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# Graph each polynomial function. Factor first if the expression is not in factored form.f(x)=x^{3} + 3x^{2} - 13x - 15 PSZ

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asked 2020-11-08

Graph each polynomial function. Factor first if the expression is not in factored form. $$f(x)=x^{3}\ +\ 3x^{2}\ -\ 13x\ -\ 15$$

## Answers (1)

2020-11-09

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

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