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},

Zoe Oneal 2021-01-02 Answered

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.

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Expert Answer

opsadnojD
Answered 2021-01-03 Author has 9667 answers

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.
image
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.
image
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: \(\displaystyle{\left({x}-{1}+\sqrt{{3}}\right)}{\left({x}-{2}\right)}{\left({x}-{1}-\sqrt{{3}}\right)}\)

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