Question

question if f(x)=x³ - 13x² + 36x + 50 and f(-1)=0, then find all of the zeros of f(x) algebraically.

Explanation:

Step1: Use factor - theorem

Since \(f(-1)=0\), \((x + 1)\) is a factor of \(f(x)=x^{3}-13x^{2}+36x + 50\). Use polynomial long - division or synthetic division.
Using synthetic division:
-1 | 1 -13 36 50
| -1 14 -50
|________________
1 -14 50 0
So, \(f(x)=(x + 1)(x^{2}-14x + 50)\).

Step2: Solve the quadratic equation

Set \(x^{2}-14x + 50=0\). Use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for the quadratic equation \(ax^{2}+bx + c = 0\). Here, \(a = 1\), \(b=-14\), and \(c = 50\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(-14)^{2}-4\times1\times50=196 - 200=-4\).
Then \(x=\frac{14\pm\sqrt{-4}}{2}=\frac{14\pm2i}{2}=7\pm i\).
Also, since \(x + 1 = 0\) gives \(x=-1\).

Answer:

\(x=-1,7 + i,7 - i\)