Polynomial Inequalities

Solving polynomial inequalities requires that we divide the polynomial into discrete intervals.

For example, if we have $x^2 +6x - 16 > 0$ , we can first find the places where the expression is equal to 0. Factoring, we can see that $(x+8)(x-2) = x^2 +6x -16$ , so the polynomial is equal to 0 at $x = -8$ and at $x = 2$ . This means that on the intervals $(-\infty, -8), (-8,2),$ and $(2, \infty)$ , the expression will either be only $>0$ or $< 0$ (because it can't pass through zero on those intervals). Testing each interval, we find that the expression is positive on the outer intervals and is negative on $(-8,2)$ .

Thus the solution is $x < -8$ and $x > 2$ .

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

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Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$