$\"\"$

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The function is $\"\"$ .

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Let $\"\"$ .

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$\"\"$

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$\"\"$

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Therefore $\"\"$ has $\"\"$ real zeros at $\"\"$ , $\"\"$ and $\"\"$ .

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Create a sign chart using these values.

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$\"\"$

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Now subustiute an $\"\"$ value in each test interval into the factored form of the polynomial to determine if $\"\"$ is positive or negative at that point.

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If $\"\"$ , then $\"\"$ .

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If $\"\"$ , then $\"\"$ .

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If $\"\"$ , then $\"\"$ .

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If $\"\"$ , then $\"\"$ .

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The solutions of $\"\"$ are $\"\"$ values such that $\"\"$ is negative.

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Because $\"\"$ is negative in the interval $\"\"$ .

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The solution is $\"\"$ .

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$\"\"$

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The solution set of $\"\"$ is $\"\"$ .