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To determine the possible rational zeros of the polynomial [tex]\( f(x) = 4x^3 - 12x^2 - 35x - 12 \)[/tex], we use the Rational Root Theorem. This theorem states that any rational zero, [tex]\(\frac{p}{q}\)[/tex], of a polynomial with integer coefficients is such that:
- [tex]\(p\)[/tex] is a factor of the constant term (the term without [tex]\(x\)[/tex]), which is [tex]\(-12\)[/tex].
- [tex]\(q\)[/tex] is a factor of the leading coefficient (the coefficient of the term with the highest power of [tex]\(x\)[/tex]), which is [tex]\(4\)[/tex].
### Step 1: Identify the factors of the constant term and leading coefficient
- The factors of [tex]\(-12\)[/tex] (constant term) are: [tex]\(\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12\)[/tex].
- The factors of [tex]\(4\)[/tex] (leading coefficient) are: [tex]\(\pm 1, \pm 2, \pm 4\)[/tex].
### Step 2: Generate all possible rational zeros
Using the factors identified, we can list all possible rational zeros [tex]\(\frac{p}{q}\)[/tex]:
[tex]\[ \frac{p}{q} = \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{1}{4}, \pm \frac{3}{4} \][/tex]
Putting them all together in a sorted order, we get the following possible rational zeros:
[tex]\[ -12, -6, -4, -3, -2, -\frac{3}{2}, -1, -\frac{3}{4}, -\frac{1}{2}, -\frac{1}{4}, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, 1, \frac{3}{2}, 2, 3, 4, 6, 12 \][/tex]
### Step 3: Test each potential rational zero to find the actual rational zeros
By evaluating [tex]\( f(x) \)[/tex] at each of the possible rational zeros, we determine which ones are actual zeros:
[tex]\[ f(x) = 4x^3 - 12x^2 - 35x - 12 \][/tex]
After testing each potential rational zero, we find that the actual rational zeros are:
[tex]\[ -\frac{3}{2} \][/tex]
So, the only rational zero of the polynomial [tex]\( f(x) = 4x^3 - 12x^2 - 35x - 12 \)[/tex] is [tex]\(-\frac{3}{2}\)[/tex].
### Step 4: Find all zeros of the polynomial
To find all zeros of the polynomial, we solve [tex]\( f(x) = 0 \)[/tex]. Besides the rational zero [tex]\(-\frac{3}{2}\)[/tex], there might be other zeros. Solving the polynomial equation, we obtain:
[tex]\[ -\frac{3}{2}, \frac{9}{4} - \frac{\sqrt{113}}{4}, \frac{9}{4} + \frac{\sqrt{113}}{4} \][/tex]
### Summary
- Possible rational zeros: [tex]\([-12, -6, -4, -3, -2, -3/2, -1, -3/4, -1/2, -1/4, 1/4, 1/2, 3/4, 1, 3/2, 2, 3, 4, 6, 12]\)[/tex]
- Actual rational zeros: [tex]\([-3/2]\)[/tex]
- All zeros: [tex]\([-3/2, 9/4 - \sqrt{113}/4, 9/4 + \sqrt{113}/4]\)[/tex]
- [tex]\(p\)[/tex] is a factor of the constant term (the term without [tex]\(x\)[/tex]), which is [tex]\(-12\)[/tex].
- [tex]\(q\)[/tex] is a factor of the leading coefficient (the coefficient of the term with the highest power of [tex]\(x\)[/tex]), which is [tex]\(4\)[/tex].
### Step 1: Identify the factors of the constant term and leading coefficient
- The factors of [tex]\(-12\)[/tex] (constant term) are: [tex]\(\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12\)[/tex].
- The factors of [tex]\(4\)[/tex] (leading coefficient) are: [tex]\(\pm 1, \pm 2, \pm 4\)[/tex].
### Step 2: Generate all possible rational zeros
Using the factors identified, we can list all possible rational zeros [tex]\(\frac{p}{q}\)[/tex]:
[tex]\[ \frac{p}{q} = \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{1}{4}, \pm \frac{3}{4} \][/tex]
Putting them all together in a sorted order, we get the following possible rational zeros:
[tex]\[ -12, -6, -4, -3, -2, -\frac{3}{2}, -1, -\frac{3}{4}, -\frac{1}{2}, -\frac{1}{4}, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, 1, \frac{3}{2}, 2, 3, 4, 6, 12 \][/tex]
### Step 3: Test each potential rational zero to find the actual rational zeros
By evaluating [tex]\( f(x) \)[/tex] at each of the possible rational zeros, we determine which ones are actual zeros:
[tex]\[ f(x) = 4x^3 - 12x^2 - 35x - 12 \][/tex]
After testing each potential rational zero, we find that the actual rational zeros are:
[tex]\[ -\frac{3}{2} \][/tex]
So, the only rational zero of the polynomial [tex]\( f(x) = 4x^3 - 12x^2 - 35x - 12 \)[/tex] is [tex]\(-\frac{3}{2}\)[/tex].
### Step 4: Find all zeros of the polynomial
To find all zeros of the polynomial, we solve [tex]\( f(x) = 0 \)[/tex]. Besides the rational zero [tex]\(-\frac{3}{2}\)[/tex], there might be other zeros. Solving the polynomial equation, we obtain:
[tex]\[ -\frac{3}{2}, \frac{9}{4} - \frac{\sqrt{113}}{4}, \frac{9}{4} + \frac{\sqrt{113}}{4} \][/tex]
### Summary
- Possible rational zeros: [tex]\([-12, -6, -4, -3, -2, -3/2, -1, -3/4, -1/2, -1/4, 1/4, 1/2, 3/4, 1, 3/2, 2, 3, 4, 6, 12]\)[/tex]
- Actual rational zeros: [tex]\([-3/2]\)[/tex]
- All zeros: [tex]\([-3/2, 9/4 - \sqrt{113}/4, 9/4 + \sqrt{113}/4]\)[/tex]
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