If $$ f(x)=x^{3}-13 x^{2}+36 x+50 $$ and $$ x+1 $$ is a factor of $$ f(x) $$, then find all of the zeros of $$ f(x) $$ algebraically. Answer $$ \left.\qquad \begin{array}{l}\oplus \ \qquad\end{array} ight) $$

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If $$ f(x)=x^{3}-13 x^{2}+36 x+50 $$ and $$ x+1 $$ is a factor of $$ f(x) $$, then find all of the zeros of $$ f(x) $$ algebraically. Answer $$ \left.\qquad \begin{array}{l}\oplus \ \qquad\end{array}ight) $$

UpStudy AI · Accepted Answer

1. Since $$ x+1 $$ is a factor of $$ f(x)=x^{3}-13x^{2}+36x+50 $$, then $$ f(-1)=0 $$.

2. We perform synthetic division of $$ f(x) $$ by $$ x+1 $$ (using $$-1$$ as the root):
   - Coefficients: $$ 1 $$ (for $$ x^3 $$), $$ -13 $$ (for $$ x^2 $$), $$ 36 $$ (for $$ x $$), $$ 50 $$ (constant).
   - Write $$-1$$ on the left and the coefficients on the top:
     $$
     \begin{array}{c|cccc}
     -1 & 1 & -13 & 36 & 50 \
       &   & -1  & 14 & -50 \
     \hline
       & 1 & -14 & 50 & 0 \
     \end{array}
     $$
   - The bottom row (except the last 0) represents the coefficients of the quotient polynomial: $$ x^2-14x+50 $$.

3. Factorization of $$ f(x) $$ is:
   $$
   f(x)=(x+1)(x^{2}-14x+50)
   $$

4. Next, we solve the quadratic equation $$ x^{2}-14x+50=0 $$ using the quadratic formula:
   $$
   x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}.
   $$
   Here, $$ a=1 $$, $$ b=-14 $$, and $$ c=50 $$.

5. Substitute the values:
   $$
   x=\frac{-(-14)\pm\sqrt{(-14)^{2}-4(1)(50)}}{2(1)}=\frac{14\pm\sqrt{196-200}}{2}=\frac{14\pm\sqrt{-4}}{2}.
   $$

6. Since $$\sqrt{-4}=2i$$, the solutions become:
   $$
   x=\frac{14\pm2i}{2}=7\pm i.
   $$

7. The zeros of $$ f(x) $$ are:
   $$
   \boxed{-1,\quad 7+i,\quad 7-i.}
   $$
The zeros of $$ f(x) $$ are $$ -1 $$, $$ 7+i $$, and $$ 7-i $$.

If \( f(x)=x^{3}-13 x^{2}+36 x+50 \) and \( x+1 \) is a factor of \( f(x) \), then find all of the zeros of \( f(x) \) algebraically. Answer \( \left.\qquad \begin{array}{l}\oplus \\ \qquad\end{array}\right) \)

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