Chapter 2.5, Problem 62E

To determine

To find remaining zeros of the polynomial function $f(x)=4x^3+23x^2+34x-10$ if one of the zero is $-3+i$ .

Answer to Problem 62E

  $\frac{1}{4},-3+i,-3-i.$

Explanation of Solution

Given:

Polynomial function: $f(x)=4x^3+23x^2+34x-10$

Zero of polynomial function: $-3+i$

Calculation:

As $-3+i$ is a zero of the polynomial, the conjugate of $-3+i$ is also a zero of the polynomial.

Conjugate of $-3+i$ is $-3-i$ .

So, the zeros are $-3+i,-3-i$ .

Now consider,

  $\begin{array}{l}\Rightarrow (x-(-3+i))(x-(-3-i))\\ \Rightarrow (x+3-i)(x+3+i)\\ \Rightarrow x^2+3x+xi+3x+9+3i-xi-3i-i^2\\ \Rightarrow x^2+3x+3x+9-i^2\\ \Rightarrow x^2+6x+9-(-1)[\because i^2=-1]\\ \Rightarrow x^2+6x+9+1\\ \Rightarrow x^2+6x+10\end{array}$

Now, divide given polynomial function with $x^2+6x+10$ .

Dividend: $4x^3+23x^2+34x-10$

Divisor: $x^2+6x+10$

  $x^2+6x+10\begin{array}{c}\hfill 4x-1\\ \hfill \overline{)\begin{array}{l}4x^3+23x^2+34x-10\\ \begin{array}{c}4x^3+24x^2+40x\end{array}\\ -x^2-6x-10\\ \begin{array}{c}-x^2-6x-10\end{array}\\ 0\end{array}}\end{array}$  $\begin{array}{l}\frac{4x^3}{x^2}=4x\\ \text{Multiply}4x(x^2+6x+10)\text{and subtract}\text{.}\\ \\ \frac{-x^2}{x^2}=-1\\ \text{Multiply}-1(x^2+6x+10)\text{and subtract}\text{.}\end{array}$

So, quotient is $4x-1$ and remainder is 0.

Hence, $f(x)=(x^2+6x+10)(4x-1).$

So, the other zero of the polynomial function is

  $\begin{array}{l}\Rightarrow 4x-1=0\\ \Rightarrow 4x=1\\ \Rightarrow x=\frac{1}{4}\end{array}$

Conclusion:

Therefore, the zeros of given polynomial function are $x=\frac{1}{4},-3+i\text{and}-3-i.$