Chapter P.7, Problem 16E

Solution

Solve the inequality. Use algebra to solve the equation.

  $6\le x\le 0orx\le -5$ .

Given:

The inequality is, $x^3-x^2-30x\le 0$

Concept Used:

• The relationship between two values that are not equal is defined by inequalities.
• Inequality means not equal.
• Generally, if two values are not equal, we use “not equal symbol (?)”. But to compare the values, whether it is less than or greater than, different inequalities are used.
• Factorization is the process of reducing the bracket of a quadratic equation, instead of expanding the bracket and converting the equation to a product of factors which cannot be reduced further. 

Calculation:

  $\begin{array}{c}{x}^3-x^2-30x\le 0\\ x(x^2-x-30)\le 0\\ x\cdot (x^2-(6-5)x-30)\le 0\\ x\cdot (x^2-6x+5x-30)\le 0\\ x\cdot (x(x-6)+5(x-6))\le 0\\ x\cdot (x+5)(x-6)\le 0\\ x=0\\ (x+5)=0\\ x=-5\\ (x-6)=0\\ x=6\\ 6\le x\le 0orx\le -5\end{array}$

Note that the given inequality is true for the values of $6\le xorx\le 0$ and for the values of $x\le -5$

Thus, the solution of the given inequality is $6\le xorx\le 0$ or $x\le -5$ .

Interval notation $(-\infty ,-5]\cup [0,6]$