2.2E Exercises
- Page ID
- 152900
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Factor out the greatest common factor from all terms.
- \( 2x^2 + 4x + 16 \)
- \( 12x^3 + 36x^2 + 24x \)
- \( m^2n + mn^2 + mn \)
- \( 5a + 25 ab + 50a^2 \)
- Answer
-
- \( 2(x^2 + 2x + 8 )\)
- \( 12x( x^2 + 3x + 2) \)
- \( mn(m+n+1) \)
- \( 5a( 1 + 5b + 10a) \)
Factor.
- \( x^2 + 8x + 15\)
- \( x^2 - 5x - 24 \)
- \( b^2 + 12 b + 35 \)
- \( x^2 + 96x - 400 \)
- \( x^4 - 5x^2 + 6 \)
- \( y^6 - 3y^3 +2 \)
- Answer
-
Hint: When a math problem just says "Factor," it means your final answer should be in a form like (____)(____). You shouldn't just factor something out of a few terms and leave other terms added/subtracted. Go try them and come back.
- \( (x+3)(x+5) \)
- \( (x-8)(x+3) \)
- \( (b + 7)(b+5) \)
- \( (x - 4)(x+100) \)
- Hint: nickname trick... \( (x^2 - 3)(x^2 - 2) \)
- \( (y^3 - 1)(y^3 - 2) \)
Factor.
- \( 2x^2 + 7x - 15 \)
- \( 3x^2 - 5x - 2 \)
- \( 4x^2 - 4x - 3 \)
- \( 3 + 13 x + 4 x^2 \)
- Answer
-
- \( (x+5)(2x-3) \)
- \( (x-2)(3x+1) \) Hint: if you're stuck at the grouping stage, realize that \( ... + x -2 \) can be written as \(... +1(x-2) \).
- \( (2x-3)(2x+1) \)
- \( (4x + 1)(x+3) \)
Factor.
- \( 3x^3 - 6x^2 + 5x - 10 \)
- \( 2x^3 + 3x^2 - 4x - 6 \)
- \( x^4 + 2x^3 - 3x^2 - 6x \)
- \( 2x^3 + 3x^2 + 2x + 3 \)
- Answer
-
- \( (x-2)(3x^2+5) \)
- \((2x+3)(x^2-2) \)
- \( (x+2)(x^3-3x) \) or \( x(x+2)(x^2-3)\)
- \( (x^2 + 1)(2x + 3)\)
Complete the square to write in the form \( (x-h)^2 + k \), where \(h,k\) are constants.
- \( x^2 - 6x - 1 \)
- \( x^2 - 10x + 26 \)
- \( x^2 + 2x + 4 \)
- \( x^2 + 4x + 8\)
- Answer
-
- \( (x-3)^2 - 10 \)
- \( (x-5)^2 + 1\)
- \( (x-(-1))^2 + 3\) (The factors look like \( (x+1)\), but we wanted a subtraction, so we write \(x + 1 = x - (-1)\) with "minus a minus.")
- \( (x+2)^2 + 4 \) or \( (x-(-2))^2 + 4 \)