Find a polynomial of the specified degree that satisfies the given conditions.

Question

Find a polynomial of the specified degree that satisfies the given conditions.
Degree $4$; zeros $-1$, $0$, $3$, $1/3$ ; coefficient of $x^3$ is $7$

My answer is... $$ P(x)=3x^4 - 7x^3 - 7x^2 + 3x. $$ When I entered this answer into the software (MindTap) for my class it was marked incorrect. mymathportal.com and wolframalpha.com both agree with my answer.

Is there another format that could be also correct or something I am missing?

Thanks for your time.

Answer 1

$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Since $\ds{\braces{\color{#f00}{-1},\color{#f00}{0},\color{#f00}{3},\color{#f00}{{1 \over 3}}}}$ are the polynomial roots, the polynomial we are looking for is proportional to

\begin{align} \bracks{x - \color{#f00}{\pars{-1}}}\pars{x - \color{#f00}{0}} \pars{x - \color{#f00}{3}}\pars{x - \color{#f00}{{1 \over 3}}} = x^{4} - {7 \over 3}\,x^{3} - {7 \over 3}\,x^{2} + x \end{align} In order to 'adjust' the $\ds{x^{3}\!}$-coefficient, multiply it by $\ds{-3 \implies \bbox[8px,#efe,border:0.1em groove navy]{-3x^{4} + \color{#f00}{7}x^{3} + 7x^{2} - 3x}}$

Answer 2

The answer was found in the comments to be $$P(x)=-3x^4+7x^3+7x^2-3x$$