Question

solve the equation x^3 - 13x^2 + 47x - 35 = 0 given that 1 is a zero of f(x)=x^3 - 13x^2 + 47x - 35. the solution set is { } (use a comma to separate answers as needed.)

Explanation:

Step1: Perform polynomial long - division

Since 1 is a zero of $f(x)=x^{3}-13x^{2}+47x - 35$, we divide $x^{3}-13x^{2}+47x - 35$ by $x - 1$.
Using polynomial long - division:
\[

$$\begin{align*} \frac{x^{3}-13x^{2}+47x - 35}{x - 1}&=x^{2}-12x + 35 \end{align*}$$

\]

Step2: Solve the quadratic equation

We now solve the quadratic equation $x^{2}-12x + 35=0$.
The quadratic formula for $ax^{2}+bx + c=0$ is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 1$, $b=-12$, and $c = 35$.
First, calculate the discriminant $\Delta=b^{2}-4ac=(-12)^{2}-4\times1\times35=144 - 140 = 4$.
Then, $x=\frac{12\pm\sqrt{4}}{2}=\frac{12\pm2}{2}$.
For the plus - sign: $x=\frac{12 + 2}{2}=7$.
For the minus - sign: $x=\frac{12-2}{2}=5$.
Combined with the known root $x = 1$.

Answer:

$1,5,7$