# How do you solve the inequality 2x^3 + 13x^2 -8x - 46 >=6?

Dec 15, 2015

$- \frac{13}{2} \le x \le - 2 \mathmr{and} 2 \le x$

#### Explanation:

1) Simplyfying

First of all, bring all the terms to the left side and simplify:

$2 {x}^{3} + 13 {x}^{2} - 8 x - 46 \ge 6$

$\iff 2 {x}^{3} + 13 {x}^{2} - 8 x - 52 \ge 0$

Now, your inequality has the form

$\text{polynomial} \ge 0$

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2) Computing the roots of the polynomial

As next, you need to find the roots of your polynomial $2 {x}^{3} + 13 {x}^{2} - 8 x - 52$.

Let me show you two ways how to do this:

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a) It is not easy in all the cases, but here you can factorize the expression:

$2 {x}^{3} + 13 {x}^{2} - 8 x - 52 = 2 {x}^{3} + 13 {x}^{2} - 4 \cdot 2 x - 4 \cdot 13$

$\textcolor{w h i t e}{\times \times \times \times \times \times \times x} = {x}^{2} \cdot 2 x + {x}^{2} \cdot 13 - 4 \cdot 2 x - 4 \cdot 13$

$\textcolor{w h i t e}{\times \times \times \times \times \times \times x} = {x}^{2} \cdot \left(2 x + 13\right) - 4 \cdot \left(2 x + 13\right)$

$\textcolor{w h i t e}{\times \times \times \times \times \times \times x} = \left({x}^{2} - 4\right) \cdot \left(2 x + 13\right)$

$\textcolor{w h i t e}{\times \times \times \times \times \times \times x} = \left(x - 2\right) \left(x + 2\right) \left(2 x + 13\right)$

A product is equal to zero if and only if at least one of its factors is equal to zero.

Thus, the roots of the expression are $x = 2$, $x = - 2$ and $x = - \frac{13}{2}$.

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b) I would recommend the second way if you don't "see" the factorization above.

Try and find the first root by plugging different values into the function. I always start with $1$, $- 1$, $2$ and $- 2$ since they are the easiest.

Here, $2$ works:

$2 \cdot {2}^{3} + 13 \cdot {2}^{2} - 8 \cdot 2 - 52 = 0$

Thus, $x = 2$ is your first root.

To find the other roots, you should first perform a polynomial division to gain a quadratic term:

$\textcolor{w h i t e}{\xi i} \left(2 {x}^{3} + 13 {x}^{2} - 8 x - 52\right) \div \left(x - 2\right) = 2 {x}^{2} + 17 x + 26$
$- \left(2 {x}^{3} - 4 {x}^{2}\right)$
$\textcolor{w h i t e}{x} \frac{\textcolor{w h i t e}{\times \times \times \times}}{}$
$\textcolor{w h i t e}{\times \times \times} 17 {x}^{2} - 8 x$
$\textcolor{w h i t e}{\times \xi i} - \left(17 {x}^{2} - 34 x\right)$
$\textcolor{w h i t e}{\times \times x} \frac{\textcolor{w h i t e}{\times \times \times \times x}}{}$
$\textcolor{w h i t e}{\times \times \times \times \times x} 26 x - 52$
$\textcolor{w h i t e}{\times \times \times \times i} - \left(26 x - 52\right)$
$\textcolor{w h i t e}{\times \times \times \times \times} \frac{\textcolor{w h i t e}{\times \times \times \times}}{}$
$\textcolor{w h i t e}{\times \times \times \times \times \times \times \times} 0$

Now, you can solve the equation

$2 {x}^{2} + 17 x + 26 = 0$

via e.g. the quadratic formula. Here, $a = 2$, $b = 17$ and $c = 26$.

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{- 17 \pm \sqrt{{17}^{2} - 4 \cdot 2 \cdot 26}}{4}$

$\text{ } = \frac{- 17 \pm 9}{4}$

$x = - 2$ and $x = - \frac{13}{2}$

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3) Finding positive intervals of the polynomial

Now that you have the three roots, you know that the function intercepts the $x$ axis three times, at $- \frac{13}{2}$, $- 2$ and $2$.

As a polynomial function, it is continuous and defined for all $x \in \mathbb{R}$. So, the only thing left to do is to identify in which of the intervals between and beyond the roots the function has positive values and in which it has negative values.

To do so, you can take a look at the end behaviour of the function:

${\lim}_{x \to \infty} 2 {x}^{3} = + \infty$, ${\lim}_{x \to - \infty} 2 {x}^{3} = - \infty$

or you can pick a value from the middle of an interval and plug it into the function to test if it is positive or negative.

An obvious choice would be evaluating the polynomial for $x = 0$:

$2 \cdot {0}^{3} + 13 \cdot {0}^{2} - 8 \cdot 0 - 52 = - 52 < 0$

Thus, we recognize the following behaviour:

$x \in \left(- \infty , - \frac{13}{2}\right) : \text{ } 2 {x}^{3} + 13 {x}^{2} - 8 x - 52 \textcolor{red}{< 0}$

$x \in \left[- \frac{13}{2} , - 2\right] : \text{ } 2 {x}^{3} + 13 {x}^{2} - 8 x - 52 \textcolor{g r e e n}{\ge 0}$

$x \in \left(- 2 , 2\right) : \text{ } 2 {x}^{3} + 13 {x}^{2} - 8 x - 52 \textcolor{red}{< 0}$

$x \in \left[2 , \infty\right) : \text{ } 2 {x}^{3} + 13 {x}^{2} - 8 x - 52 \textcolor{g r e e n}{\ge 0}$

You can also validate the result by graphing the polynomial and inspecting in which intervals the graph is above the $x$-axis or intercepts the $x$-axis (beware when looking at the graph, the scale of the $y$ axis is very different from the scale of the $x$ axis):

graph{2x^3 + 13x^2 - 8x - 52 [-8, 8, -80, 80]}

Thus, the result of this inequality is

$x \in \left[- \frac{13}{2} , - 2\right] \cup x \in \left[2 , \infty\right)$

or, depending on your notation preference,

$- \frac{13}{2} \le x \le - 2 \mathmr{and} 2 \le x$