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Currently, we are learning how to solve inequalities of degrees greater than 2 such as $x^3+9x^2+26x+24 < 0$.

Our teacher told us to graph and find the intervals to solve for $x$ but is there an efficient method to find the $x$'s rather than graphing them? Our teacher also told us to find each case, but that is also inefficient.

What is the best method to find the $x$'s of any inequality greater than degree 2?

Thanks.

NOTE: We aren't going to be finding inequalities for degrees higher than 4. Therefore, if there does exist such a method that works more efficiently with relatively low degrees, then that is fine also.

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3 Answers 3

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This is one of very interesting and deep questions. There are methods to solve polynomials upto 4th degree. But for higher degree, it has been proved that it cant be solved by method of radicals(what you use to solve lower degree polynomials)(This has been proved by Galios)

The problem your teacher has given might be based on the intent to explain how the equations are solved in real life(as in any sophisticated programs like mathematica, matlab etc.,).

The branch of mathematics dealing these kinds of problems like how to find solutions numerically is called "NUMERICAL METHODS"

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Suppose your inequality is $P(x) < 0$. Let $r_1 < \ldots < r_k$ be the distinct real roots of $P(x)$, with corresponding multiplicities $d_1, \ldots, d_k$. Thus $P(x) = (x - r_1)^{d_1} (x - r_2)^{d_2} \ldots (x - r_k)^{d_k} Q(x)$ where $Q(x)$ has the same sign for all $x$ (the sign of the leading coefficient of $P$). As $x$ increases or decreases, the sign of $P(x)$ changes at $x = r_j$ if $d_j$ is odd but stays the same if $d_j$ is even. To find the $r_j$, you generally will need to use numerical methods.

In your example, the roots happen to be integers $-2,-3,-4$ and the multiplicities are all $1$. Obviously $P(x) > 0$ for $x > 0$, and thus also for $x > -2$. At each root the sign changes, so $P(x) < 0$ for $-3 < x < -2$, $P(x) > 0$ for $-4 < x < -3$, and $P(x) < 0$ for $x < -4$.

On the other hand, if it had been $R(x) = {x}^{4}+12\,{x}^{3}+53\,{x}^{2}+102\,x+72 = (x + 3) P(x)$, the roots would be the same except that the multiplicity of the root $-3$ is $2$ instead of $1$. Then at $-3$ the sign doesn't change, so $R(x) < 0$ for both $-3 < x < -2$ and $-4 < x < -3$, and $R(x) > 0$ for $x < -4$.

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For this example, you might try the rational roots test to get a first root $r$, synthetic division to divide out $x-r$, then the quadratic formula on the resulting quadratic.

Once you have all the roots, you can test a number in each interval as you have been doing before.

(I think this is all probably less efficient than just graphing, however.)

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