8.4: Exercises
- Page ID
- 48999
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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
-
- \(x^{2}-2 x-2-\dfrac{3}{x-2}\)
- \(x^{2}+3 x-2+\dfrac{4}{x+3}\)
- \(x+6-\dfrac{10}{x+1}\)
- \(x^{2}+x+\dfrac{5}{x+2}\)
- \(2 x^{2}+3 x+6+\dfrac{11}{x-1}\)
- \(2 x^{3}-3 x^{2}+15 x-74+\dfrac{373}{x+5}\)
- \(2 x^{3}+8 x^{2}+x+4+\dfrac{3}{x-4}\)
- \(x^{2}-3 x+9\)
- \(x^{3}+2 x^{2}+x+2+\dfrac{2}{3 x+1}\)
- \(4 x^{2}+3 x+6\)
- \(x+1-\dfrac{7 x+6}{x^{2}+2 x+1}\)
- \(x^{3}+3 x^{2}-3 x-9+\dfrac{9 x+7}{x^{2}+3}\)
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 \(r = 15\)
- \(r = 20\)
- \(r = -2\)
- \(r = 12\)
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 \(x = −1\)
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
-
- \(f(x)=(x-2)(x-1)(x+1)\)
- \(f(x)=(x-1)(x-2)(x-3)\)
- \(f(x)=(x-3)(x-i)(x+i)\)
- \(f(x)=(x+2)^{3}\)
- \(f(x)=(x+2)(x+4)(x+7)\)
- \(f(x)=(x-4)(x+3)(x+4)\)
- \(f(x)=(x-2)(x-1)(x+1)(x+2)(x+5)\)
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
-
- \(2 x^{2}+7 x+9+\dfrac{25}{x-2}\)
- \(4 x^{2}-9 x+12-\dfrac{18}{x+3}\)
- \(x^{2}+2 x-7+\dfrac{15}{x+2}\)
- \(x^{3}+2 x^{2}+2 x+2+\dfrac{3}{x-1}\)
- \(x^{4}-2 x^{3}+4 x^{2}-8 x+16\)
- \(x^{2}-3+\dfrac{5}{x+5}\)