Math Problem Statement

Consider the following polynomial inequality: x^2(x + 4)(x - 1) > 0. Write the polynomial inequality in the form p(x) < 0, p(x) ≤ 0, p(x) > 0, or p(x) ≥ 0, then find the real zeros of p(x).

Solution

To solve the inequality given in the image:

x2(x+4)(x1)>0x^2(x + 4)(x - 1) > 0

we need to analyze the inequality by finding the intervals where this expression is positive.

Step 1: Identify the Zeros

The inequality is a product of three terms. We start by finding the zeros of each term:

  1. x2=0x^2 = 0 implies x=0x = 0
  2. x+4=0x + 4 = 0 implies x=4x = -4
  3. x1=0x - 1 = 0 implies x=1x = 1

These zeros are x=0x = 0, x=4x = -4, and x=1x = 1.

Step 2: Analyze Intervals

Using these zeros, we divide the real line into intervals:

  1. (,4)(-\infty, -4)
  2. (4,0)(-4, 0)
  3. (0,1)(0, 1)
  4. (1,)(1, \infty)

For each interval, we test a point within the interval to see if the expression x2(x+4)(x1)x^2(x + 4)(x - 1) is positive or negative.

Step 3: Sign Analysis

  1. Interval (,4)(-\infty, -4): Pick x=5x = -5. x2=25(positive),(x+4)=1(negative),(x1)=6(negative)x^2 = 25 \quad (positive), \quad (x+4) = -1 \quad (negative), \quad (x-1) = -6 \quad (negative) The product 25×(1)×(6)=15025 \times (-1) \times (-6) = 150 (positive).

  2. Interval (4,0)(-4, 0): Pick x=1x = -1. x2=1(positive),(x+4)=3(positive),(x1)=2(negative)x^2 = 1 \quad (positive), \quad (x+4) = 3 \quad (positive), \quad (x-1) = -2 \quad (negative) The product 1×3×(2)=61 \times 3 \times (-2) = -6 (negative).

  3. Interval (0,1)(0, 1): Pick x=0.5x = 0.5. x2=0.25(positive),(x+4)=4.5(positive),(x1)=0.5(negative)x^2 = 0.25 \quad (positive), \quad (x+4) = 4.5 \quad (positive), \quad (x-1) = -0.5 \quad (negative) The product 0.25×4.5×(0.5)=0.56250.25 \times 4.5 \times (-0.5) = -0.5625 (negative).

  4. Interval (1,)(1, \infty): Pick x=2x = 2. x2=4(positive),(x+4)=6(positive),(x1)=1(positive)x^2 = 4 \quad (positive), \quad (x+4) = 6 \quad (positive), \quad (x-1) = 1 \quad (positive) The product 4×6×1=244 \times 6 \times 1 = 24 (positive).

Step 4: Solution

The intervals where x2(x+4)(x1)>0x^2(x + 4)(x - 1) > 0 (positive) are:

  • (,4)(-\infty, -4)
  • (1,)(1, \infty)

Thus, the solution to the inequality is: x(,4)(1,)x \in (-\infty, -4) \cup (1, \infty)

Let me know if you need further details or explanations on any of the steps.

Here are five related questions for further exploration:

  1. How would the solution change if the inequality were 0\geq 0 instead of >0> 0?
  2. What would the solution be if the inequality were x2(x+4)(x1)<0x^2(x + 4)(x - 1) < 0?
  3. Can this polynomial inequality be solved graphically, and how?
  4. How do changes to the coefficients affect the intervals where the inequality holds?
  5. How can interval notation help represent the solution clearly in complex inequalities?

Tip: For polynomial inequalities, always find the zeros first, then use interval testing to determine the sign of the expression within each interval.

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Math Problem Analysis

Mathematical Concepts

Polynomial Inequalities
Interval Testing
Real Zeros

Formulas

Product of factors p(x) = x^2(x + 4)(x - 1)

Theorems

Sign Testing in Intervals for Polynomial Inequalities

Suitable Grade Level

Grades 10-12