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To analyze the polynomial [tex]\( f(x) = x^3 + 2x^2 - 16x - 32 \)[/tex] using Descartes' Rule of Signs, we follow these steps:
1. Determine the number of positive real zeros by examining the sign changes in [tex]\( f(x) \)[/tex].
2. Determine the number of negative real zeros by examining the sign changes in [tex]\( f(-x) \)[/tex].
3. List possible combinations of positive, negative, and imaginary zeros.
4. Identify all real zeros by solving the polynomial.
### Step 1: Positive Real Zeros
First, observe the sign changes in [tex]\( f(x) \)[/tex].
[tex]\[ f(x) = x^3 + 2x^2 - 16x - 32 \][/tex]
The coefficients are [tex]\( 1, 2, -16, -32 \)[/tex]. Analyze the sequence of signs:
- Going from [tex]\( 1 \)[/tex] to [tex]\( 2 \)[/tex], the sign does not change (positive to positive).
- Going from [tex]\( 2 \)[/tex] to [tex]\( -16 \)[/tex], the sign changes (positive to negative).
- Going from [tex]\( -16 \)[/tex] to [tex]\( -32 \)[/tex], the sign does not change (negative to negative).
Thus, there is exactly 1 sign change. According to Descartes' Rule of Signs, there is exactly one positive real zero.
### Step 2: Negative Real Zeros
Next, we need [tex]\( f(-x) \)[/tex]. Substitute [tex]\(-x\)[/tex] for [tex]\(x\)[/tex]:
[tex]\[ f(-x) = (-x)^3 + 2(-x)^2 - 16(-x) - 32 = -x^3 + 2x^2 + 16x - 32 \][/tex]
Now, observe the sign changes in [tex]\( f(-x) \)[/tex]:
- Going from [tex]\( -1 \)[/tex] to [tex]\( 2 \)[/tex], the sign changes (negative to positive).
- Going from [tex]\( 2 \)[/tex] to [tex]\( 16 \)[/tex], the sign does not change (positive to positive).
- Going from [tex]\( 16 \)[/tex] to [tex]\( -32 \)[/tex], the sign changes (positive to negative).
Thus, there are 2 sign changes. According to Descartes' Rule of Signs, there could be 2 or 0 negative real zeros (each sign change corresponds to a possible pair of negative real zeros).
### Step 3: Possible Combinations of Real and Imaginary Zeros
The degree of [tex]\( f(x) \)[/tex] is 3, so the total number of zeros (real and imaginary) should add up to 3.
Since we have determined:
- 1 positive real zero
- Either 2 or 0 negative real zeros,
- The rest must be imaginary.
Thus, the possible combinations are:
1. [tex]\(1\)[/tex] positive, [tex]\(2\)[/tex] negative, and [tex]\(0\)[/tex] imaginary zeros.
2. [tex]\(1\)[/tex] positive, [tex]\(0\)[/tex] negative, and [tex]\(2\)[/tex] imaginary zeros.
### Step 4: Identify All Zeros
To determine the exact zeros, solve the polynomial equation [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ x^3 + 2x^2 - 16x - 32 = 0 \][/tex]
To do this analytically, one approach is to apply the Rational Root Theorem to find potential rational zeros, and then verify them by synthetic division or direct substitution.
Possible Rational Roots are [tex]\( \pm 1, \pm 2, \pm 4, \pm 8, \pm 16, \pm 32 \)[/tex].
Test these:
- Trying [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = 2^3 + 2(2)^2 - 16(2) - 32 \][/tex]
[tex]\[ = 8 + 8 - 32 - 32 \][/tex]
[tex]\[ = 16 - 64 \][/tex]
[tex]\[ = -48 \quad \text{(not a zero)} \][/tex]
- Trying [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = (-2)^3 + 2(-2)^2 - 16(-2) - 32 \][/tex]
[tex]\[ = -8 + 8 + 32 - 32 \][/tex]
[tex]\[ = 0 \quad \text{(a zero)} \][/tex]
So [tex]\( x = -2 \)[/tex] is a root. Use synthetic division to factor [tex]\( f(x) \)[/tex]:
[tex]\[ \begin{array}{r|rrr} -2 & 1 & 2 & -16 & -32 \\ & & -2 & 0 & 32 \\ \hline & 1 & 0 & -16 & 0 \\ \end{array} \][/tex]
Resulting in:
[tex]\[ f(x) = (x + 2)(x^2 - 16) \][/tex]
Further, factor [tex]\( x^2 - 16 \)[/tex]:
[tex]\[ x^2 - 16 = (x - 4)(x + 4) \][/tex]
Hence:
[tex]\[ f(x) = (x + 2)(x - 4)(x + 4) \][/tex]
### Zeros
Thus, the zeros of [tex]\( f(x) \)[/tex] are:
[tex]\[ x = -2, \quad x = 4, \quad x = -4 \][/tex]
So the final result is:
- There is 1 positive real zero: [tex]\( x = 4 \)[/tex].
- There are 2 negative real zeros: [tex]\( x = -2, -4 \)[/tex].
- There are no imaginary zeros.
Therefore, the zeros are [tex]\( \boxed{-2, 4, -4} \)[/tex].
1. Determine the number of positive real zeros by examining the sign changes in [tex]\( f(x) \)[/tex].
2. Determine the number of negative real zeros by examining the sign changes in [tex]\( f(-x) \)[/tex].
3. List possible combinations of positive, negative, and imaginary zeros.
4. Identify all real zeros by solving the polynomial.
### Step 1: Positive Real Zeros
First, observe the sign changes in [tex]\( f(x) \)[/tex].
[tex]\[ f(x) = x^3 + 2x^2 - 16x - 32 \][/tex]
The coefficients are [tex]\( 1, 2, -16, -32 \)[/tex]. Analyze the sequence of signs:
- Going from [tex]\( 1 \)[/tex] to [tex]\( 2 \)[/tex], the sign does not change (positive to positive).
- Going from [tex]\( 2 \)[/tex] to [tex]\( -16 \)[/tex], the sign changes (positive to negative).
- Going from [tex]\( -16 \)[/tex] to [tex]\( -32 \)[/tex], the sign does not change (negative to negative).
Thus, there is exactly 1 sign change. According to Descartes' Rule of Signs, there is exactly one positive real zero.
### Step 2: Negative Real Zeros
Next, we need [tex]\( f(-x) \)[/tex]. Substitute [tex]\(-x\)[/tex] for [tex]\(x\)[/tex]:
[tex]\[ f(-x) = (-x)^3 + 2(-x)^2 - 16(-x) - 32 = -x^3 + 2x^2 + 16x - 32 \][/tex]
Now, observe the sign changes in [tex]\( f(-x) \)[/tex]:
- Going from [tex]\( -1 \)[/tex] to [tex]\( 2 \)[/tex], the sign changes (negative to positive).
- Going from [tex]\( 2 \)[/tex] to [tex]\( 16 \)[/tex], the sign does not change (positive to positive).
- Going from [tex]\( 16 \)[/tex] to [tex]\( -32 \)[/tex], the sign changes (positive to negative).
Thus, there are 2 sign changes. According to Descartes' Rule of Signs, there could be 2 or 0 negative real zeros (each sign change corresponds to a possible pair of negative real zeros).
### Step 3: Possible Combinations of Real and Imaginary Zeros
The degree of [tex]\( f(x) \)[/tex] is 3, so the total number of zeros (real and imaginary) should add up to 3.
Since we have determined:
- 1 positive real zero
- Either 2 or 0 negative real zeros,
- The rest must be imaginary.
Thus, the possible combinations are:
1. [tex]\(1\)[/tex] positive, [tex]\(2\)[/tex] negative, and [tex]\(0\)[/tex] imaginary zeros.
2. [tex]\(1\)[/tex] positive, [tex]\(0\)[/tex] negative, and [tex]\(2\)[/tex] imaginary zeros.
### Step 4: Identify All Zeros
To determine the exact zeros, solve the polynomial equation [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ x^3 + 2x^2 - 16x - 32 = 0 \][/tex]
To do this analytically, one approach is to apply the Rational Root Theorem to find potential rational zeros, and then verify them by synthetic division or direct substitution.
Possible Rational Roots are [tex]\( \pm 1, \pm 2, \pm 4, \pm 8, \pm 16, \pm 32 \)[/tex].
Test these:
- Trying [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = 2^3 + 2(2)^2 - 16(2) - 32 \][/tex]
[tex]\[ = 8 + 8 - 32 - 32 \][/tex]
[tex]\[ = 16 - 64 \][/tex]
[tex]\[ = -48 \quad \text{(not a zero)} \][/tex]
- Trying [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = (-2)^3 + 2(-2)^2 - 16(-2) - 32 \][/tex]
[tex]\[ = -8 + 8 + 32 - 32 \][/tex]
[tex]\[ = 0 \quad \text{(a zero)} \][/tex]
So [tex]\( x = -2 \)[/tex] is a root. Use synthetic division to factor [tex]\( f(x) \)[/tex]:
[tex]\[ \begin{array}{r|rrr} -2 & 1 & 2 & -16 & -32 \\ & & -2 & 0 & 32 \\ \hline & 1 & 0 & -16 & 0 \\ \end{array} \][/tex]
Resulting in:
[tex]\[ f(x) = (x + 2)(x^2 - 16) \][/tex]
Further, factor [tex]\( x^2 - 16 \)[/tex]:
[tex]\[ x^2 - 16 = (x - 4)(x + 4) \][/tex]
Hence:
[tex]\[ f(x) = (x + 2)(x - 4)(x + 4) \][/tex]
### Zeros
Thus, the zeros of [tex]\( f(x) \)[/tex] are:
[tex]\[ x = -2, \quad x = 4, \quad x = -4 \][/tex]
So the final result is:
- There is 1 positive real zero: [tex]\( x = 4 \)[/tex].
- There are 2 negative real zeros: [tex]\( x = -2, -4 \)[/tex].
- There are no imaginary zeros.
Therefore, the zeros are [tex]\( \boxed{-2, 4, -4} \)[/tex].
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