Question

Find all the zeros of f(x)= x³-14x²+74x-136 given that 5-3i is a zero.

Answer 1

Answer:

5-3i, 5+3i and 4.

Explanation:

Given the function f(x) defined as follows:

$f\mleft(x\mright)=x^3-14x^2+74x-136$

Given that 5-3i is a zero of f(x).

Then, the conjugate of 5-3i must also be a root of f(x).

$\text{Conjugate of 5-3i}=5+3i$

We expand the two roots.

$\begin{gathered} x=5+3i\text{ or x=5-3i} \\ x-(5+3i)=0\text{ or }x-(5-3i)=0 \\ (x-(5+3i))(x-(5-3i))=0 \\ \implies x^2-10x+34=0 \end{gathered}$

The next step is to divide f(x) by the quadratic polynomial derived above.

$\frac{x^3-14x^2+74x-136}{x^2-10x+34}=x-4$

Therefore:

$\begin{gathered} x-4=0 \\ x=4 \end{gathered}$

The zeros of f(x) are: 5-3i, 5+3i and 4.