Question

Information is given about a complex polynomial function f(x) whose coefficients are real numbers. Find the remaining zeros of f. Then find a polynomial function with real coefficients that has the zeros.

Degree 3, Zeroes:$4+i,6$

Short answer

The remaining zero is$4-i$and the polynomial is$x^3-14x^2+65x-102$

Step-by-step solution

Step 1. Given Information

The given data is Degree 3, Zeroes:$4+i,6$

Step 2. Explanation

For a complex polynomial function with real coefficients, if zero is a complex number then its conjugate will also be a zero of the function.

Since $4+i$is complex, its conjugate will be the remaining zero.

$\overline{4+i}=4-i$

Thus,$4-i$is the remaining zero.

Step 3. Calculation

Use the distributive property to find the polynomial using the factors.

$$\begin{gathered} f\left(x\right)=(x-6)[x-(4+i\left)\right][x-(4-i\left)\right] \\ =(x-6)[x^2-x(4+i)-x(4-i)+(4+i\left)\right(4-i\left)\right] \\ =(x-6)(x^2-8x+17) \\ =(x^3-8x^2+17x-6x^2+48x-102) \\ =x^3-14x^2+65x-102 \end{gathered}$$