Question

find the zeros of the following function.
$f(x) = x^3 + 4x^2 + 22x - 68$
2,
$-3 \pm 5i$
$-2,$
$-3 \pm 5i$
2,
$3 \pm 5i$

Explanation:

Step1: Test possible rational zeros

Using Rational Root Theorem, possible rational zeros are factors of 68 over factors of 1, so ±1, ±2, ±4, ±17, ±34, ±68. Test \(x = 2\): \(f(2)=2^3 + 4(2)^2+22(2)-68 = 8 + 16 + 44 - 68 = 0\). So \(x - 2\) is a factor.

Step2: Perform polynomial division or use synthetic division

Divide \(f(x)\) by \(x - 2\) using synthetic division:
\[

$$\begin{array}{r|rrrr} 2 & 1 & 4 & 22 & -68 \\ & & 2 & 12 & 68 \\ \hline & 1 & 6 & 34 & 0 \\ \end{array}$$

\]
So \(f(x)=(x - 2)(x^2 + 6x + 34)\).

Step3: Solve the quadratic equation

Solve \(x^2 + 6x + 34 = 0\) using quadratic formula \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 6\), \(c = 34\).
Discriminant: \(b^2 - 4ac = 36 - 136 = -100\).
So \(x=\frac{-6\pm\sqrt{-100}}{2}=\frac{-6\pm10i}{2}=-3\pm5i\).

Answer:

2, \(-3 \pm 5i\) (corresponding to the first option: 2, \(-3 \pm 5i\))