QuestionAugust 27, 2025

12 Which binomial is a factor of g^3+6g^2+g-14 (1) g-1 (3) g+1 (2) g-2 (4) g+2

12 Which binomial is a factor of g^3+6g^2+g-14 (1) g-1 (3) g+1 (2) g-2 (4) g+2
12 Which binomial is a factor of g^3+6g^2+g-14 (1) g-1 (3) g+1 (2) g-2 (4) g+2

Solution
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Answer

g-2 Explanation 1. Use the Factor Theorem According to the Factor Theorem, if g-c is a factor of a polynomial, then g(c) = 0. Evaluate g(c) for each option. 2. Evaluate g(1) Substitute g = 1: g^{3} + 6g^{2} + g - 14 = 1^3 + 6 \cdot 1^2 + 1 - 14 = -6. Not zero, so g-1 is not a factor. 3. Evaluate g(-1) Substitute g = -1: g^{3} + 6g^{2} + g - 14 = (-1)^3 + 6 \cdot (-1)^2 + (-1) - 14 = -10. Not zero, so g+1 is not a factor. 4. Evaluate g(2) Substitute g = 2: g^{3} + 6g^{2} + g - 14 = 2^3 + 6 \cdot 2^2 + 2 - 14 = 0. Zero, so g-2 is a factor. 5. Evaluate g(-2) Substitute g = -2: g^{3} + 6g^{2} + g - 14 = (-2)^3 + 6 \cdot (-2)^2 + (-2) - 14 = 18. Not zero, so g+2 is not a factor.

Explanation

1. Use the Factor Theorem<br /> According to the Factor Theorem, if $g-c$ is a factor of a polynomial, then $g(c) = 0$. Evaluate $g(c)$ for each option.<br /><br />2. Evaluate $g(1)$<br /> Substitute $g = 1$: $g^{3} + 6g^{2} + g - 14 = 1^3 + 6 \cdot 1^2 + 1 - 14 = -6$. Not zero, so $g-1$ is not a factor.<br /><br />3. Evaluate $g(-1)$<br /> Substitute $g = -1$: $g^{3} + 6g^{2} + g - 14 = (-1)^3 + 6 \cdot (-1)^2 + (-1) - 14 = -10$. Not zero, so $g+1$ is not a factor.<br /><br />4. Evaluate $g(2)$<br /> Substitute $g = 2$: $g^{3} + 6g^{2} + g - 14 = 2^3 + 6 \cdot 2^2 + 2 - 14 = 0$. Zero, so $g-2$ is a factor.<br /><br />5. Evaluate $g(-2)$<br /> Substitute $g = -2$: $g^{3} + 6g^{2} + g - 14 = (-2)^3 + 6 \cdot (-2)^2 + (-2) - 14 = 18$. Not zero, so $g+2$ is not a factor.
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