How do you solve m^4-2m^3-35m^2<=0 using a sign chart?

Sep 18, 2017

${m}^{2} \left({m}^{2} - 2 m - 35\right) \le 0$ when $- 5 \le m \le 7$

Explanation:

Let us factorize The LHS of ${m}^{4} - 2 {m}^{3} - 35 {m}^{2} \le 0$

i.e. ${m}^{2} \left({m}^{2} - 2 m - 35\right) \le 0$

or ${m}^{2} \left(m - 7\right) \left(m + 5\right) \le 0$

Observe that as ${m}^{2} \ge 0$, the above will satisfy only if

$\left(m - 7\right) \left(m + 5\right) < = 0$

Leaving equality for the time being,the product $\left(m - 7\right) \left(m + 5\right)$ is desired to be negative.

It is apparent that sign of binomials $\left(m - 7\right)$ and $\left(m + 5\right)$ will change around the values $- 5$ and $7$ respectively. In sign chart we divide the real number line using these values, i.e. below $- 5$, between $- 5$ and $7$ and above $7$ and see how the sign of $\left(m - 7\right) \left(m + 5\right)$ changes.

Sign Chart

$\textcolor{w h i t e}{X X X X X X X X X X X} - 5 \textcolor{w h i t e}{X X X X X} 7$

$\left(m + 5\right) \textcolor{w h i t e}{X X X} - i v e \textcolor{w h i t e}{X X X X} + i v e \textcolor{w h i t e}{X X X X} + i v e$

$\left(m - 7\right) \textcolor{w h i t e}{X X X} - i v e \textcolor{w h i t e}{X X X X} - i v e \textcolor{w h i t e}{X X X X} + i v e$

$\left(m + 5\right) \left(m - 7\right) \textcolor{w h i t e}{} + i v e \textcolor{w h i t e}{X X X} - i v e \textcolor{w h i t e}{X X X X} + i v e$

It is observed that $\left(m + 5\right) \left(m - 7\right) < 0$ when $m$ is between $- 5$ and $7$, which is the solution for the inequality. Wealso addequality sign as we have ${m}^{2} \left({m}^{2} - 2 m - 35\right) \le 0$

Hence ${m}^{2} \left({m}^{2} - 2 m - 35\right) \le 0$ when $- 5 \le m \le 7$