QuestionAugust 27, 2025

One factor of f(x)=x^3-14x^2+61x-84 is (x-7) What are the zeros of the function? 7,4,-3 7,4,3 7,-4,-3 7,-4,3

One factor of f(x)=x^3-14x^2+61x-84 is (x-7) What are the zeros of the function? 7,4,-3 7,4,3 7,-4,-3 7,-4,3
One factor of f(x)=x^3-14x^2+61x-84 is (x-7) What are the zeros of the function? 7,4,-3 7,4,3 7,-4,-3 7,-4,3

Solution
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Answer

7, 4, 3 Explanation 1. Use Factor Theorem Since (x-7) is a factor, x=7 is a zero. 2. Perform Polynomial Division Divide f(x) by (x-7) to find the quotient polynomial. 3. Simplify Quotient The quotient is x^2 - 7x + 12. 4. Find Zeros of Quotient Solve x^2 - 7x + 12 = 0 using **quadratic formula**: x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. Here, a=1, b=-7, c=12. Calculate discriminant: (-7)^2 - 4(1)(12) = 49 - 48 = 1. Roots are x = \frac{7 \pm 1}{2}. Solutions: x = 4 and x = 3.

Explanation

1. Use Factor Theorem<br /> Since $(x-7)$ is a factor, $x=7$ is a zero.<br />2. Perform Polynomial Division<br /> Divide $f(x)$ by $(x-7)$ to find the quotient polynomial.<br />3. Simplify Quotient<br /> The quotient is $x^2 - 7x + 12$.<br />4. Find Zeros of Quotient<br /> Solve $x^2 - 7x + 12 = 0$ using **quadratic formula**: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.<br /> Here, $a=1$, $b=-7$, $c=12$. Calculate discriminant: $(-7)^2 - 4(1)(12) = 49 - 48 = 1$.<br /> Roots are $x = \frac{7 \pm 1}{2}$. Solutions: $x = 4$ and $x = 3$.
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