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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].
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