Question

For each polynomial function, one zero is given. Find all rational zeros and factor the polynomial. Then graph the function. f(x)=3x^{3}+x^{2}-10x-8ZSK, zero:2

Answer 1

Step 1 Since 2 is a zero of \(\displaystyle{f{{\left({x}\right)}}}={3}{x}^{{{3}}}+{x}^{{{2}}}-{10}{x}-{8}\) The quotient is \(\displaystyle{3}{x}^{{{2}}}+{7}{x}+{4}\)
\(\displaystyle{f{{\left({x}\right)}}}={\left({x}-{2}\right)}{\left({3}{x}^{{{2}}}+{7}{x}+{4}\right)}\) Factor the trinomial \(\displaystyle{3}{x}^{{{2}}}+{7}{x}+{4}={\left({3}{x}+{4}\right)}{\left({x}+{1}\right)}\) So, the function in factored form is \(\displaystyle{f{{\left({x}\right)}}}={\left({x}-{2}\right)}{\left({3}{x}+{4}\right)}{\left({x}+{1}\right)}\) Set each factor equal to zero \(\displaystyle{x}-{2}={0}\) or \(\displaystyle{3}{x}+{4}={0}\) or \(\displaystyle{x}+{1}={0}\) So, all rational zeros are \(\displaystyle{2},-{\frac{{{4}}}{{{2}}}},-{1}\) Step 2 Since, all rational zeros are \(\displaystyle{2},-{\frac{{{4}}}{{{3}}}},-{1}\) So, the graph of the function crosses the x-axis at \(\displaystyle{\left({2},{0}\right)},{\left(-{\frac{{{4}}}{{{3}}}},{0}\right)},{\left(-{1},{0}\right)}\) To find the y-intercept evaluate f(0) \(\displaystyle{f{{\left({x}\right)}}}={3}{x}^{{{3}}}+{x}^{{{2}}}-{10}{x}-{8}\)
\(\displaystyle{f{{\left({0}\right)}}}={3}{\left({0}\right)}^{{{3}}}+{\left({0}\right)}^{{{2}}}-{10}{\left({0}\right)}-{8}\) Substitute 0 for x \(\displaystyle=-{8}\) The leading coefficient is 3 (positive), and the function of degree 3 (odd) So, the end behavior is \(\displaystyle{x}\rightarrow\infty,{f{{\left({x}\right)}}}\rightarrow\infty\) \(\displaystyle{x}\rightarrow-\infty,{f{{\left({x}\right)}}}\rightarrow-\infty\) See the graph of f(x)