The rational Root Theorem states that a rational root of a polynomial equation \(\displaystyle{a}_{{{n}}}\ {x}^{{{n}}}\ +\ {a}_{{{n}\ -\ {1}}}\ {x}^{{{n}\ -\ {1}}}\ +\ \dot{{s}}{c}\ +\ {a}_{{{2}}}\ {x}^{{{2}}}\ +\ {a}_{{{1}}}\ {x}\ +\ {a}_{{{0}}}\ {x}^{{{0}}}={0}\ \text{with integer coeffcients is of the form}\ {\frac{{{p}}}{{{q}}}}\ \text{weher}\ {p}\ \text{is a factor of the constant term,}\ {a}_{{{0}}},\ \text{and}\ {q}\ \text{is a factor of the leading coefficient,}\ {a}_{{{n}}}\)

\(\displaystyle{f{{\left({x}\right)}}}={x}^{{{3}}}\ +\ {3}{x}^{{{2}}}\ -\ {13}{x}\ -\ {15}\)

\(\displaystyle{p}\ \text{could equal any factors of}\ -{15}\)

\(\displaystyle\text{So,}\ \pm\ {1},\ \pm\ {3},\ \pm\ {5},\ \pm\ {15}\) q could equal any foctors of 1 So, \(\displaystyle\pm\ {1}\) Therefore, possible rational zeros are \(\displaystyle\pm\ {1},\ \pm\ {3},\ \pm\ {5},\ \pm\ {15}\) Since the polynomial is \(\displaystyle{x}^{{{3}}}\ +\ {3}{x}^{{{2}}}\ -\ {13}{x}\ -\ {15}\) so, the cooefficient are 1, 3, -13, -15 Let 1 is an actual zero of the polynomial and use synthetic division The remainder is 0 so, -1 is an actual zero and \(\displaystyle{\left({x}\ +\ {1}\right)}\ \text{is a factor of}\ {x}^{{{3}}}\ +\ {3}{x}^{{{2}}}\ -\ {13}{x}\ -\ {15}\) The equatient is \(\displaystyle{x}^{{{2}}}\ +\ {2}{x}\ -\ {15}\)

\(\displaystyle{f{{\left({x}\right)}}}={\left({x}\ +\ {1}\right)}{\left({x}^{{{2}}}\ +\ {2}{x}\ -\ {15}\right)}\)

\(\displaystyle{x}^{{{2}}}\ +\ {2}{x}\ -\ {15}={\left({x}\ -\ {3}\right)}{\left({x}\ +\ {5}\right)}\) Factor the trinomial So, \(\displaystyle{f{{\left({x}\right)}}}={\left({x}\ +\ {1}\right)}{\left({x}\ -\ {3}\right)}{\left({x}\ +\ {5}\right)}\) The function in factored form The function has three zeros -1, 3, and -5 So, the graph of f(x) crosses the x-axis at (-1, 0), (3, 0), and (-5, 0) To find the y-intercept, substitute 0 for x in f(x) \(\displaystyle{f{{\left({x}\right)}}}={x}^{{{3}}}\ +\ {3}{x}^{{{2}}}\ -\ {13}{x}\ -\ {15}\)

\(\displaystyle{f{{\left({0}\right)}}}={0}^{{{3}}}\ +\ {3}{\left({0}\right)}^{{{2}}}\ -\ {13}{\left({0}\right)}\ -\ {15}\) Substitute 0 for x \(\displaystyle=\ -{15}\) So, the function f(x) crosses the y-axis at (0, -15) The leading coefficient is 1 Since the leading coefficient is positive and the function f(x) of degree 3 (odd degree) 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 below

\(\displaystyle{f{{\left({x}\right)}}}={x}^{{{3}}}\ +\ {3}{x}^{{{2}}}\ -\ {13}{x}\ -\ {15}\)

\(\displaystyle{p}\ \text{could equal any factors of}\ -{15}\)

\(\displaystyle\text{So,}\ \pm\ {1},\ \pm\ {3},\ \pm\ {5},\ \pm\ {15}\) q could equal any foctors of 1 So, \(\displaystyle\pm\ {1}\) Therefore, possible rational zeros are \(\displaystyle\pm\ {1},\ \pm\ {3},\ \pm\ {5},\ \pm\ {15}\) Since the polynomial is \(\displaystyle{x}^{{{3}}}\ +\ {3}{x}^{{{2}}}\ -\ {13}{x}\ -\ {15}\) so, the cooefficient are 1, 3, -13, -15 Let 1 is an actual zero of the polynomial and use synthetic division The remainder is 0 so, -1 is an actual zero and \(\displaystyle{\left({x}\ +\ {1}\right)}\ \text{is a factor of}\ {x}^{{{3}}}\ +\ {3}{x}^{{{2}}}\ -\ {13}{x}\ -\ {15}\) The equatient is \(\displaystyle{x}^{{{2}}}\ +\ {2}{x}\ -\ {15}\)

\(\displaystyle{f{{\left({x}\right)}}}={\left({x}\ +\ {1}\right)}{\left({x}^{{{2}}}\ +\ {2}{x}\ -\ {15}\right)}\)

\(\displaystyle{x}^{{{2}}}\ +\ {2}{x}\ -\ {15}={\left({x}\ -\ {3}\right)}{\left({x}\ +\ {5}\right)}\) Factor the trinomial So, \(\displaystyle{f{{\left({x}\right)}}}={\left({x}\ +\ {1}\right)}{\left({x}\ -\ {3}\right)}{\left({x}\ +\ {5}\right)}\) The function in factored form The function has three zeros -1, 3, and -5 So, the graph of f(x) crosses the x-axis at (-1, 0), (3, 0), and (-5, 0) To find the y-intercept, substitute 0 for x in f(x) \(\displaystyle{f{{\left({x}\right)}}}={x}^{{{3}}}\ +\ {3}{x}^{{{2}}}\ -\ {13}{x}\ -\ {15}\)

\(\displaystyle{f{{\left({0}\right)}}}={0}^{{{3}}}\ +\ {3}{\left({0}\right)}^{{{2}}}\ -\ {13}{\left({0}\right)}\ -\ {15}\) Substitute 0 for x \(\displaystyle=\ -{15}\) So, the function f(x) crosses the y-axis at (0, -15) The leading coefficient is 1 Since the leading coefficient is positive and the function f(x) of degree 3 (odd degree) 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 below