2.4 Algebravaganza (Chapter 2 Study Guide)
- Page ID
- 152935
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In this section, you will:
- bring together all of the techniques we have learned from Chapter 1 and Chapter 2 to perform multi-step simplifications of algebraic expressions, especially expressions commonly seen in Calculus.
This section is a big ol' algebra skills extravaganza (algebravaganza, if you will) that is also going to serve as the Chapter 2 study guide, because it is synthesizing everything we've just learned into frankenproblems. Your aim at this point is to become so fluent in algebra skills that you can FOIL and factor in your sleep, and your future professors call me in a few months and say, "Great work on these students, De Wolf! Let me send you an edible arrangement!" To which I'll say, "Aw shucks, it's all the students' hard work and dedication, plus also I don't like cantaloupe." First, a quick highlights reel from this chapter. You will also need to remember all the rules about exponents and radicals from Chapter 1, and general arithmetic with numbers of course.
- A polynomial in variable \(x\) is an expression that looks like \(a_nx^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0\), having real number coefficients and positive integer powers of \(x\). The degree is the highest power \(n\). The leading coefficient is \(a_n\). The constant term is \(a_0\).
- You can add and subtract polynomials by combining like terms, i.e. terms that have the same exact variables to the same exact powers in them.
- You can distribute multiplication through a sum by the distributive property of addition: \( a(b+c) = ab+ac \). This allows you to multiply polynomials by expanding, or FOILing: \( (a+b)(c+d) = ac + ad + bc + bd \).
- Un-FOILing or un-distributing is factoring. We covered several techniques.
- Factoring \(x^2 + bx + c\) by finding two numbers that multiply to give \(c\) and add to give \(b\).
- Factoring \(ax^2 + bx + c\) using the \( ac\) method.
- Factoring by grouping.
- Completing the square.
- Special common situations like "difference of squares" \( A^2 - B^2 = (A+B)(A-B) \) or "perfect squares" \(A^2 + 2AB + B^2 = (A+B)^2\).
- A rational expression is a fraction with a polynomial in the numerator and a polynomial in the denominator. Because of the division, a rational expression could have a domain issue; i.e. there might be restrictions on what can be plugged in for the variable. Talking about domain is just saying what values are allowable for \(x\), and there are two things to watch out for so far:
- Any number(s) that, when plugged in for \(x\), would cause division by zero.
- Any number(s) that, when plugged in for \(x\), would cause us to take the square root of a negative number.
- Multiplying rational expressions follows fraction multiplication rules, it just requires FOILing and factoring sometimes. Division is done by multiplying by the reciprocal.
- Adding and subtracting rational expressions requires a common denominator, which means some factoring and multiplications, but essentially follows the addition and subtractions rules for basic fractions.
- Rationalizing is when you see an expression like \(A + B\sqrt{C}\) and multiply it by its conjugate, \(A - B \sqrt{C}\). If you see \(A - B \sqrt{C}\) instead, the conjugate is \(A + B\sqrt{C}\). The point of this technique is to make radicals go away.
Be on your guard against these common errors!
| YES | NO!!!!! |
| \( \textcolor{green}{ A - (B+C) = A - B - C } \) | \( \textcolor{red}{ A - (B+C) = A - B + C }\) |
| \( \textcolor{green}{(AB)^2 = A^2 B^2} \) | \( \textcolor{red}{ (A+B)^2 \neq A^2 + B^2 }\) |
| \( \textcolor{green}{\dfrac{A+B}{C} = \dfrac{A}{C} + \dfrac{B}{C} } \) | \( \textcolor{red}{\dfrac{A}{B+C} \neq \dfrac{A}{B} + \dfrac{A}{C}} \) |
| \(\textcolor{green}{ \sqrt{AB} = \sqrt{A}\sqrt{B}} \) | \( \textcolor{red}{\sqrt{A+B} \neq \sqrt{A} + \sqrt{B} } \) |
| \( \textcolor{green}{\dfrac{AB}{AC} = \dfrac{A}{B} } \) | \( \textcolor{red}{ \dfrac{A+B}{A} \neq B }\) |
| \( \textcolor{green}{ \sqrt{a^2b^2} = \sqrt{(ab)^2} = ab, \quad \text{a,b \geq 0} } \) | \( \textcolor{red}{ \sqrt{a^2 + b^2} \neq a + b } \) |
I recommend that you go through each line and say out loud to yourself in caveman-speak what is going on. For example, the first line is
Now let's get into practice problems. I'm taking these from actual literal expressions that arose in my past courses that students needed to simplify as part of doing calculus or applied math problems. This way, when you get to those later classes, you'll be like, "In the immortal words of Dr. Taylor Swift...I think I've seen this film before." And everyone will look impressed!
Drainage culverts are made from large cylindrical shells like the one shown below, cast in concrete. If \(h\) represents the height of the cylinder, \(r\) represents the radius from the center to the inside of the shell, and \(R\) represents the outer radius, then you can calculate the volume of concrete needed using the formula \( \pi R^2 h - \pi r^2 h \). Factor this expression.
Another way to compute the volume of concrete needed is to use the formula \( 2\pi \cdot \dfrac{(R+r)}{2} \cdot h \cdot (R-r) \). Expand and simplify this expression to confirm that the results are equivalent!
Solution
For the first formula, observe that both terms have the factors \(\pi\) and \(h\) in common, so pull those out to get \(\pi h(R^2 - r^2)\). Then notice that the parentheses contain a difference of squares! So the final answer is \( \pi h (R+r)(R-r) \).
Next, we need to simplify the second option. We first see that there is a multiplication by 2 out front and later a division by 2, so we can cancel those. Then we just FOIL (or remember the difference of squares identity) and simplify.
\[ 2\pi \cdot \dfrac{(R+r)}{2} \cdot h \cdot (R-r) = \pi h (R+r)(R-r) \notag \]
This matches the factored form we found for the first expression.
If an expression is in factored form, expand it completely. If it is in expanded form, factor it completely.
- \( x(20-2x)(10-2x) \)
- \( (x - 5)^2 + 3 \)
- \( 1600 + 40x - 2x^2 \)
- \( 2x^2 + 4xh \)
- Answer
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- \( 4x^3 - 60x^2 + 200x \). This expression calculates the volume of an open box created by cutting the corners from a \(20 \times 10\) rectangle of cardboard, and then folding up the sides.
- \( x^2 - 10x + 28\). This expression appears in an equation whose graph is a parabola, in what is called vertex form. Expanding takes the equation from vertex form to what is called standard form.
- \( -2 (x+20)(x-40) \). This expression could appear in the equation of a downward-opening parabola and could be useful for determining where it crosses the \(x\)-axis.
- \( 2x(x+2h) \). This expression calculates the surface area of a closed box with square \( x \times x \) base and height \(h\).
- FOIL and simplify: \( (3x+\sqrt{x})(3x - \sqrt{x}) \)
- Simplify: \( \dfrac{x^2 y^{-1}}{x^{-5}} \)
- Write as \(x\) to a single power: \( \sqrt[3]{x\sqrt{x}} \)
Solution
- FOIL or notice that this is a difference of squares! \( (3x)^2 - (\sqrt{x})^2 = 9x^2 - x \).
- We get rid of negative powers by moving from numerator to denominator or vice versa as needed, then use the power rules. Remember "multiplication in the base translates to addition in the exponent." We get \( \dfrac{x^2 y^{-1}}{x^{-5}} = \dfrac{x^2x^5}{y} = \dfrac{x^7}{y} \).
- This is all about writing the cube root and square root as fractional powers and being very careful with power rules. Remember that you can bring an exponent inside a product, and "power to a power translates to multiplication in the exponent," and "multiplication in the base translates to addition in the exponent." Use parentheses!
\[ \sqrt[3]{x\sqrt{x}} = (x\sqrt{x})^{\frac{1}{3}} = (x(x)^{\frac{1}{2}})^{\frac{1}{3}} = x^{\frac{1}{3}} (x^{\frac{1}{2}})^{\frac{1}{3}} = x^{\frac{1}{3}} x^{\frac{1}{6}} = x^{\frac{1}{3} + \frac{1}{6}} = x^{\frac{1}{2}} \notag \]
Another valid path is
\[ \sqrt[3]{x\sqrt{x}} = (x\sqrt{x})^{\frac{1}{3}} = (x(x)^{\frac{1}{2}})^{\frac{1}{3}} = ( x^{1+\frac{1}{2}})^{\frac{1}{3}} = (x^{\frac{3}{2}})^{\frac{1}{3}} = x^{\frac{1}{2}} \notag \]
Simplify.
- \( \dfrac{4x^2 y^{-3}}{(2x)^3 y} \)
- \( (10 + \sqrt{9x})^2 \)
- \( x\sqrt{x} + \left( \dfrac{x}{\sqrt[4]{x}} \right)^2 \)
- Answer
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- \( \dfrac{1}{2xy^4} \)
- \( 100 + 60\sqrt{x} + 9x \)
- \( 2x^{\frac{3}{2}} \)
Simplify the rational expression \( \dfrac{ 2x^2 - 2x - 4}{x^2 + 2x - 8} \).
Solution
We need to factor top and bottom completely and cancel any common factors. We will need the \(ac\) method (or guess and check) to handle the numerator since it's in the form \(ax^2 + bx + c\). Let's do that scratch work first.
The product \(ac\) is \(-8\). We need two numbers that multiply to \(-8\) and add to \(-2\). These would be \(-4\) and \(2\), so we split the middle term using that fact and then factor by grouping:
\[ 2x^2 +(- 2)x - 4 = 2x^2 + (-4 + 2)x - 4 = 2x^2 - 4x + 2x - 4 = 2x(x - 2) + 2(x - 2) = (2x+2)(x-2) \notag \]
Are we done? No! That was just the numerator, lol. The denominator factors as \( (x+4)(x-2) \). So we have
\[ \dfrac{(2x+2)(x-2)}{(x+4)(x-2)} = \dfrac{2x+2}{x+4} \notag \]
Okay, psych! I just wanted to make you practice the \(ac\) method. I could have noticed from the beginning that every term in the numerator has a factor of \(2\) in it, and pulled that out first. Then I would have done this:
\[ \dfrac{ 2(x^2 - x - 2)}{x^2 + 2x - 8} = \dfrac{ 2(x-2)(x+1)}{(x+4)(x-2)}= \dfrac{ 2(x+1)}{(x+4)} \notag \]
Simplify the rational expression \( \dfrac{3x^2-3x}{x^2-2x+1} \).
- Answer
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\( \dfrac{3x}{x-1} \)
The following expression comes from the equation of an ellipse in standard form. Perform the addition, expand the products, and simplify if possible.
\[ \dfrac{ (x-2)^2}{2^2} + \dfrac{ (y-1)^2}{3^2} \notag \]
Solution
The least common denominator for the two fractions is \( 9 \cdot 4 = 36 \). We multiply top and bottom by what is needed for each fraction and add:
\[ \dfrac{ (x-2)^2\cdot 9}{4 \cdot 9} + \dfrac{ (y-1)^2\cdot 4}{9 \cdot 4} = \dfrac{ 9(x-2)^2 + 4(y-1)^2}{36} \notag \]
Then it's just a matter of FOILing and simplifying to get \( \dfrac{9x^2 - 36 x + 4y^2 - 8y + 40}{36} \). You could also split the fraction over the pluses and minuses and simplify like so: \( \frac{1}{4}x^2 - x + \frac{1}{9} y^2 - \frac{2}{9} y + \frac{10}{9} \notag \).
The following expression comes from the equation of a hyperbola in standard form. Perform the subtraction, expand the products, and simplify if possible.
\[ \dfrac{ (x+1)^2}{1^2} - \dfrac{ (y-1)^2}{2^2} \notag \]
- Answer
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\( \dfrac{4x^2 + 8x - y^2 + 2y +3}{4} \) or \( x^2 + 2x - \frac{1}{4} y^2 + \frac{1}{2}y + \frac{3}{4} \).
Complete the square to write \( x^2 + 6x + 3\) in the form \( (x+a)^2 + b\).
Solution
This is part of the process needed to find a useful form of an equation whose graph is a circle. We write \( x^2 + 6x + \quad \quad + 3 \), leaving a space. Then we take the coefficient from the \(x\) term, divide it by 2, and square the result. That is, \(6 \overset{\div 2}{\longrightarrow} 3 \overset{\text{squared}}{\longrightarrow} 9 \). Add this in the blank, and subtract it from the end of the expression so that we have effectively not changed the expression. Then factor the perfect square we created.
\[ (x^2 + 6x + 9) + 3 - 9 = (x+3)^2 - 6 \notag \]
Complete the square to write \( y^2 - 2y + 3 \) in the form \( (y-a)^2 + b\).
- Answer
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\( (y-1)^2 +2 \)
Practice clearing denominators and rationalizing the numerator by simplifying the difference quotient \( \dfrac{ \dfrac{1}{\sqrt{x+h}} - \dfrac{1}{\sqrt{x}}}{h} \).
Solution
These difference quotients appear when you study the limit definition of a derivative in Calc I. The division by \(h\) is a problem child because we want to be see what happens if \(h\) becomes 0, so the goal is generally to get him to go away. We start by clearing the denominators.
\[ \dfrac{ \left( \dfrac{1}{\sqrt{x+h}} - \dfrac{1}{\sqrt{x}} \right) }{h} \dfrac{\cdot \sqrt{x}\sqrt{x+h}}{\cdot \sqrt{x}\sqrt{x+h}} =\dfrac{ \dfrac{\sqrt{x}\sqrt{x+h}}{\sqrt{x+h}} - \dfrac{\sqrt{x}\sqrt{x+h}}{\sqrt{x}}}{h\sqrt{x}\sqrt{x+h}} = \dfrac{ \sqrt{x} - \sqrt{x+h}}{h\sqrt{x}\sqrt{x+h}} \notag \]
Now, I told you the goal is to get rid of the \(h\) factor in the denominator, so let's try rationalizing the numerator and see if that helps. We multiply top and bottom by the conjugate \( \sqrt{x} + \sqrt{x+h} \).
\[\dfrac{ (\sqrt{x} - \sqrt{x+h})}{h\sqrt{x}\sqrt{x+h}}\dfrac{ \cdot (\sqrt{x} + \sqrt{x+h})}{\cdot(\sqrt{x} + \sqrt{x+h})} = \dfrac{ x - (x+h)}{h\sqrt{x}\sqrt{x+h}(\sqrt{x} + \sqrt{x+h})} = \dfrac{ -h}{h\sqrt{x}\sqrt{x+h}(\sqrt{x} + \sqrt{x+h})} \notag \]
Now I can cancel the \(h\)'s! You can distribute in the denominator and simplify as well if you want.
\[ -\dfrac{ 1}{\sqrt{x}\sqrt{x+h}(\sqrt{x} + \sqrt{x+h})} = -\dfrac{ 1}{x \sqrt{x+h} + (x+h) \sqrt{x} } \notag \]
I did that one with you because it's lowkey a challenge problem. The one below is more chill.
Practice clearing denominators to simplify the difference quotient \( \dfrac{ \dfrac{1}{(x+h)^2} - \dfrac{1}{x^2}}{h} \).
- Answer
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\( - \dfrac{2x+h}{x^2(x+h)^2} \).
I did some simplifications but I think I goofed up somewhere. Identify the error in my work. No need to find the correct answer, just find the mistake and I'll fix it later.
1. \( \dfrac{ (x+2)^2 - (x+1)(x+2)}{3x-2} = \dfrac{ x^2 + 4 - (x^2 + 3x +2)}{3x-2} = \dfrac{ x^2 + 4 - x^2 - 3x - 2}{3x-2} = \dfrac{ -3x + 2}{3x - 2} = -1 \)
2. \( (x+1)(x-1) - (3x-2)(x+1) = x^2 - 1 - (3x^2 + x - 2) = x^2 - 1 - 3x^2 + x - 2 = -2x^2 + x - 3 \)
- Answer
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1. I made the error right out of the gate with saying that \( (x+2)^2 = x^2 + 4^2 \). That's illegal! It's \( (x+2)(x+2) = x^2 + 4x + 4 \).
2. Everything looks okay until I cross over the second equals sign. I didn't apply the subtraction to all terms in the parentheses! It should be \(x^2 - 1 - 3x^2 \textcolor{red}{-} x \textcolor{red}{+} 2 \).