Algebra and Trigonometry (6th Edition)
6th Edition
ISBN: 9780134463216
Author: Robert F. Blitzer
Publisher: PEARSON
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![**Problem Statement:**
Find the solution of the following polynomial inequality. Express your answer in interval notation.
\[ x(x + 4)(x - 4) \geq 0 \]
**Solution Box:**
The solution involves identifying the intervals where the polynomial expression is greater than or equal to zero. Consider the critical points (roots) from the inequality: \( x = 0 \), \( x = -4 \), and \( x = 4 \).
1. **Test Intervals:** Break down the real number line into intervals based on the critical points:
- \((-∞, -4)\)
- \((-4, 0)\)
- \((0, 4)\)
- \((4, ∞)\)
2. **Sign Analysis:**
- For \( (-∞, -4) \), test with \( x = -5 \): the expression \((-5)(-1)(-9) = -45\) is negative.
- For \((-4, 0)\), test with \( x = -2 \): the expression \((-2)(2)(-6) = 24\) is positive.
- For \((0, 4)\), test with \( x = 2 \): the expression \((2)(6)(-2) = -24\) is negative.
- For \((4, ∞)\), test with \( x = 5 \): the expression \((5)(9)(1) = 45\) is positive.
3. **Include Critical Points:** Since the inequality is “greater than or equal to zero”, include points where the expression equals zero: \( x = -4 \), \( x = 0 \), \( x = 4 \).
4. **Solution in Interval Notation:**
The solution is \( [-4, 0] \cup [4, ∞) \).
Ensure each interval where the expression is non-negative is correctly identified and that edges of those intervals reflect equality conditions appropriately.](https://content.bartleby.com/qna-images/question/f8111b40-f806-4f4f-8a96-c8bed7e93551/5494790a-f316-4dd1-b5a0-1dfa06233226/e5pns7g_processed.jpeg)
Transcribed Image Text:**Problem Statement:**
Find the solution of the following polynomial inequality. Express your answer in interval notation.
\[ x(x + 4)(x - 4) \geq 0 \]
**Solution Box:**
The solution involves identifying the intervals where the polynomial expression is greater than or equal to zero. Consider the critical points (roots) from the inequality: \( x = 0 \), \( x = -4 \), and \( x = 4 \).
1. **Test Intervals:** Break down the real number line into intervals based on the critical points:
- \((-∞, -4)\)
- \((-4, 0)\)
- \((0, 4)\)
- \((4, ∞)\)
2. **Sign Analysis:**
- For \( (-∞, -4) \), test with \( x = -5 \): the expression \((-5)(-1)(-9) = -45\) is negative.
- For \((-4, 0)\), test with \( x = -2 \): the expression \((-2)(2)(-6) = 24\) is positive.
- For \((0, 4)\), test with \( x = 2 \): the expression \((2)(6)(-2) = -24\) is negative.
- For \((4, ∞)\), test with \( x = 5 \): the expression \((5)(9)(1) = 45\) is positive.
3. **Include Critical Points:** Since the inequality is “greater than or equal to zero”, include points where the expression equals zero: \( x = -4 \), \( x = 0 \), \( x = 4 \).
4. **Solution in Interval Notation:**
The solution is \( [-4, 0] \cup [4, ∞) \).
Ensure each interval where the expression is non-negative is correctly identified and that edges of those intervals reflect equality conditions appropriately.
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- Some of the following functions are quadratic and others are not. For each function, determine the zeros of the function. f(w)=2w^2+6w+1arrow_forwardFind the domain of each a. f(x) = √2x + 4 b. f(x) = 2x-8arrow_forwardFor this item, show matches by dragging the answers into the appropriate boxes. Complete the table by matching each function to its zeros. f (x) = 3x + 24 g (x) = 2x² – 8x | h (x) = x² – 6x + 9 j (x) = x² – 2x – 15 :: -8 only :: -5 and 3 :: -4 only :: -3 and 3 :: -3 and 5 :: 0 and 4 : 3 only :: 4 only :: 8 onlyarrow_forward
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