How to solve inequalities: (cubic, polynomial)?

**Sorena** · September 18th 2011, 06:50 PM

h3 - 40h2 + 375h - 1000 > 0 (h3 is h to the power of 3) - (h2 is h to the power of 2)

I know the general method - finding factors of 1000 and substituting them instead of h (factor theorem) - but the numbers are way too high. Is there any other method or do I just have to keep on substituting numbers?

(x2 + x - 12) / (x2+5x+6) < 0

(x+1)(x-2)(2x) >= 0 (2x is 2 to the power of x

**TheChaz** · September 18th 2011, 06:55 PM

Originally Posted by Sorena

$h^3 - 40h^2 + 375h - 1000$ ...

I know the general method - finding factors of 1000 and substituting them instead of h (factor theorem) - but the numbers are way too high. Is there any other method or do I just have to keep on substituting numbers?...

First, this isn't an inequality...!
But to factor this, we can use synthetic division.
Have you learned that?

**Sorena** · September 18th 2011, 06:57 PM

yea my bad it's > 0. And how do u solve this usiong synthenic division?

**pickslides** · September 18th 2011, 06:58 PM

Originally Posted by Sorena

I know the general method - finding factors of 1000 and substituting them instead of h (factor theorem) - but the numbers are way too high. Is there any other method or do I just have to keep on substituting numbers?

There are other numeric methods like newton's method. I would suggest using technology (spreadsheet or graphical calculator) to find one linear factor and then use polynomial long division to find the others.

Originally Posted by Sorena

(x2 + x - 12) / (x2+5x+6) < 0

(x+1)(x-2)(2x) >= 0 (2x is 2 to the power of x

2^x assumes you mean $2^x$ if you can't use latex.

Are you trying to solve for x here? What have you tried? What do you know about inequalities?

**Sorena** · September 18th 2011, 07:28 PM

Originally Posted by pickslides

There are other numeric methods like newton's method. I would suggest using technology (spreadsheet or graphical calculator) to find one linear factor and then use polynomial long division to find the others.

2^x assumes you mean $2^x$ if you can't use latex.

Are you trying to solve for x here? What have you tried? What do you know about inequalities?

The question is: "Solve the inequality (x+1)(x-2)(2^x) >= 0 Algebraically."

**Prove It** · September 18th 2011, 07:54 PM

Originally Posted by Sorena

The question is: "Solve the inequality (x+1)(x-2)(2^x) >= 0 Algebraically."

First of all, $\displaystyle 2^x > 0$ for all $\displaystyle x$ , so dividing both sides by it won't change the inequality sign.

$\displaystyle \begin{align*} (x + 1)(x - 2)2^x &\geq 0 \\ (x + 1)(x - 2) &\geq 0 \\ x^2 - x - 2 &\geq 0 \\ x^2 - x + \left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right)^2 - 2 &\geq 0 \\ \left(x - \frac{1}{2}\right)^2 - \frac{1}{4} - \frac{8}{4} &\geq 0 \\ \left(x - \frac{1}{2}\right)^2 &\geq \frac{9}{4} \\ \left|x - \frac{1}{2}\right| &\geq \frac{3}{2} \\ x - \frac{1}{2} \leq -\frac{3}{2} \textrm{ or }x - \frac{1}{2} &\geq \frac{3}{2} \\ x \leq -1 \textrm{ or }x &\geq 2 \end{align*}$

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