Solve the inequality. Write the solution set in interval notation.
-$$x ^ { 3 } + 4 x ^ { 2 } - 9 x - 36 \geq 0$$
A) $[ - 4 , \infty )$
B) $[ - 4 , - 3 ] \cup [ 3 , \infty )$
C) $[ - 3,3 ] \cup [ 4 , \infty )$
D) $[ - 4,3 ]$

Question

Quizplus · Accepted Answer

To solve the inequality $x^3 + 4x^2 - 9x - 36 \geq 0$, we factor the cubic polynomial. Factoring by grouping, we get:$$x^2(x + 4) - 9(x + 4) = (x^2 - 9)(x + 4) = (x + 3)(x - 3)(x + 4)$$Setting each factor equal to zero gives the roots $x = -4$, $x = -3$, and $x = 3$. These roots divide the real number line into intervals. Test points in each interval to determine where the inequality is satisfied:1. For $x < -4$, choose $x = -5$: $(-5 + 3)(-5 - 3)(-5 + 4) < 0$, so this interval does not satisfy the inequality.2. For $-4 < x < -3$, choose $x = -3.5$: $(-3.5 + 3)(-3.5 - 3)(-3.5 + 4) > 0$, so this interval satisfies the inequality.3. For $-3 < x < 3$, choose $x = 0$: $(0 + 3)(0 - 3)(0 + 4) < 0$, so this interval does not satisfy the inequality.4. For $x > 3$, choose $x = 4$: $(4 + 3)(4 - 3)(4 + 4) > 0$, so this interval satisfies the inequality.Therefore, the solution set in interval notation is $[ - 4 , - 3 ] \cup [ 3 , \infty )$.

Solve the Inequality x3+4x2−9x−36≥0x ^ { 3 } + 4 x ^ { 2 } - 9 x - 36 \geq 0x3+4x2−9x−36≥0

Solve the Inequality $x ^ { 3 } + 4 x ^ { 2 } - 9 x - 36 \geq 0$