2.4E Exercises
- Page ID
- 152945
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)For each expression, identify whether it is a polynomial, and if so, tell its degree and leading coefficient, and factor it completely.
- \( x^2 + 8x - 20 \)
- \( 6x - 2\sqrt{x} + 4 \)
- \( x^3 + 3^x \)
- \( 2x^2 - 5x - 7 \)
- \( -x^4 + 16\)
- \( 4x^4 - 9 \)
- Answer
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- Yes, degree 2 and leading coefficient 1, and factored form is \( (x+10)(x-2) \).
- No, the \( \sqrt{x}\) term is not an \(x\) to a positive integer power.
- No, the \(3^x\) is the problem.
- Yes, degree 2 and leading coefficient 2, and factored form is \( (2x -7 )(x + 1) \).
- Yes, degree 4 and leading coefficient \(-1\), and seeing this as a difference of squares, \( 4^2 - (x^2)^2 = (4+x^2)(4-x^2) = (4+x^2)(2-x)(2+x) \).
- Yes, degree 4 and leading coefficient \(4\), and \( (2x^2 + 3)(2x^2 - 3) \).
Expand and simplify.
- \( (2-7y)(y+3) - 6 \)
- \( a(a + 1) + (a+2)(a-5) + 11 \)
- \( (x + 3)(x+5) - (x -1)^2 \)
- \( (x^4 - x^3) + x^2(x+1)(x-1) \)
- \( (a + b)^4 \)
- \( x^3(x^2 + 3x) - 2x(x^4 - 3x^2) \)
- Answer
-
- \( -7y^2 - 19y \)
- \( 2a^2 - 2a + 1\)
- \( 10x + 14\)
- \( 2x^4 - x^3 - x^2 \)
- \( a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 \)
- \( -x^5 + 3x^4 + 6x^3 \)
Factor completely. Remember the technique of factoring by grouping, and the \(ac\) method!
- \( x^2 + 130x + 3000 \)
- \( 3m^3 + 6m^2 + 3m \)
- \( x^3 + 2x^2 - x-2 \)
- \( (a+b)^2 + 2(a+b) + 1\)
- \( 5x^2 - 7x - 6 \)
- Answer
-
- \( (x+30)(x+100) \)
- \( 3m(m+1)^2 \)
- \( (x-1)(x+1)(x+2) \)
- \( ((a+b)+1)^2 = (a+b+1)^2 \)
- \( (5x+3)(x-2) \)
Complete the square to write the degree 2 polynomials in the form \( (x + a)^2 + b\).
- \( x^2 + 4x - 10 \)
- \( x^2 - 3x + \frac{5}{2} \)
- Answer
-
- \( (x+2)^2 - 14 \)
- \( \left( x- \frac{3}{2} \right)^2 + \frac{1}{4} \)
Simplify.
1. \( \dfrac{ x^2 + 6x}{x^2 + x - 30} \)
2. \( \dfrac{2x^2 - 4x - 6}{x^2 - 1} \)
3. \( \dfrac{x^2 + 8x - 20}{x^2 + 20x + 100} \)
4. \( \dfrac{ (x+h)^2 + 3(x+h) - x^2 - 3x }{h} \)
- Answer
-
1. \( \dfrac{x}{x-5} \)
2. \( \dfrac{ 2(x-3)}{(x-1)} \)
3. \( \dfrac{x-2}{x+10} \)
4. \( 2x + h + 3 \)
Simplify into a single rational expression.
1. \( \dfrac{h}{h+1} + \dfrac{h^2}{h-1} \)
2. \( \dfrac{2}{x} - \dfrac{2}{x+1} + \dfrac{2}{x^2 +x} \)
3. \( \dfrac{3x +6}{16 - x^4} \cdot \dfrac{8x^2 + 32}{9} \)
- Answer
-
1. \( \dfrac{h^3 + 2h^2 - h}{h^2 - 1} \)
2. \( \dfrac{4}{x(x+1)} \) or \( \dfrac{4}{x^2 + x} \)
3. \(\dfrac{8}{3(2-x)}\)
Simplify.
1. \( \dfrac{ \dfrac{x+1}{x^2 + 5x + 6}} { \dfrac{2x^2 - 2}{x+3} } \)
2. \( \dfrac{ \dfrac{1}{x+h} - \dfrac{1}{x}}{h} \)
- Answer
-
1. \( \dfrac{1}{2(x-1)(x+2)} \) (Hint: multiplying by the reciprocal works best here.)
2. \( -\dfrac{1}{x(x+h)} \) (Hint: clearing the denominators is quick.)
Simplify.
1. \( \dfrac{ (x+h)^3 - x^3}{h} \)
2. \( \dfrac{ (3x^2 + 1) - 28}{x-3} \)
3. \( \dfrac{ \frac{1}{2}(x+h)^2 - \frac{1}{2}x^2 }{h} \)
- Answer
-
Hint: What does "simplify" mean for this type of problem? Remember that in "difference quotient" problems, the goal is to make the denominator go away. This means first simplifying the numerator, which may involve expanding and/or combining like terms, and then re-factoring to see if anything cancels.
- \( 3x^2 + 3xh + h^2 \)
- \( 3(x+3) \)
- \( x + \frac{1}{2}h \)