Question 4.7.4: Solving a Polynomial Inequality OBJECTIVE Solve a polynomial...
Solving a Polynomial Inequality
OBJECTIVE
Solve a polynomial inequality
P(x) > 0, P(x) ≥ 0, P(x) < 0 or P(x) ≤ 0.
Step 1 Arrange the inequality so that 0 is on the right-hand side.
Step 2 Solve the associated equation P(x) = 0. The real solutions of this equation form the boundary points.
Step 3 Locate the boundary points on the number line. These boundary points divide the number line into intervals. Write these intervals.
Step 4 Select a test point in each interval from Step 3 and evaluate P(x) at each test point to determine the sign of P(x) on that interval.
Step 5 Write the solution set by selecting the intervals in Step 4 that satisfy the given inequality. If the inequality is P(x) ≤ 0 (or ≥ 0) include the corresponding boundary points. Graph the solution set.
Solve: x³ + 2x + 7 ≤ 3x² + 6x – 5
\begin{array}{ll}x^{3}+2 x+7 \leq 3 x^{2}+6 x-5 & \text { Original inequality } \\x^{3}-3 x^{2}-4 x+12 \leq 0 & \text { Subtract } \\& 3 x^{2}+6 x-5 \text { from } \\& \text { both sides simplify. }\end{array}
Solve: x³-3 x²-4 x+12=0
\begin{array}{cl}x^{2}(x-3)-4(x-3)=0 & \text { Factor by grouping.} \\(x-3)\left(x^{2}-4\right)=0 & \text { Distributive property } \\(x-3)(x-2)(x+2)=0 & \text { Difference of two squares} \\x-3=0 \text { or } x-2=0 \text { or } x+2=0 & \text { Zero-product property}\\ x=3 \text { or } \quad x=2 \text { or } \quad x=-2 & \text { Solve for x.} \end{array}
The boundary points are -2, 2, and 3.
The boundary points divide the number line into four intervals: (-\infty,-2),(-2,2),(2,3), \text { and }(3, \infty)
\begin{array}{|c|c|c|c|}\hline \text { Interval } & \text { Point } & P ( x )= x ^{ 3 }- 3 x ^{ 2 }- 4 x + 1 2 & \text { Result } \\\hline(-\infty,-2) & -3 & (-3)^{3}-3(-3)^{2}-4(-3)+12=-30 & \text { Negative } \\(-2,2) & 0 & (0)^{3}-3(0)^{2}-4(0)+12=12 & \text { Positive } \\(2,3) & 2.5 & (2.5)^{3}-3(2.5)^{2}-4(2.5)+12=-1.125 & \text { Negative } \\(3, \infty) & 4 & (4)^{3}-3(4)^{2}-4(4)+12=12 & \text { Positive } \\\hline\end{array}
The Solution set is (-\infty,-2] \cup[2,3].
