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To determine the possible rational zeros of the polynomial [tex]\( f(x) = 2x^3 + x^2 - 2x - 1 \)[/tex] and then find all actual rational zeros, we will use the Rational Root Theorem and synthetic division. The Rational Root Theorem states that any possible rational root, [tex]\( \frac{p}{q} \)[/tex], of a polynomial is a factor of the constant term (constant term = [tex]\(-1\)[/tex]) divided by a factor of the leading coefficient (leading coefficient = [tex]\(2\)[/tex]).
### Step 1: Identify factors of the constant term and the leading coefficient
- Factors of [tex]\(-1\)[/tex]: [tex]\(\pm 1\)[/tex]
- Factors of [tex]\(2\)[/tex]: [tex]\(\pm 1, \pm 2\)[/tex]
### Step 2: List all possible rational zeros
The possible rational zeros are the fractions formed by dividing each factor of the constant term by each factor of the leading coefficient:
[tex]\[ \text{Possible rational zeros} = \left\{ \pm 1, \pm \frac{1}{2} \right\} \][/tex]
### Step 3: Test each possible rational zero
We will substitute each possible rational zero into the polynomial [tex]\( f(x) \)[/tex] to see if it equals zero.
1. Test [tex]\( x = 1 \)[/tex]
[tex]\[ f(1) = 2(1)^3 + (1)^2 - 2(1) - 1 = 2 + 1 - 2 - 1 = 0 \][/tex]
So, [tex]\( x = 1 \)[/tex] is a rational zero.
2. Test [tex]\( x = -1 \)[/tex]
[tex]\[ f(-1) = 2(-1)^3 + (-1)^2 - 2(-1) - 1 = -2 + 1 + 2 - 1 = 0 \][/tex]
So, [tex]\( x = -1 \)[/tex] is a rational zero.
3. Test [tex]\( x = \frac{1}{2} \)[/tex]
[tex]\[ f\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 + \left(\frac{1}{2}\right)^2 - 2\left(\frac{1}{2}\right) - 1 = 2\left(\frac{1}{8}\right) + \left(\frac{1}{4}\right) - 1 - 1 \][/tex]
[tex]\[ = \frac{2}{8} + \frac{2}{8} - 1 - 1 = \frac{1}{4} + \frac{1}{4} - 1 - 1 = \frac{1}{2} - 2 = -\frac{3}{2} \neq 0 \][/tex]
So, [tex]\( x = \frac{1}{2} \)[/tex] is not a rational zero.
4. Test [tex]\( x = -\frac{1}{2} \)[/tex]
[tex]\[ f\left(-\frac{1}{2}\right) = 2\left(-\frac{1}{2}\right)^3 + \left(-\frac{1}{2}\right)^2 - 2\left(-\frac{1}{2}\right) - 1 = -2\left(\frac{1}{8}\right) + \left(\frac{1}{4}\right) + 1 - 1 \][/tex]
[tex]\[ = -\frac{1}{4} + \frac{1}{4} + 1 - 1 = 0 \][/tex]
So, [tex]\( x = -\frac{1}{2} \)[/tex] is a rational zero.
### Step 4: List all rational zeros
We found the rational zeros by plugging the possible zeros into the polynomial:
[tex]\[ \text{Rational zeros} = \left\{ 1, -1, -\frac{1}{2} \right\} \][/tex]
Thus, the possible rational zeros for the polynomial [tex]\( f(x) = 2x^3 + x^2 - 2x - 1 \)[/tex] are [tex]\(\pm 1\)[/tex] and [tex]\(\pm \frac{1}{2}\)[/tex], and the actual rational zeros are [tex]\( x = 1, x = -1, \)[/tex] and [tex]\( x = -\frac{1}{2} \)[/tex].
### Step 1: Identify factors of the constant term and the leading coefficient
- Factors of [tex]\(-1\)[/tex]: [tex]\(\pm 1\)[/tex]
- Factors of [tex]\(2\)[/tex]: [tex]\(\pm 1, \pm 2\)[/tex]
### Step 2: List all possible rational zeros
The possible rational zeros are the fractions formed by dividing each factor of the constant term by each factor of the leading coefficient:
[tex]\[ \text{Possible rational zeros} = \left\{ \pm 1, \pm \frac{1}{2} \right\} \][/tex]
### Step 3: Test each possible rational zero
We will substitute each possible rational zero into the polynomial [tex]\( f(x) \)[/tex] to see if it equals zero.
1. Test [tex]\( x = 1 \)[/tex]
[tex]\[ f(1) = 2(1)^3 + (1)^2 - 2(1) - 1 = 2 + 1 - 2 - 1 = 0 \][/tex]
So, [tex]\( x = 1 \)[/tex] is a rational zero.
2. Test [tex]\( x = -1 \)[/tex]
[tex]\[ f(-1) = 2(-1)^3 + (-1)^2 - 2(-1) - 1 = -2 + 1 + 2 - 1 = 0 \][/tex]
So, [tex]\( x = -1 \)[/tex] is a rational zero.
3. Test [tex]\( x = \frac{1}{2} \)[/tex]
[tex]\[ f\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 + \left(\frac{1}{2}\right)^2 - 2\left(\frac{1}{2}\right) - 1 = 2\left(\frac{1}{8}\right) + \left(\frac{1}{4}\right) - 1 - 1 \][/tex]
[tex]\[ = \frac{2}{8} + \frac{2}{8} - 1 - 1 = \frac{1}{4} + \frac{1}{4} - 1 - 1 = \frac{1}{2} - 2 = -\frac{3}{2} \neq 0 \][/tex]
So, [tex]\( x = \frac{1}{2} \)[/tex] is not a rational zero.
4. Test [tex]\( x = -\frac{1}{2} \)[/tex]
[tex]\[ f\left(-\frac{1}{2}\right) = 2\left(-\frac{1}{2}\right)^3 + \left(-\frac{1}{2}\right)^2 - 2\left(-\frac{1}{2}\right) - 1 = -2\left(\frac{1}{8}\right) + \left(\frac{1}{4}\right) + 1 - 1 \][/tex]
[tex]\[ = -\frac{1}{4} + \frac{1}{4} + 1 - 1 = 0 \][/tex]
So, [tex]\( x = -\frac{1}{2} \)[/tex] is a rational zero.
### Step 4: List all rational zeros
We found the rational zeros by plugging the possible zeros into the polynomial:
[tex]\[ \text{Rational zeros} = \left\{ 1, -1, -\frac{1}{2} \right\} \][/tex]
Thus, the possible rational zeros for the polynomial [tex]\( f(x) = 2x^3 + x^2 - 2x - 1 \)[/tex] are [tex]\(\pm 1\)[/tex] and [tex]\(\pm \frac{1}{2}\)[/tex], and the actual rational zeros are [tex]\( x = 1, x = -1, \)[/tex] and [tex]\( x = -\frac{1}{2} \)[/tex].
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