------|---------|--------------|------------------------|
| $x-2$ | $x=2$ | 1 (odd) | crosses the x-axis |
| $x+1$ | $x=-1$ | 1 (odd) | crosses the x-axis |

1. **The graph crosses the x-axis at $x=2$.**
2. **The graph crosses the x-axis at $x=-1$.**

#### End Behavior

$$x^2-x-2 \geq 0$$

- Leading term is $x^2$.
- Leading coefficient is $1$ (positive).
- Degree is $2$ (even).

3. **The graph rises on the left.**
4. **The graph rises on the right.**

#### Sketch

Draw an upward‐opening parabola that crosses the x-axis at $2$ and $-1$:

1. **The graph crosses the x-axis at $x=2$.**
2. **The graph crosses the x-axis at $x=-1$.**
3. **The graph rises on the left.**
4. **The graph rises on the right.**


Since we're solving $(x-2)(x+1) \geq 0$ we want to find when the graph is above or on the x-axis.
The graph is above or on the x-axis on the intervals $\left(-\infty,-1\right] \cup \left[2,\infty\right)$

Answer: $$\left(-\infty,-1\right] \cup \left[2,\infty\right)$$

---

### Example. Solve $x^3-4x^2-5x>0$.

## Rational inequalities.

- Here, we plot the points which make the numerator {\bf or} denominator zero.

- These are the places where the function can change sign.

### Example. Solve $$\frac{2x+1}{x-3}>0$$

### Example. Solve $$\frac{x+1}{4x-3}<0$$