Question

Using the tools you learned in this lesson, find all solutions (real and non-real) for the polynomial: 2x^3-3x^2+32x+17 To earn full credit please share all work, calculations and thinking. If you prefer you can do the work by hand on a piece of paper, take a picture of that work and upload it.

Answer 1

The given polynomial function is;

$f(x)=2x^3-3x^2+32x+17$

Firstly, we would find a factor of the function by a trial and error method.

Thus, we have:

$\begin{gathered} \text{when x=-1} \\ f(-1)=2(-1)^3-3(-1)^2+32(-1)+17 \\ f(-1)=2(-1)-3(1)-32+17 \\ f(-1)=-2-3-32+17 \\ f(-1)=-20 \\ \text{Thus,(x}+1)\text{ is not factor} \end{gathered}$$\begin{gathered} \text{when x=}\frac{-1}{2} \\ f(-\frac{1}{2})=2(-\frac{1}{2})^3-3(-\frac{1}{2})^2+32(-\frac{1}{2})+17 \\ f(-\frac{1}{2})=2(-\frac{1}{8})-3(\frac{1}{4})-\frac{32}{2}+17 \\ f(-\frac{1}{2})=-\frac{2}{8}-\frac{3}{4}-\frac{32}{2}+\frac{17}{1} \\ f(-\frac{1}{2})=\frac{-2-6-128+136}{8} \\ f(-\frac{1}{2})=\frac{0}{8}=0 \\ \text{Since, this is equal to zero, it implies that;} \\ 2x+1\text{ is a factor of the given polynomial} \end{gathered}$

We are going to use the long division method to get the other factor(s).

Thus, we have:

The final remainder of the long division is 0.

Thus, the factors of the polynomial are:

$(2x+1)and(x^2-2x+17)$

Find the zeros of these factors, we have:

$\begin{gathered} 2x+1=0 \\ 2x=-1 \\ x=-\frac{1}{2} \end{gathered}$$\begin{gathered} x^2-2x+17 \\ \text{This is not factorizable, thus we would make use of the quadratic formula to solve it} \end{gathered}$

The quadratic formula is given by the equation;

$x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}$$\begin{gathered} \text{From the equation;} \\ x^2-2x+17 \\ a=1;b=-2;c=17 \\ \text{Thus, we have} \\ x=\frac{-(-2)\pm\sqrt[]{(-2)^2-4(1)(17)}}{2(1)} \\ x=\frac{2\pm\sqrt[]{4-68}}{2} \\ x=\frac{2\pm\sqrt[]{-64}}{2} \\ \text{Recall from complex number;} \\ i=\sqrt[]{-1} \\ \text{Thus, }\sqrt[]{-64}\text{ can be further expressed as }\sqrt[]{64}\text{ }\times\sqrt[]{-1} \\ \Rightarrow8i \\ \text{Thus, } \\ x=\frac{2\pm8i}{2} \\ x=\frac{2+8i}{2}\text{ OR }\frac{2-8i}{2} \\ x=\frac{2(1+4i)}{2}\text{ OR }\frac{2(1-4i)}{2} \\ x=1+4i_{} \\ OR \\ x=1-4i \end{gathered}$

Hence, the solutions are:

$x=-\frac{1}{2},1+4i\text{ and 1-4i}$