# a) Find the rational zeros and then the other zeros of the polynomial function displaystyle{left({x}right)}={x}^{3}-{4}{x}^{2}+{2}{x}+{4},

a) Find the rational zeros and then the other zeros of the polynomial function $$\displaystyle{\left({x}\right)}={x}^{3}-{4}{x}^{2}+{2}{x}+{4}, \text{that is, solve}\ \displaystyle f{{\left({x}\right)}}={0}.$$
b) Factor $$f(x)$$ into linear factors.

• Questions are typically answered in as fast as 30 minutes

### Plainmath recommends

• Get a detailed answer even on the hardest topics.
• Ask an expert for a step-by-step guidance to learn to do it yourself.

Step 1
a) We graph $$\displaystyle f{{\left({x}\right)}}={x}^{3}-{4}{x}^{2}+{2}{x}+{4}$$ using a graphing calculator to find the rational zeros.

It cuts the graph at $$\displaystyle{x}={2}$$
Means $$x = 2$$ is a zero.
Step 2
Then we use $$x = 2$$ to find the other factor. We use synthetic division.

Another factor is $$\displaystyle{x}^{2}-{2}{x}-{2}.$$
Step 3
Then we use the quadratic formula to find the zeros fromit
$$\displaystyle{x}^{2}-{2}{x}-{2}={0}$$
$$\displaystyle{x}=\frac{{{2}\pm\sqrt{{{\left(-{2}\right)}^{2}-{4}{\left({1}\right)}{\left(-{2}\right)}}}}}{{{2}{\left({1}\right)}}}$$
$$\displaystyle{x}=\frac{{{2}\pm\sqrt{{12}}}}{{2}}$$
$$\displaystyle{x}=\frac{{{2}\pm{2}\sqrt{{3}}}}{{2}}$$
$$\displaystyle{x}={1}\pm\sqrt{{3}}$$
Answer(a): $$\displaystyle{1}-\sqrt{{3}},{2},{1}\pm\sqrt{{3}}$$
Step 4
b) Linear factored form of f is:
$$\displaystyle f{{\left({x}\right)}}={\left[{x}-{\left({1}-\sqrt{{3}}\right)}\right]}{\left({x}-{2}\right)}{\left[{x}-{\left({1}+\sqrt{{3}}\right)}\right]}$$
$$\displaystyle f{{\left({x}\right)}}={\left({x}-{1}+\sqrt{{3}}\right)}{\left({x}-{2}\right)}{\left({x}-{1}-\sqrt{{3}}\right)}$$
Answer: $$\displaystyle{\left({x}-{1}+\sqrt{{3}}\right)}{\left({x}-{2}\right)}{\left({x}-{1}-\sqrt{{3}}\right)}$$