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

##### 1 Answer

#### Answer:

#### Explanation:

**1) Simplyfying**

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

#2x^3 + 13x^2 - 8x - 46 >= 6#

Now, your inequality has the form

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**2) Computing the roots of the polynomial**

As next, you need to find the roots of your polynomial

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:

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

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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

Here,

#2 * 2^3 + 13 * 2^2 - 8 * 2 - 52 = 0#

Thus,

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

Now, you can solve the equation

#2x^2 + 17x + 26 = 0#

via e.g. the quadratic formula. Here,

#x = (-b +- sqrt(b^2 - 4ac))/(2a) = (-17 +- sqrt(17^2 - 4*2*26))/(4)#

#" "= (-17 +- 9)/4#

Thus, your second and your third roots are

#x = -2# and#x = - 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

As a polynomial function, it is continuous and defined for all

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

#lim_(x-> oo) 2x^3 = +oo# ,#lim_(x-> - oo) 2x^3 = - oo#

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

#2 * 0^3 + 13 * 0^2 - 8 * 0 - 52 = -52 < 0#

Thus, we recognize the following behaviour:

You can also validate the result by graphing the polynomial and inspecting in which intervals the graph is above the

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

Thus, the result of this inequality is

or, depending on your notation preference,