# How do you solve and write the following in interval notation:  x^4+4x^3-21x^2>=0?

Jun 9, 2016

$x \le 7 \cup x \ge 3$

#### Explanation:

${x}^{4} + 4 {x}^{3} - 21 {x}^{2} \ge 0 \to {x}^{2} \left({x}^{2} + 4 x - 21\right)$

but

${x}^{2} \ge 0 \forall x \in \mathbb{R}$ so
${x}^{2} + 4 x - 21 = \left(x + 7\right) \left(x - 3\right) \ge 0$ or finally
$x \le 7 \cup x \ge 3$ Jun 9, 2016

$\left(- \infty , - 7\right] \cup \left[3 , + \infty\right)$

#### Explanation:

Factor this to find when the expression is equal to $0$. From there, we will create a sign chart to see when the expression is greater than $0$.

Factor ${x}^{2}$ from each term.

${x}^{2} \left({x}^{2} + 4 x - 21\right) \ge 0$

Factor with $7$ and $- 3$.

${x}^{2} \left(x + 7\right) \left(x - 3\right) \ge 0$

Thus the expression equals $0$ at $x = 0$, $x = - 7$, and $x = 3$.

Use one test point in every interval between two pairs of zeros, and another test point between a zero and positive/negative infinities.

Before $x = - 7$:

We can plug in $x = - 8$ into ${x}^{2} \left(x + 7\right) \left(x - 3\right)$.
${\left(- 8\right)}^{2}$ is positive, $\left(- 8 + 7\right)$ is negative, and $\left(- 8 - 3\right)$ is negative, so their product will be positive. Thus the entire interval $\left(- \infty , - 7\right]$ will be $\ge 0$.

Between $x = - 7$ and $x = 0$:

Plugging in $x = - 1$ yields:

${\overbrace{{\left(- 1\right)}^{2}}}^{+} {\overbrace{\left(- 1 + 7\right)}}^{+} {\overbrace{\left(- 1 - 3\right)}}^{-} \le 0$

This is negative, so the interval is not a part of the solution set.

Between $x = 0$ and $x = 3$:

${\overbrace{{\left(1\right)}^{2}}}^{+} {\overbrace{\left(1 + 7\right)}}^{+} {\overbrace{\left(1 - 3\right)}}^{-} \le 0$

This is also negative, and not a part of the solution set.

From $x = 3$ to infinity:

Plugging in $x = 4$ yields:

${\overbrace{{\left(4\right)}^{2}}}^{+} {\overbrace{\left(4 + 7\right)}}^{+} {\overbrace{\left(4 - 3\right)}}^{+} \ge 0$

Since this is positive, $\left[3 , + \infty\right)$ is also a part of the solution set.

Thus the entire solution is the union of the two positive intervals:

$\left(- \infty , - 7\right] \cup \left[3 , + \infty\right)$