# Chapter 3 - Polynomial and Rational Functions - Section 3.2 The Real Zeros of a Polynomial Function - 3.2 Assess Your Understanding - Page 224: 46

$f(x)=(x-1)(x+5)(x+4)$ Real Zeros: $-5,-4,1$ and all with multiplicity $1$.

#### Work Step by Step

Let us consider that $m$ is a factor of the constant term and $n$ is a factor of the leading coefficient. Then the potential zeros can be expressed by the possible combinations as: $\dfrac{m}{n}$. We see from the given polynomial function that it has at most $3$ real zeros as degree is $3$. The possible factors $m$ of the constant term and $n$ of the leading coefficient are: $m=\pm 1, \pm 2, \pm 3, \pm 6$ and $n=\pm 1$ Therefore, the possible rational roots of $f(x)$ are: $\dfrac{m}{n}=\pm 1, \pm 2, \pm 3, \pm 6$ We test with synthetic division: we will try $x-1$. $\left.\begin{array}{l} 1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr} 1 & 8 &11&-20\\\hline &1&9 &20\\\hline 1&9 &20& |\ \ 0\end{array}$ Thus, we have: $f(x)=(x-1)(x^2+9x+20) \\=(x-1)(x+5)(x+4)$ Real Zeros: $-5,-4,1$ and all with multiplicity $1$.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.