# How do you solve the polynomial inequality and state the answer in interval notation given x^6+x^3>=6?

Mar 12, 2018

The inequality is Quadratic in form.

#### Explanation:

Step 1: We require zero on one side.
${x}^{6} + {x}^{3} - 6 \ge 0$
Step 2: Since the left side consists of a constant term, a middle term, and a term whose exponent is exactly double that on the middle term, this equation is quadratic "in form." We either factor it like a quadratic, or we use the Quadratic Formula. In this case we are able to factor.

Just as ${y}^{2} + y - 6 = \left(y + 3\right) \left(y - 2\right)$, we now have
${x}^{6} + {x}^{3} - 6 = \left({x}^{3} + 3\right) \left({x}^{3} - 2\right)$.
We treat ${x}^{3}$ as though it were a simple variable, y.
If it is more helpful, you may substitute $y = {x}^{3}$, then solve for y, and finally substitute back into x.

Step 3: Set each factor equal to zero separately, and solve the equation ${x}^{6} + {x}^{3} - 6 = 0$. We find where the left side equals zero because these values will be the boundaries of our inequality.

${x}^{3} + 3 = 0$
${x}^{3} = - 3$
$x = - \sqrt[3]{3}$

${x}^{3} - 2 = 0$
${x}^{3} = - 2$
$x = \sqrt[3]{2}$

These are the two real roots of the equation.
They separate the real line into three intervals:
(-oo, -root(3)3); (-root(3)3, root(3)2); and (root(3)2, oo).

Step 4: Determine the sign of the left side of the inequality on each of the above intervals.

Using test points is the usual method. Select a value from each interval, and substitute it for x in the left side of the inequality. We might choose -2, then 0, and then 2.
You will discover that the Left Hand Side is
positive on $\left(- \infty , - \sqrt[3]{3}\right)$;
negative on $\left(- \sqrt[3]{3} , \sqrt[3]{2}\right)$;
and positive on $\left(\sqrt[3]{2} , \infty\right)$.

Step 5: Complete the problem.
We are interested in knowing where ${x}^{6} + {x}^{3} - 6 \ge 0$.
We know now where the left side equals 0, and we know where it is positive. Write this information in interval form as:

$\left(- \infty , - \sqrt[3]{3}\right] \cup \left[\sqrt[3]{2} , \infty\right)$.

NOTE: We have the brackets because the two sides of the inequality are equal at those points, and the original problem requires for us to include those values. Had the problem used $>$ instead of $\ge$, we would have used parentheses.