Solve and Graph Polynomial Inequalities - Pre-Calculus

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Question

What is the solution to the following inequality?

Answer

First, we must solve for the roots of the cubic polynomial equation.

We obtain that the roots are $-\frac{2}{3}, 1, 2$ .

Now there are four regions created by these numbers:

$x < -\frac{2}{3}$ . In this region, the values of the polynomial are negative (i.e.plug in and you obtain

$-\frac{2}{3} < x < 1$ . In this region, the values of the polynomial are positive (when , polynomial evaluates to )

. In this region the polynomial switches again to negative.

. In this region the values of the polynomial are positive

Hence the two regions we want are $x < -\frac{2}{3}$ and .

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Question

What is the solution to the following inequality?

Answer

First, we must solve for the roots of the cubic polynomial equation.

We obtain that the roots are $-\frac{2}{3}, 1, 2$ .

Now there are four regions created by these numbers:

$x < -\frac{2}{3}$ . In this region, the values of the polynomial are negative (i.e.plug in and you obtain

$-\frac{2}{3} < x < 1$ . In this region, the values of the polynomial are positive (when , polynomial evaluates to )

. In this region the polynomial switches again to negative.

. In this region the values of the polynomial are positive

Hence the two regions we want are $x < -\frac{2}{3}$ and .

Compare your answer with the correct one above

Question

What is the solution to the following inequality?

Answer

First, we must solve for the roots of the cubic polynomial equation.

We obtain that the roots are $-\frac{2}{3}, 1, 2$ .

Now there are four regions created by these numbers:

$x < -\frac{2}{3}$ . In this region, the values of the polynomial are negative (i.e.plug in and you obtain

$-\frac{2}{3} < x < 1$ . In this region, the values of the polynomial are positive (when , polynomial evaluates to )

. In this region the polynomial switches again to negative.

. In this region the values of the polynomial are positive

Hence the two regions we want are $x < -\frac{2}{3}$ and .

Compare your answer with the correct one above

Question

What is the solution to the following inequality?

Answer

First, we must solve for the roots of the cubic polynomial equation.

We obtain that the roots are $-\frac{2}{3}, 1, 2$ .

Now there are four regions created by these numbers:

$x < -\frac{2}{3}$ . In this region, the values of the polynomial are negative (i.e.plug in and you obtain

$-\frac{2}{3} < x < 1$ . In this region, the values of the polynomial are positive (when , polynomial evaluates to )

. In this region the polynomial switches again to negative.

. In this region the values of the polynomial are positive

Hence the two regions we want are $x < -\frac{2}{3}$ and .

Compare your answer with the correct one above

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