Search

Text Color

Margin Size

Font Type

Enable Dyslexic Font

8.4: Exercises

Last updated

May 2, 2022
Save as PDF
- 8.3: Optional section- Synthetic division
- 9: Graphing Polynomials

Thomas Tradler and Holly Carley
CUNY New York City College of Technology via New York City College of Technology at CUNY Academic Works

$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$

$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$

$\newcommand{\id}{\mathrm{id}}$ $\newcommand{\Span}{\mathrm{span}}$

( \newcommand{\kernel}{\mathrm{null}\,}\) $\newcommand{\range}{\mathrm{range}\,}$

$\newcommand{\RealPart}{\mathrm{Re}}$ $\newcommand{\ImaginaryPart}{\mathrm{Im}}$

$\newcommand{\Argument}{\mathrm{Arg}}$ $\newcommand{\norm}[1]{\| #1 \|}$

$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$\newcommand{\Span}{\mathrm{span}}$

$\newcommand{\id}{\mathrm{id}}$

$\newcommand{\Span}{\mathrm{span}}$

$\newcommand{\kernel}{\mathrm{null}\,}$

$\newcommand{\range}{\mathrm{range}\,}$

$\newcommand{\RealPart}{\mathrm{Re}}$

$\newcommand{\ImaginaryPart}{\mathrm{Im}}$

$\newcommand{\Argument}{\mathrm{Arg}}$

$\newcommand{\norm}[1]{\| #1 \|}$

$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$\newcommand{\Span}{\mathrm{span}}$ $\newcommand{\AA}{\unicode[.8,0]{x212B}}$

$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$

$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$

$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$

$\newcommand{\vectorC}[1]{\textbf{#1}}$

$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$

$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$

$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$

$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$

$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$

$\newcommand{\avec}{\mathbf a}$

$\newcommand{\bvec}{\mathbf b}$

$\newcommand{\cvec}{\mathbf c}$

$\newcommand{\dvec}{\mathbf d}$

$\newcommand{\dtil}{\widetilde{\mathbf d}}$

$\newcommand{\evec}{\mathbf e}$

$\newcommand{\fvec}{\mathbf f}$

$\newcommand{\nvec}{\mathbf n}$

$\newcommand{\pvec}{\mathbf p}$

$\newcommand{\qvec}{\mathbf q}$

$\newcommand{\svec}{\mathbf s}$

$\newcommand{\tvec}{\mathbf t}$

$\newcommand{\uvec}{\mathbf u}$

$\newcommand{\vvec}{\mathbf v}$

$\newcommand{\wvec}{\mathbf w}$

$\newcommand{\xvec}{\mathbf x}$

$\newcommand{\yvec}{\mathbf y}$

$\newcommand{\zvec}{\mathbf z}$

$\newcommand{\rvec}{\mathbf r}$

$\newcommand{\mvec}{\mathbf m}$

$\newcommand{\zerovec}{\mathbf 0}$

$\newcommand{\onevec}{\mathbf 1}$

$\newcommand{\real}{\mathbb R}$

$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$

$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$

$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$

$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$

$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$

$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$

$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$

$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$

$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$

$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$

$\newcommand{\bcal}{\cal B}$

$\newcommand{\ccal}{\cal C}$

$\newcommand{\scal}{\cal S}$

$\newcommand{\wcal}{\cal W}$

$\newcommand{\ecal}{\cal E}$

$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$

$\newcommand{\gray}[1]{\color{gray}{#1}}$

$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$

$\newcommand{\rank}{\operatorname{rank}}$

$\newcommand{\row}{\text{Row}}$

$\newcommand{\col}{\text{Col}}$

$\renewcommand{\row}{\text{Row}}$

$\newcommand{\nul}{\text{Nul}}$

$\newcommand{\var}{\text{Var}}$

$\newcommand{\corr}{\text{corr}}$

$\newcommand{\len}[1]{\left|#1\right|}$

$\newcommand{\bbar}{\overline{\bvec}}$

$\newcommand{\bhat}{\widehat{\bvec}}$

$\newcommand{\bperp}{\bvec^\perp}$

$\newcommand{\xhat}{\widehat{\xvec}}$

$\newcommand{\vhat}{\widehat{\vvec}}$

$\newcommand{\uhat}{\widehat{\uvec}}$

$\newcommand{\what}{\widehat{\wvec}}$

$\newcommand{\Sighat}{\widehat{\Sigma}}$

$\newcommand{\lt}{<}$

$\newcommand{\gt}{>}$

$\newcommand{\amp}{&}$

$\definecolor{fillinmathshade}{gray}{0.9}$

Exercise $\PageIndex{1}$

Divide by long division.

$\dfrac{x^3-4x^2+2x+1}{x-2}$
$\dfrac{x^3+6x^2+7x-2}{x+3}$
$\dfrac{x^2+7x-4}{x+1}$
$\dfrac{x^3+3x^2+2x+5}{x+2}$
$\dfrac{2x^3+x^2+3x+5}{x-1}$
$\dfrac{2x^4+7x^3+x+3}{x+5}$
$\dfrac{2x^4-31x^2-13}{x-4}$
$\dfrac{x^3+27}{x+3}$
$\dfrac{3x^4+7x^3+5x^2+7x+4}{3x+1}$
$\dfrac{8x^3+18x^2+21x+18}{2x+3}$
$\dfrac{x^3+3x^2-4x-5}{x^2+2x+1}$
$\dfrac{x^5+3x^4-20}{x^2+3}$

Answer

$x^{2}-2 x-2-\dfrac{3}{x-2}$
$x^{2}+3 x-2+\dfrac{4}{x+3}$
$x+6-\dfrac{10}{x+1}$
$x^{2}+x+\dfrac{5}{x+2}$
$2 x^{2}+3 x+6+\dfrac{11}{x-1}$
$2 x^{3}-3 x^{2}+15 x-74+\dfrac{373}{x+5}$
$2 x^{3}+8 x^{2}+x+4+\dfrac{3}{x-4}$
$x^{2}-3 x+9$
$x^{3}+2 x^{2}+x+2+\dfrac{2}{3 x+1}$
$4 x^{2}+3 x+6$
$x+1-\dfrac{7 x+6}{x^{2}+2 x+1}$
$x^{3}+3 x^{2}-3 x-9+\dfrac{9 x+7}{x^{2}+3}$

Exercise $\PageIndex{2}$

Find the remainder when dividing $f(x)$ by $g(x)$ .

$f(x)=x^3+2x^2+x-3, \quad g(x)=x-2$
$f(x)=x^3-5x+8, \quad g(x)=x-3$
$f(x)=x^5-1, \quad g(x)=x+1$
$f(x)=x^5+5x^2-7x+10, \quad g(x)=x+2$

Answer

remainder $r = 15$
$r = 20$
$r = -2$
$r = 12$

Exercise $\PageIndex{3}$

Determine whether the given $g(x)$ is a factor of $f(x)$ . If so, name the corresponding root of $f(x)$ .

$f(x)=x^2+5x+6, \quad g(x)=x+3$
$f(x)=x^3-x^2-3x+8, \quad g(x)=x-4$
$f(x)=x^4+7x^3+3x^2+29x+56, \quad g(x)=x+7$
$f(x)=x^{999}+1, \quad g(x)=x+1$

Answer

yes, $g(x)$ is a factor of $f(x)$ , the root of $f(x)$ is $x = −3$
$g(x)$ is not a factor of $f(x)$
$g(x)$ is a factor of $f(x)$ , the root of $f(x)$ is $x = −7$
$g(x)$ is a factor of $f(x)$ , the root of $f(x)$ is $x = −1$

Exercise $\PageIndex{4}$

Check that the given numbers for $x$ are roots of $f(x)$ (see Observation). If the numbers $x$ are indeed roots, then use this information to factor $f(x)$ as much as possible.

$f(x)=x^3-2x^2-x+2, \quad x=1$
$f(x)=x^3-6x^2+11x-6, \quad x=1, x=2, x=3$
$f(x)=x^3-3x^2+x-3, \quad x=3$
$f(x)=x^3+6x^2+12x+8, \quad x=-2$
$f(x)=x^3+13x^2+50x+56, \quad x=-3, x=-4$
$f(x)=x^3+3x^2-16x-48, \quad x=2, x=-4$
$f(x)=x^5+5x^4-5x^3-25x^2+4x+20, \quad x=1, x=-1, \quad x=2, x=-2$

Answer

$f(x)=(x-2)(x-1)(x+1)$
$f(x)=(x-1)(x-2)(x-3)$
$f(x)=(x-3)(x-i)(x+i)$
$f(x)=(x+2)^{3}$
$f(x)=(x+2)(x+4)(x+7)$
$f(x)=(x-4)(x+3)(x+4)$
$f(x)=(x-2)(x-1)(x+1)(x+2)(x+5)$

Exercise $\PageIndex{5}$

Divide by using synthetic division.

$\dfrac{2x^3+3x^2-5x+7}{x-2}$
$\dfrac{4x^3+3x^2-15x+18}{x+3}$
$\dfrac{x^3+4x^2-3x+1}{x+2}$
$\dfrac{x^4+x^3+1}{x-1}$
$\dfrac{x^5+32}{x+2}$
$\dfrac{x^3+5x^2-3x-10}{x+5}$

Answer

$2 x^{2}+7 x+9+\dfrac{25}{x-2}$
$4 x^{2}-9 x+12-\dfrac{18}{x+3}$
$x^{2}+2 x-7+\dfrac{15}{x+2}$
$x^{3}+2 x^{2}+2 x+2+\dfrac{3}{x-1}$
$x^{4}-2 x^{3}+4 x^{2}-8 x+16$
$x^{2}-3+\dfrac{5}{x+5}$

Exercise 8.4.1\PageIndex{1}

Exercise 8.4.2\PageIndex{2}

Exercise 8.4.3\PageIndex{3}

Exercise 8.4.4\PageIndex{4}

Exercise 8.4.5\PageIndex{5}

Support Center

How can we help?

Exercise $\PageIndex{1}$

Exercise $\PageIndex{2}$

Exercise $\PageIndex{3}$

Exercise $\PageIndex{4}$

Exercise $\PageIndex{5}$