Use the zero $5 + 2 i$ of $f ( x ) = x ^ { 3 } - 7 x ^ { 2 } - x + 87$ to find all the zeros of f.
A) $- 3,5 \pm 2 i$
B) $3,5 + 2 i$
C) $3,5 - 2 i$
D) $- 3,5 - 2 i$
E) $\pm 3,5 - 2 i$

Question

Quizplus · Accepted Answer

Given a polynomial with real coefficients, if $5 + 2i$ is a zero, then its complex conjugate $5 - 2i$ must also be a zero. The remaining zero can be found by dividing the polynomial by $(x - (5 + 2i))(x - (5 - 2i))$ or by using the fact that the sum of the zeros equals the coefficient of $x^2$ (with opposite sign) divided by the leading coefficient. Here, the sum of the zeros is $7$, and we already have $5 + 2i + 5 - 2i = 10$, so the remaining zero must be $-3$ to make the sum equal to $7$. Therefore, the zeros are $-3$, $5 + 2i$, and $5 - 2i$.

Use the Zero 5+2i5 + 2 i5+2i Of f(x)=x3−7x2−x+87f ( x ) = x ^ { 3 } - 7 x ^ { 2 } - x + 87f(x)=x3−7x2−x+87

Use the Zero $5 + 2 i$ Of $f ( x ) = x ^ { 3 } - 7 x ^ { 2 } - x + 87$