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Sure! To determine which of the given factors is a factor of the polynomial [tex]\(15x^2 - 32x - 7\)[/tex], we can use the Remainder Theorem. According to the theorem, if a polynomial [tex]\(f(x)\)[/tex] is divided by a binomial of the form [tex]\(x - c\)[/tex], the remainder of this division is [tex]\(f(c)\)[/tex]. For a binomial to be a factor of the polynomial, the remainder must be zero.
Given the polynomial [tex]\(15x^2 - 32x - 7\)[/tex], let's check each factor by substituting potential roots into the polynomial and seeing if it equals zero.
1. Checking [tex]\(5x - 1\)[/tex]:
- Rewrite as [tex]\(5x - 1 = 0 \Rightarrow x = \frac{1}{5}\)[/tex].
- Substitute [tex]\(x = \frac{1}{5}\)[/tex] into the polynomial:
[tex]\[ 15\left(\frac{1}{5}\right)^2 - 32\left(\frac{1}{5}\right) - 7 = 15 \cdot \frac{1}{25} - 32 \cdot \frac{1}{5} - 7 = \frac{15}{25} - \frac{32}{5} - 7 = \frac{3}{5} - \frac{32}{5} - 7 \neq 0 \][/tex]
- Therefore, [tex]\(5x - 1\)[/tex] is not a factor.
2. Checking [tex]\(3x + 1\)[/tex]:
- Rewrite as [tex]\(3x + 1 = 0 \Rightarrow x = -\frac{1}{3}\)[/tex].
- Substitute [tex]\(x = -\frac{1}{3}\)[/tex] into the polynomial:
[tex]\[ 15\left(-\frac{1}{3}\right)^2 - 32\left(-\frac{1}{3}\right) - 7 = 15 \cdot \frac{1}{9} + 32 \cdot \frac{1}{3} - 7 = \frac{15}{9} + \frac{32}{3} - 7 = \frac{5}{3} + \frac{32}{3} - 7 \neq 0 \][/tex]
- Therefore, [tex]\(3x + 1\)[/tex] is not a factor.
3. Checking [tex]\(3x - 7\)[/tex]:
- Rewrite as [tex]\(3x - 7 = 0 \Rightarrow x = \frac{7}{3}\)[/tex].
- Substitute [tex]\(x = \frac{7}{3}\)[/tex] into the polynomial:
[tex]\[ 15\left(\frac{7}{3}\right)^2 - 32\left(\frac{7}{3}\right) - 7 = 15 \cdot \frac{49}{9} - 32 \cdot \frac{7}{3} - 7 = \frac{15 \cdot 49}{9} - \frac{224}{3} - 7 = \frac{735}{9} - \frac{224}{3} - 7 = 81 - 81 = 0 \][/tex]
- Therefore, [tex]\(3x - 7\)[/tex] is a factor.
4. Checking [tex]\(3x + 7\)[/tex]:
- Rewrite as [tex]\(3x + 7 = 0 \Rightarrow x = -\frac{7}{3}\)[/tex].
- Substitute [tex]\(x = -\frac{7}{3}\)[/tex] into the polynomial:
[tex]\[ 15\left(-\frac{7}{3}\right)^2 - 32\left(-\frac{7}{3}\right) - 7 = 15 \cdot \frac{49}{9} + 32 \cdot \frac{7}{3} - 7 = \frac{15 \cdot 49}{9} + \frac{224}{3} - 7 = \frac{735}{9} + \frac{224}{3} - 7 \neq 0 \][/tex]
- Therefore, [tex]\(3x + 7\)[/tex] is not a factor.
5. Checking [tex]\(15x - 1\)[/tex]:
- Rewrite as [tex]\(15x - 1 = 0 \Rightarrow x = \frac{1}{15}\)[/tex].
- Substitute [tex]\(x = \frac{1}{15}\)[/tex] into the polynomial:
[tex]\[ 15\left(\frac{1}{15}\right)^2 - 32\left(\frac{1}{15}\right) - 7 = 15 \cdot \frac{1}{225} - 32 \cdot \frac{1}{15} - 7 = \frac{15}{225} - \frac{32}{15} - 7 = \frac{1}{15} - \frac{32}{15} - 7 \neq 0 \][/tex]
- Therefore, [tex]\(15x - 1\)[/tex] is not a factor.
Thus, among the given binomials, the factor of the polynomial [tex]\(15x^2 - 32x - 7\)[/tex] is [tex]\(3x - 7\)[/tex].
Given the polynomial [tex]\(15x^2 - 32x - 7\)[/tex], let's check each factor by substituting potential roots into the polynomial and seeing if it equals zero.
1. Checking [tex]\(5x - 1\)[/tex]:
- Rewrite as [tex]\(5x - 1 = 0 \Rightarrow x = \frac{1}{5}\)[/tex].
- Substitute [tex]\(x = \frac{1}{5}\)[/tex] into the polynomial:
[tex]\[ 15\left(\frac{1}{5}\right)^2 - 32\left(\frac{1}{5}\right) - 7 = 15 \cdot \frac{1}{25} - 32 \cdot \frac{1}{5} - 7 = \frac{15}{25} - \frac{32}{5} - 7 = \frac{3}{5} - \frac{32}{5} - 7 \neq 0 \][/tex]
- Therefore, [tex]\(5x - 1\)[/tex] is not a factor.
2. Checking [tex]\(3x + 1\)[/tex]:
- Rewrite as [tex]\(3x + 1 = 0 \Rightarrow x = -\frac{1}{3}\)[/tex].
- Substitute [tex]\(x = -\frac{1}{3}\)[/tex] into the polynomial:
[tex]\[ 15\left(-\frac{1}{3}\right)^2 - 32\left(-\frac{1}{3}\right) - 7 = 15 \cdot \frac{1}{9} + 32 \cdot \frac{1}{3} - 7 = \frac{15}{9} + \frac{32}{3} - 7 = \frac{5}{3} + \frac{32}{3} - 7 \neq 0 \][/tex]
- Therefore, [tex]\(3x + 1\)[/tex] is not a factor.
3. Checking [tex]\(3x - 7\)[/tex]:
- Rewrite as [tex]\(3x - 7 = 0 \Rightarrow x = \frac{7}{3}\)[/tex].
- Substitute [tex]\(x = \frac{7}{3}\)[/tex] into the polynomial:
[tex]\[ 15\left(\frac{7}{3}\right)^2 - 32\left(\frac{7}{3}\right) - 7 = 15 \cdot \frac{49}{9} - 32 \cdot \frac{7}{3} - 7 = \frac{15 \cdot 49}{9} - \frac{224}{3} - 7 = \frac{735}{9} - \frac{224}{3} - 7 = 81 - 81 = 0 \][/tex]
- Therefore, [tex]\(3x - 7\)[/tex] is a factor.
4. Checking [tex]\(3x + 7\)[/tex]:
- Rewrite as [tex]\(3x + 7 = 0 \Rightarrow x = -\frac{7}{3}\)[/tex].
- Substitute [tex]\(x = -\frac{7}{3}\)[/tex] into the polynomial:
[tex]\[ 15\left(-\frac{7}{3}\right)^2 - 32\left(-\frac{7}{3}\right) - 7 = 15 \cdot \frac{49}{9} + 32 \cdot \frac{7}{3} - 7 = \frac{15 \cdot 49}{9} + \frac{224}{3} - 7 = \frac{735}{9} + \frac{224}{3} - 7 \neq 0 \][/tex]
- Therefore, [tex]\(3x + 7\)[/tex] is not a factor.
5. Checking [tex]\(15x - 1\)[/tex]:
- Rewrite as [tex]\(15x - 1 = 0 \Rightarrow x = \frac{1}{15}\)[/tex].
- Substitute [tex]\(x = \frac{1}{15}\)[/tex] into the polynomial:
[tex]\[ 15\left(\frac{1}{15}\right)^2 - 32\left(\frac{1}{15}\right) - 7 = 15 \cdot \frac{1}{225} - 32 \cdot \frac{1}{15} - 7 = \frac{15}{225} - \frac{32}{15} - 7 = \frac{1}{15} - \frac{32}{15} - 7 \neq 0 \][/tex]
- Therefore, [tex]\(15x - 1\)[/tex] is not a factor.
Thus, among the given binomials, the factor of the polynomial [tex]\(15x^2 - 32x - 7\)[/tex] is [tex]\(3x - 7\)[/tex].
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