Westonci.ca is the premier destination for reliable answers to your questions, provided by a community of experts. Join our platform to connect with experts ready to provide detailed answers to your questions in various areas. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
Let's solve the given polynomial function step-by-step.
The polynomial function is given by:
[tex]\[ P(x) = x^4 + 4x^3 - 3x^2 - 18x \][/tex]
which is already factored as:
[tex]\[ P(x) = x(x - 2)(x + 3)^2 \][/tex]
### (a) Solve [tex]\( P(x) = 0 \)[/tex]
To find the values of [tex]\( x \)[/tex] for which [tex]\( P(x) = 0 \)[/tex], we set each factor equal to zero:
[tex]\[ x = 0 \][/tex]
[tex]\[ x - 2 = 0 \implies x = 2 \][/tex]
[tex]\[ (x + 3)^2 = 0 \implies x + 3 = 0 \implies x = -3 \][/tex]
Therefore, the solution set for [tex]\( P(x) = 0 \)[/tex] is:
[tex]\[ \{ -3, 0, 2 \} \][/tex]
### (b) Solve [tex]\( P(x) \geq 0 \)[/tex]
To solve [tex]\( P(x) \geq 0 \)[/tex], we need to determine where the polynomial [tex]\( P(x) \)[/tex] is non-negative. We will analyze the sign of [tex]\( P(x) \)[/tex] over the intervals determined by the roots [tex]\(-3\)[/tex], [tex]\(0\)[/tex], and [tex]\(2\)[/tex]. The critical points divide the real line into the following intervals:
[tex]\[ (-\infty, -3), (-3, 0), (0, 2), (2, \infty) \][/tex]
We will test the sign of [tex]\( P(x) \)[/tex] within each interval by selecting a test point from each interval and determining the corresponding sign of [tex]\( P(x) \)[/tex].
1. Interval [tex]\((- \infty, -3)\)[/tex]: Choose [tex]\( x = -4 \)[/tex]
[tex]\[ P(-4) = (-4)(-6)(-1)^2 < 0 \][/tex]
2. Interval [tex]\((-3, 0)\)[/tex]: Choose [tex]\( x = -1 \)[/tex]
[tex]\[ P(-1) = (-1)(-3)(2)^2 > 0 \][/tex]
3. Interval [tex]\((0, 2)\)[/tex]: Choose [tex]\( x = 1 \)[/tex]
[tex]\[ P(1) = (1)(-1)(4) < 0 \][/tex]
4. Interval [tex]\((2, \infty)\)[/tex]: Choose [tex]\( x = 3 \)[/tex]
[tex]\[ P(3) = (3)(1)(6)^2 > 0 \][/tex]
Additionally, we need to consider the values at the critical points:
[tex]\[ P(-3) = 0, \quad P(0) = 0, \quad P(2) = 0 \][/tex]
Based on the signs and the fact that the polynomial is zero at the critical points, we have:
[tex]\[ P(x) \geq 0 \text{ on the intervals } [-3, 0] \cup [2, \infty) \][/tex]
### (c) Solve [tex]\( P(x) \leq 0 \)[/tex]
Since we have already analyzed the intervals, [tex]\( P(x) \)[/tex] is non-positive on the remaining intervals:
[tex]\[ P(x) \leq 0 \text{ on the intervals } (-\infty, -3) \cup (0, 2) \][/tex]
### Summary
(a) The solution set for [tex]\( P(x) = 0 \)[/tex] is [tex]\(\{-3, 0, 2\}\)[/tex].
(b) The solution set for [tex]\( P(x) \geq 0 \)[/tex] is [tex]\([ -3, 0 ] \cup [ 2, \infty )\)[/tex].
(c) The solution set for [tex]\( P(x) \leq 0 \)[/tex] is [tex]\(( - \infty, -3) \cup ( 0, 2 )\)[/tex].
The polynomial function is given by:
[tex]\[ P(x) = x^4 + 4x^3 - 3x^2 - 18x \][/tex]
which is already factored as:
[tex]\[ P(x) = x(x - 2)(x + 3)^2 \][/tex]
### (a) Solve [tex]\( P(x) = 0 \)[/tex]
To find the values of [tex]\( x \)[/tex] for which [tex]\( P(x) = 0 \)[/tex], we set each factor equal to zero:
[tex]\[ x = 0 \][/tex]
[tex]\[ x - 2 = 0 \implies x = 2 \][/tex]
[tex]\[ (x + 3)^2 = 0 \implies x + 3 = 0 \implies x = -3 \][/tex]
Therefore, the solution set for [tex]\( P(x) = 0 \)[/tex] is:
[tex]\[ \{ -3, 0, 2 \} \][/tex]
### (b) Solve [tex]\( P(x) \geq 0 \)[/tex]
To solve [tex]\( P(x) \geq 0 \)[/tex], we need to determine where the polynomial [tex]\( P(x) \)[/tex] is non-negative. We will analyze the sign of [tex]\( P(x) \)[/tex] over the intervals determined by the roots [tex]\(-3\)[/tex], [tex]\(0\)[/tex], and [tex]\(2\)[/tex]. The critical points divide the real line into the following intervals:
[tex]\[ (-\infty, -3), (-3, 0), (0, 2), (2, \infty) \][/tex]
We will test the sign of [tex]\( P(x) \)[/tex] within each interval by selecting a test point from each interval and determining the corresponding sign of [tex]\( P(x) \)[/tex].
1. Interval [tex]\((- \infty, -3)\)[/tex]: Choose [tex]\( x = -4 \)[/tex]
[tex]\[ P(-4) = (-4)(-6)(-1)^2 < 0 \][/tex]
2. Interval [tex]\((-3, 0)\)[/tex]: Choose [tex]\( x = -1 \)[/tex]
[tex]\[ P(-1) = (-1)(-3)(2)^2 > 0 \][/tex]
3. Interval [tex]\((0, 2)\)[/tex]: Choose [tex]\( x = 1 \)[/tex]
[tex]\[ P(1) = (1)(-1)(4) < 0 \][/tex]
4. Interval [tex]\((2, \infty)\)[/tex]: Choose [tex]\( x = 3 \)[/tex]
[tex]\[ P(3) = (3)(1)(6)^2 > 0 \][/tex]
Additionally, we need to consider the values at the critical points:
[tex]\[ P(-3) = 0, \quad P(0) = 0, \quad P(2) = 0 \][/tex]
Based on the signs and the fact that the polynomial is zero at the critical points, we have:
[tex]\[ P(x) \geq 0 \text{ on the intervals } [-3, 0] \cup [2, \infty) \][/tex]
### (c) Solve [tex]\( P(x) \leq 0 \)[/tex]
Since we have already analyzed the intervals, [tex]\( P(x) \)[/tex] is non-positive on the remaining intervals:
[tex]\[ P(x) \leq 0 \text{ on the intervals } (-\infty, -3) \cup (0, 2) \][/tex]
### Summary
(a) The solution set for [tex]\( P(x) = 0 \)[/tex] is [tex]\(\{-3, 0, 2\}\)[/tex].
(b) The solution set for [tex]\( P(x) \geq 0 \)[/tex] is [tex]\([ -3, 0 ] \cup [ 2, \infty )\)[/tex].
(c) The solution set for [tex]\( P(x) \leq 0 \)[/tex] is [tex]\(( - \infty, -3) \cup ( 0, 2 )\)[/tex].
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.
