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Question Solve the polynomial inequality x3−x2−12x>0x^{3} - x^{2} - 12x > 0 using the test-point method.

Studdy Solution

STEP 1

Assumptions
1. We are given a polynomial inequality x3−x2−12x>0 x^{3} - x^{2} - 12x > 0 .
2. We need to solve for the values of x x that satisfy this inequality.
3. The test-point method involves finding the critical points of the polynomial, determining the intervals they create on the number line, and then testing points within those intervals to see if they satisfy the inequality.

STEP 2

First, we factor the polynomial to find its critical points (the values of x x where the polynomial equals zero).
x3−x2−12x=x(x2−x−12) x^{3} - x^{2} - 12x = x(x^{2} - x - 12)

STEP 3

Next, we factor the quadratic expression x2−x−12 x^{2} - x - 12 .
x2−x−12=(x−4)(x+3) x^{2} - x - 12 = (x - 4)(x + 3)

STEP 4

Now we have the factored form of the polynomial:
x(x−4)(x+3) x(x - 4)(x + 3)

STEP 5

The critical points are the values of x x that make each factor zero:
x=0,x=4, and x=−3 x = 0, x = 4, \text{ and } x = -3

STEP 6

These critical points divide the number line into four intervals. We will choose test points from each interval to determine if the inequality holds within that interval. The intervals are:
1. (−∞,−3) (-\infty, -3)
2. (−3,0) (-3, 0)
3. (0,4) (0, 4)
4. (4,∞) (4, \infty)

STEP 7

Choose a test point from each interval:
1. For (−∞,−3) (-\infty, -3) , choose x=−4 x = -4 .
2. For (−3,0) (-3, 0) , choose x=−1 x = -1 .
3. For (0,4) (0, 4) , choose x=2 x = 2 .
4. For (4,∞) (4, \infty) , choose x=5 x = 5 .

STEP 8

Evaluate the polynomial at each test point:
1. For x=−4 x = -4 : (−4)3−(−4)2−12(−4)=−64−16+48=−32 (-4)^{3} - (-4)^{2} - 12(-4) = -64 - 16 + 48 = -32
2. For x=−1 x = -1 : (−1)3−(−1)2−12(−1)=−1−1+12=10 (-1)^{3} - (-1)^{2} - 12(-1) = -1 - 1 + 12 = 10
3. For x=2 x = 2 : 23−22−12(2)=8−4−24=−20 2^{3} - 2^{2} - 12(2) = 8 - 4 - 24 = -20
4. For x=5 x = 5 : 53−52−12(5)=125−25−60=40 5^{3} - 5^{2} - 12(5) = 125 - 25 - 60 = 40

STEP 9

Determine the sign of the polynomial at each test point:
1. For x=−4 x = -4 , the polynomial is negative.
2. For x=−1 x = -1 , the polynomial is positive.
3. For x=2 x = 2 , the polynomial is negative.
4. For x=5 x = 5 , the polynomial is positive.

STEP 10

Based on the test points, we can determine the sign of the polynomial in each interval:
1. For (−∞,−3) (-\infty, -3) , the polynomial is negative.
2. For (−3,0) (-3, 0) , the polynomial is positive.
3. For (0,4) (0, 4) , the polynomial is negative.
4. For (4,∞) (4, \infty) , the polynomial is positive.

STEP 11

Since we are looking for the intervals where the polynomial is greater than zero, we select the intervals where the test points gave us a positive result:
1. (−3,0) (-3, 0)
2. (4,∞) (4, \infty)

STEP 12

Combine the intervals to express the solution of the inequality:
The solution to the polynomial inequality x3−x2−12x>0 x^{3} - x^{2} - 12x > 0 is:
x∈(−3,0)∪(4,∞) x \in (-3, 0) \cup (4, \infty)

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