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Intermediate Algebra: Functions and Graphs

Katherine Yoshiwara

Section A.8 Chapter 8 Polynomial and Rational Functions

Subsection A.8.1 Polynomial Functions

Subsubsection A.8.1.1 Compute sums and products

Compare the rules for simplilfying products to the rules for simplifying sums.
Example A.8.1.
Simplify each expression if possible.
  1. \(\displaystyle 3x^2-5x^3\)
  2. \(\displaystyle 3x^2(-5x^3)\)
Solution.
  1. This expression is a difference of terms, but they are not like terms (because the variable has different exponents), so we cannot combine them.
  2. This expression is a product, and the powers have the same base, so we can apply the first law of exponents to get \(~3x^2(-5x^3)=-15x^5\text{.}\)
Example A.8.2.
Simplify each expression if possible.
  1. \(\displaystyle -6t^4-8t^4\)
  2. \(\displaystyle -6t^4(-8t^4)\)
Solution.
  1. This expression is a difference of like terms, so we can combine their coefficients to get \(~-6t^4-8t^4=-14t^4\)
  2. This expression is a product, and the powers have the same base, so we can apply the first law of exponents to get \(~-6t^4(-8t^4)=48t^8\)
Checkpoint A.8.3.
Simplify each expression if possible.
  1. \(\displaystyle 2a^2-9a^3+a^2\)
  2. \(\displaystyle 2a^2(-9a^3+a^2)\)
Answer.
  1. \(\displaystyle 3a^2-9a^3\)
  2. \(\displaystyle -18a^5+2a^4\)
Checkpoint A.8.4.
Simplify each expression if possible.
  1. \(\displaystyle 7-4a^3+2a^3\)
  2. \(\displaystyle 7-4a^2(2a^3)\)
Answer.
  1. \(\displaystyle 7-2a^3\)
  2. \(\displaystyle 7-8a^5\)

Subsubsection A.8.1.2 Use formulas

There are several useful formulas for simplifying polynomials.
Example A.8.5.
If \(~a=5t^4,~\) find \(a^3\) and \(3a^2\text{.}\)
Solution.
We substitute \(5t^4\) for \(a\) to find
\begin{align*} a^3 \amp =(\blert{5t^4})^3 = 5^3(t^4)^3 = 125t^{12} \amp \amp \blert{\text{Apply the third law of exponents.}}\\ 3a^2 \amp = 3(\blert{5t^4})^2 = 3 \cdot 5^2(t^4)^2 = 75t^8 \end{align*}
Example A.8.6.
If \(a=2y\) and \(b=-3z^2\text{,}\) find \(b^3\) and \(3a^2b\text{.}\)
Solution.
We substitute \(2y\) for \(a\) and \(-3z^2\) for \(b\) to find
\begin{align*} b^3 \amp =(\blert{-3z^2})^3 = (-3)^3(z^2)^3 = -27z^6 \\ 3a^2b \amp = 3 (\blert{2y})^2(\blert{-3z^2}) = 3(4y^2)(-3z^2)=-36y^2z^2 \end{align*}
Checkpoint A.8.7.
If \(~a=-4x^3~\) and \(~b=3h\text{,}\) find \(a^3\) and \(ab^2\text{.}\)
Answer.
\(-64x^9;~~-36x^3h^2\)
Checkpoint A.8.8.
If \(x=6p^2\) and \(y=mq^2\text{,}\) find \(y^3\) and \(x^2y\text{.}\)
Answer.
\(m^3q^6;~~-36mp^4q^2\)

Subsubsection A.8.1.3 Square binomials

Sometimes it is easier to use formulas to square binomials.
Special Products of Binomials.
\begin{align*} \amp(a + b)^2 = (a + b) (a + b) = a^2 + 2ab + b^2\\ \amp(a - b)^2 = (a - b) (a - b) = a^2 - 2ab + b^2\\ \amp(a + b) (a - b)= a^2 -b^2 \end{align*}
Example A.8.9.
Use the identity \(~(a+b)^2=a^2+2ab+b^2~\) to expand \(~(3h^2+4k^3)^2\text{.}\)
Solution.
We substitute \(3h^2\) for \(a\) and \(4k^3\) for \(b\) into the identity.
\begin{align*} (3h^2+4k^3)^2 \amp =(3h^2)^2 +2(3h^2)(4k^3) +(4k^3)^2\\ \amp = 9h^4+24h^2k^3+16k^6 \end{align*}
Example A.8.10.
Use the identity \(~(a-b)^2=a^2-2ab+b^2~\) to expand \(~(2xy^2-5)^2\text{.}\)
Solution.
We substitute \(2xy^2\) for \(a\) and \(5\) for \(b\) into the identity.
\begin{align*} (2xy^2-5)^2 \amp =(2xy^2)^2 -2(2xy^2)(5) +5^2\\ \amp = 4x^2y^4 - 20xy^2+25 \end{align*}
Checkpoint A.8.11.
Expand \(~(8w^4-3w^3)^2\)
Answer.
\(64w^8-48w^7+9w^6\)
Checkpoint A.8.12.
Expand \(~(a^3b+9ab^3)^2\)
Answer.
\(a^6b^2+18a^4b^4+81a^2b^6\)

Subsection A.8.2 Algebraic Fractions

Subsubsection A.8.2.1 Factor a polynomial

To reduce an algebraic fraction, we must factor its numerator and denominator.
Example A.8.13.
Factor.
  1. \(\displaystyle 4x^2-4x\)
  2. \(\displaystyle 4x^2-1\)
  3. \(\displaystyle 4x^2-4x+2\)
Solution.
  1. We factor out a common factor of \(4x\) to get \(4x(x-1)\text{.}\)
  2. This is a difference of two squares that factors as \((2x-1)(2x+1)\text{.}\)
  3. This is the square of a binomial, \((2x-1)^2\text{.}\)
Example A.8.14.
Factor.
  1. \(\displaystyle 27a^2-3\)
  2. \(\displaystyle 27a^3-1\)
  3. \(\displaystyle 81a^3-a\)
Solution.
  1. We first factor out 3 to find \(3(9a^2-1)\text{,}\) then factor the difference of two squares to get \(3(3a-1)(3a+1)\text{.}\)
  2. This is a differece of two cubes, which factors as \((3a-1)(9a^2+3a+1)\text{.}\)
  3. We first factor out \(a\) to get \(a(81a^2-1)\text{,}\) then factor the difference of two squares to get \(a(9a-1)(9a+1)\) .
Checkpoint A.8.15.
Factor completely \(2x^3+16y^3\)
Answer.
\(2(x+2y)(x^2-2xy+4y^2)\)
Checkpoint A.8.16.
Factor completely \(4x^2y-36y^3\)
Answer.
\(4y(x-3y)(x+3y)\)
Checkpoint A.8.17.
Factor completely \(2b^3-6b^2-36b\)
Answer.
\(2b(b-6)(b+3)\)
Checkpoint A.8.18.
Factor completely \(9b^4+9b^2\)
Answer.
\(9b^2(b^2+1)\)

Subsubsection A.8.2.2 Find the opposite of a binomial

To find the opposite or negative of a binomial we multiply by \(-1\text{.}\)
Example A.8.19.
Which of these is the opposite of \(m^2-p~\text{?}\)
  1. \(\displaystyle m^2+p\)
  2. \(\displaystyle m-p^2\)
  3. \(\displaystyle p-m^2\)
Solution.
The opposite of \(m^2-p\) is \(-(m^2-p) = -m^2+p\text{,}\) or \(p-m^2\text{.}\)
Example A.8.20.
Which of these pairs of binomials are opposites?
  1. \(3c-5\) and \(5+3c\)
  2. \(5-3c\) and \(3-5c\)
  3. \(5c-3\) and \(3-5c\)
Solution.
The opposite of \(5c-3\) is \(-(5c-3)=-5c+3\text{,}\) or \(3-5c\text{,}\) so (c) is correct.
Checkpoint A.8.21.
Find the opposite of the binomial \(~2x+1\)
Answer.
\(-2x-1\)
Checkpoint A.8.22.
Find the opposite of the binomial \(~b^2-b\)
Answer.
\(b-b^2\)
Checkpoint A.8.23.
Find the opposite of the binomial \(~-4n+8\)
Answer.
\(4n-8\)
Checkpoint A.8.24.
Find the opposite of the binomial \(~-3z^2-2\)
Answer.
\(3z^2+2\)

Subsubsection A.8.2.3 Asymptotes

The asymptotes of rational functions are horizontal and vertical lines.
Example A.8.25.
Give the equation and slope of the line.
vertical line
Solution.
This line is a vertical line. All the points on the line have \(x\)-coordinate \(-15\text{,}\) so the equation of the line is \(x=-15\text{.}\) Because \(\Delta x=0\) between any two points on the line, its slope is undefined.
Example A.8.26.
Give the equation and slope of the line.
horizontal line
Solution.
This line is a horizontal line. All the points on the line have \(y\)-coordinate \(04\text{,}\) so the equation of the line is \(y=40\text{.}\) Because \(\Delta y=0\) between any two points on the line, its slope is \(0\text{.}\)
Checkpoint A.8.27.
Give the equation and slope of the line.
horizontal line
Answer.
\(y=-16;~ m=0\)
Checkpoint A.8.28.
Give the equation and slope of the line.
vertical line
Answer.
\(x=30\text{;}\) \(~ m~\)is undefined
Checkpoint A.8.29.
Give the equation and slope of the line.
horizontal line
Answer.
\(y=45; m=0\)
Checkpoint A.8.30.
Give the equation and slope of the line.
vertical line
Answer.
\(x=-24\text{;}\) \(~ m~\)is undefined

Subsection A.8.3 Operations on Algebraic Fractions

Subsubsection A.8.3.1 Multiply fractions

To multiply two fractions together, we multiply their numerators together, and multiply their denominators together. We can divide out any common factors in numerator and denominator before we multiply.
Example A.8.31.
Multiply \(~\dfrac{1}{2}\left(\dfrac{P}{Q}\right)\)
Solution.
\(\dfrac{1}{2}\left(\dfrac{P}{Q}\right) = \dfrac{1 \cdot P}{2 \cdot Q} = \dfrac{P}{2Q}\)
Example A.8.32.
Multiply \(~4 \left(\dfrac{b}{c}\right)\)
Solution.
\(4 \left(\dfrac{b}{c}\right) = \dfrac{4}{1} \left(\dfrac{b}{c}\right) = \dfrac{4 \cdot b}{1 \cdot c} = \dfrac{4b}{c}\)
Checkpoint A.8.33.
Multiply \(~\dfrac{3}{2a}\left(\dfrac{a^2}{6}\right)\)
Answer.
\(\dfrac{a}{9}\)
Checkpoint A.8.34.
Multiply \(~8\left(\dfrac{m}{4x^2}\right)\)
Answer.
\(\dfrac{2m}{x^2}\)

Subsubsection A.8.3.2 Divide fractions

To divide one fraction by another, we first take the reciprocal of the second fraction, then proceed as in multiplication.
Example A.8.35.
Divide \(~\dfrac{t}{w} \div \dfrac{3t^2}{w}\)
Solution.
\(\dfrac{t}{w} \cdot \dfrac{w}{3t^2} = \dfrac{\cancel{t}}{\cancel{w}} \cdot \dfrac{\cancel{w}}{3t \cdot \cancel{t}} = \dfrac{1}{3t}\)
Example A.8.36.
Divide \(~\dfrac{a}{2} \div \dfrac{b}{1-b}\)
Solution.
\(\dfrac{a}{2} \cdot \dfrac{1-b}{b} = \dfrac{a(1-b)}{2 \cdot b} = \dfrac{a-ab}{2b}\)
Checkpoint A.8.37.
Divide \(~\dfrac{3}{4} \div \dfrac{cw^2}{t^2-2}\)
Answer.
\(\dfrac{3t^2-6}{4cw^2}\)
Checkpoint A.8.38.
Divide \(~\dfrac{n}{n-1} \div \dfrac{n^2}{p+1}\)
Answer.
\(\dfrac{p+1}{n^2-n}\)

Subsubsection A.8.3.3 Use factoring

Example A.8.39.
Multiply \(~\dfrac{4y^2-1}{4-y^2} \cdot \dfrac{y^2-2y}{4y+2}\)
Solution.
We factor each numerator and denominator, and look for common factors.
\begin{align*} \amp \dfrac{4y^2-1}{4-y^2} \cdot \dfrac{y^2-2y}{4y+2} \amp \\ \amp\qquad=\dfrac{(2y-1)\cancel{(2y+1)}}{\cancel{(2-y)}(2+y)} \cdot \dfrac{y(-1)\cancel{(y-2)}}{2\cancel{(2y+1)}} \amp \amp \blert{\text{Divide out common factors.}}\\ \amp\qquad = \dfrac{-y(2y-1)}{2(y+2)} \amp \amp \blert{\text{Note:}~~y-2=-(2-y)} \end{align*}
Example A.8.40.
Divide \(~\dfrac{6ab}{2a+b} \div (4a^2b)\)
Solution.
We multiply the first fraction by the reciprocal of the second fraction.
\begin{align*} \dfrac{6ab}{2a+b} \div (4a^2b) \amp = \dfrac{\blert{\cancel{2} \cdot 3}\cancel{ab}}{2a+b} \cdot \dfrac{1}{\blert{\cancel{2} \cdot 2}a \cdot \cancel{ab}} \amp \amp \blert{\text{Divide out common factors.}}\\ \amp = \dfrac{3}{2a(2a+b)} \end{align*}
Checkpoint A.8.41.
Multiply \(~\dfrac{3xy}{4xy-6y^2} \cdot \dfrac{2x-3y}{12x}\)
Answer.
\(\dfrac{1}{8}\)
Checkpoint A.8.42.
Multiply \(~\dfrac{9x^2-25}{2x-2} \cdot \dfrac{x^2-1}{6x-10}\)
Answer.
\(\dfrac{(3x+5)(x+1)}{4}\)
Checkpoint A.8.43.
Divide \(~(x^2-9) \div \dfrac{x^2-6x+9}{3x}\)
Answer.
\(\dfrac{3x(x+3)}{x-3}\)
Checkpoint A.8.44.
Divide \(~\dfrac{x^2-1}{x+3} \div \dfrac{x^2-x-2}{x^2+5x+6}\)
Answer.
\(\dfrac{(x-1)(x+2)}{x-2}\)

Subsubsection A.8.3.4 Find an LCD

The first step in adding unlike fractions is to find the lowest common denominator, or LCD.
Example A.8.45.
Find the LCD for the fractions \(~\dfrac{2}{5} + \dfrac{3}{10}\)
Solution.
We factor each denominator and line up any common factors vertically. We use one factor from each column in the LCD.
\begin{align*} 15 \amp = \blert{3} ~~~\cdot~~~ 5\\ 10 \amp = ~~~~~~~~~~~~\blert{5}~~~\cdot~~~\blert{2} \end{align*}
The LCD is \(3 \cdot 5\cdot 2\text{,}\) or 30.
Example A.8.46.
Find the LCD for the fractions \(~\dfrac{5}{6a} - \dfrac{2}{9a^2}\)
Solution.
We factor each denominator and line up any common factors vertically. We use one factor from each column in the LCD.
\begin{align*} 6a \amp = \blert{2}~~~\cdot~~~ 3~~~\cdot~~~ a\\ 9a^2 \amp = ~~~~~~~~~~~~\blert{3}~~~\cdot~~~\blert{3}~~~\cdot~~~ \blert{a}~~~\cdot~~~ \blert{a} \end{align*}
The LCd is \(2 \cdot 3 \cdot 3 \cdot a \cdot a\text{,}\) or \(18a^2\text{.}\)
Checkpoint A.8.47.
Find the LCD for the fractions.
  1. \(\displaystyle \dfrac{7}{8} - \dfrac{1}{6}\)
  2. \(\displaystyle \dfrac{52}{75} + \dfrac{13}{24}\)
Answer.
  1. \(\displaystyle 24\)
  2. \(\displaystyle 600\)
Checkpoint A.8.48.
Find the LCD for the fractions.
  1. \(\displaystyle \dfrac{3}{4a^2b^2} + \dfrac{7}{10ab^3}\)
  2. \(\displaystyle \dfrac{3}{8t^4} - \dfrac{3}{5t^2}\)
Answer.
  1. \(\displaystyle 20a^2b^3\)
  2. \(\displaystyle 40t^4\)

Subsubsection A.8.3.5 Build fractions

Before we can add unlike fractions, we must build each fraction to an equivalent one with the LCD as denominator.
Example A.8.49.
Write an equivalent fraction with the new denominator:
\begin{equation*} ~\dfrac{3}{s} = \dfrac{?}{2s^2} \end{equation*}
Solution.
We first find the building factor for the fraction: what must we mulitply the old denominator by to get the new denominator? We factor the new denominator to see what factors are missing.
The new denominator is \(2 \cdot s \cdot 3\text{,}\) so we need to multiply the old denominator by \(2s\text{.}\) This is the building factor. We multiply top and bottom of the old fraction by the building factor:
\begin{equation*} \dfrac{3}{s} \cdot \blert{\dfrac{2s}{2s}} = \dfrac{6}{2s^2} \end{equation*}
Example A.8.50.
Write an equivalent fraction with the new denominator:
\begin{equation*} ~\dfrac{2}{t+1} = \dfrac{?}{t^2+t} \end{equation*}
Solution.
The new denominator factors as \(t(t+1)\text{,}\) so the building factor is \(t\text{.}\) We multiply top and bottom of the old fraction by \(t\) to obtain:
\begin{equation*} \dfrac{2}{t+1} \cdot \blert{\dfrac{t}{t}} = \dfrac{2t}{t^2+t} \end{equation*}
Checkpoint A.8.51.
Write an equivalent fraction with the new denominator.
  1. \(\displaystyle \dfrac{1}{n} = \dfrac{?}{n^2-n}\)
  2. \(\displaystyle \dfrac{4}{a-1} = \dfrac{?}{a^2-1}\)
Answer.
  1. \(\displaystyle \dfrac{n-1}{n^2-n}\)
  2. \(\displaystyle \dfrac{4a+4}{a^2-1}\)
Checkpoint A.8.52.
Write an equivalent fraction with the new denominator.
  1. \(\displaystyle \dfrac{b}{b-2} = \dfrac{?}{(b+3)(b-2)}\)
  2. \(\displaystyle \dfrac{x-1}{x^2+2x} = \dfrac{?}{(x^2-x)(x+2)}\)
Answer.
  1. \(\displaystyle \dfrac{b^2+3b}{(b+3)(b-2)}\)
  2. \(\displaystyle \dfrac{x^2-2x+1}{(x^2-x)(x+2)}\)

Subsubsection A.8.3.6 Add or subtract fractions

To add or subtract unlike fractions.
  1. Find the LCD for the fractions.
  2. Build each fraction to an equivalent one with the LCD as its denominator.
  3. Add or subtract the numerators. Keep the same denominator.
Example A.8.53.
Subtract \(~\dfrac{2x}{w} - \dfrac{3x}{w}\)
Solution.
These are like fractions, so we need only combine their numerators.
\begin{equation*} \dfrac{2x}{w} - \dfrac{3x}{w} = \dfrac{2x-3x}{w} = \dfrac{-x}{w} \end{equation*}
Example A.8.54.
Subtract \(~\dfrac{3}{a} - \dfrac{a+2}{a}\)
Solution.
These are like fractions, so we need only combine their numerators. Be careful to subtract both terms of the second numerator.
\begin{equation*} \dfrac{3}{a} - \dfrac{a+2}{a} = \dfrac{3-(a+2)}{a} = \dfrac{1-a}{a} \end{equation*}
Example A.8.55.
Add \(~2 + \dfrac{1}{x}\)
Solution.
We write 2 as a fraction, \(\dfrac{2}{1}\text{,}\) and build it to the LCD, \(x\text{.}\)
\begin{equation*} \dfrac{2}{1} \cdot \alert{\dfrac{x}{x}} + \dfrac{1}{x} = \dfrac{2x}{x} + \dfrac{1}{x} = \dfrac{2x+1}{x} \end{equation*}
Example A.8.56.
Subtract \(~\dfrac{a}{2b} - \dfrac{3}{b^2}\)
Solution.
We build each fraction to the LCD, \(2b^2\text{.}\)
\begin{align*} \dfrac{a}{2b} - \dfrac{3}{b^2} \amp = \dfrac{a}{2b} \cdot \alert{\dfrac{b}{b}} - \dfrac{3}{b^2} \cdot \alert{\dfrac{2}{2}}\\ \amp = \dfrac{ab}{2b^2} - \dfrac{6}{2b^2} = \dfrac{ab-6}{2b^2} \end{align*}
Checkpoint A.8.57.
Subtract \(~\dfrac{4a}{b^2} - \dfrac{c}{b^2}\)
Answer.
\(\dfrac{4a-c}{b^2}\)
Checkpoint A.8.58.
Subtract \(~\dfrac{p+2}{2q} - \dfrac{p-1}{2q}\)
Answer.
\(\dfrac{3}{2q}\)
Checkpoint A.8.59.
Add \(~N + \dfrac{2}{N}\)
Answer.
\(\dfrac{N^2+2}{N}\)
Checkpoint A.8.60.
Add \(~\dfrac{2}{xy} - \dfrac{y}{3x}\)
Answer.
\(\dfrac{6+y^2}{3xy}\)

Subsection A.8.4 More Operations on Fractions

Subsubsection A.8.4.1 Use negative exponents

Recall that a negative exponent indicates a reciprocal.
Example A.8.61.
Write each expression without negative exponents.
  1. \(\displaystyle \dfrac{1}{4}xy^{-2}\)
  2. \(\displaystyle a^{-3}b^{-2}\)
  3. \(\displaystyle \dfrac{2a^{-1}}{{bc}^{-2}}\)
  4. \(\displaystyle \dfrac{x}{y^{-2}}+\dfrac{x^{-2}}{y}\)
Solution.
We use the fact that \(~a^{-n} = \dfrac{1}{a^n}\text{,}\) and consequently that \(~\dfrac{1}{a^{-n}} = a^n\text{.}\)
  1. \(\displaystyle \dfrac{x}{4y^2}\)
  2. \(\displaystyle \dfrac{1}{a^{3}b^{2}}\)
  3. \(\displaystyle \dfrac{2c^2}{ab}\)
  4. \(\displaystyle xy^{2}+\dfrac{1}{x^2y}\)
Example A.8.62.
Simplify where possible using the laws of exponents.
  1. \(\displaystyle 3x^{-3}x^5\)
  2. \(\displaystyle \dfrac{4a^{-4}}{8a^{-8}}\)
  3. \(\displaystyle (2bc^{-3})^{-2}\)
  4. \(\displaystyle 3x^{-4}-2x^{-3}\)
Solution.
  1. Add the exponents: \(~3x^{-3}x^5 = 3x^2\)
  2. Subtract the exponents: \(~\dfrac{4a^{-4}}{8a^{-8}} = \dfrac{1}{2}a^{-4-(-8)} = \dfrac{a^4}{2}\)
  3. Raise each factor to the power \(-2\text{.}\) Multiply exponents:
    \begin{equation*} (2bc^{-3})^{-2} = 2^{-2}b^{-2}(c^{-3})^{-2} = \dfrac{c^6}{4b^2} \end{equation*}
  4. We cannot add or subtract powers with different exponents.
    \begin{equation*} 3x^{-4}-2x^{-3} = \dfrac{3}{x^4} - \dfrac{2}{x^3} \end{equation*}
Checkpoint A.8.63.
Simplify where possible. Write your answer without negative exponents.
\begin{equation*} (2x^3y^{-4})(\dfrac{3}{4}x^{-2}y^2) \end{equation*}
Answer.
\(\dfrac{3x}{2y^2}\)
Checkpoint A.8.64.
Simplify where possible. Write your answer without negative exponents.
\begin{equation*} ~\dfrac{ab^{-3}}{(3ab)^{-2}} \end{equation*}
Answer.
\(\dfrac{9a^3}{b}\)
Checkpoint A.8.65.
Simplify where possible. Write your answer without negative exponents.
\begin{equation*} 2x^{-2}-(2x)^{-2} \end{equation*}
Answer.
\(\dfrac{8x^4-1}{4x^2}\)
Checkpoint A.8.66.
Simplify where possible. Write your answer without negative exponents.
\begin{equation*} 2x^{-2}(-2x)^{-2} \end{equation*}
Answer.
\(\dfrac{1}{2}\)

Subsubsection A.8.4.2 Improper fractions

An improper fraction is one in which the numerator is larger than the denominator.
Example A.8.67.
Write an improper fraction for the sum \(~2 + \dfrac{3}{8}\)
Solution.
We write the whole number with the same denominator as the fraction. Thus,
\begin{equation*} \blert{2} + \dfrac{3}{8} = \blert{\dfrac{16}{8}}+\dfrac{3}{8} = \dfrac{16+3}{8} = \dfrac{19}{8} \end{equation*}
Example A.8.68.
Write an improper fraction for the difference \(~3 - \dfrac{1}{5}\)
Solution.
We write the whole number with the same denominator as the fraction. Thus,
\begin{equation*} \blert{3} - \dfrac{1}{5} = \blert{\dfrac{15}{5}}-\dfrac{1}{5} = \dfrac{15-1}{5} = \dfrac{14}{5} \end{equation*}
Checkpoint A.8.69.
Write an improper fraction for each sum or difference.
  1. \(\displaystyle 1 + \dfrac{3}{4}\)
  2. \(\displaystyle 2 - \dfrac{3}{10}\)
Answer.
  1. \(\displaystyle \dfrac{7}{4}\)
  2. \(\displaystyle \dfrac{17}{10}\)
Checkpoint A.8.70.
Write an improper fraction for each sum or difference.
  1. \(\displaystyle 1 - \dfrac{23}{100}\)
  2. \(\displaystyle 3 + \dfrac{3}{25}\)
Answer.
  1. \(\displaystyle \dfrac{77}{100}\)
  2. \(\displaystyle \dfrac{78}{25}\)

Subsubsection A.8.4.3 Check a division

Remember that division is the inverse operation for multiplication.
Example A.8.71.
Check that the division is correct: \(~536\div 15 = 35\dfrac{11}{15}\)
Solution.
The quotient tells us that 15 divides into 536 thirty-five times, with a remainder of 11. This in turn means that if we multiply 15 by 35, and then add 11, we should get 536 back again.
\begin{equation*} 15 \times 35 + 11 = 525 + 11 = 536 \end{equation*}
Note the pattern: divisor \(\times\) quotient \(+\) remainder \(=\) starting number
Example A.8.72.
Check that the division is correct:
\begin{equation*} (3n^2+n-6)\div (n+2) = 3n-5 + \dfrac{4}{n+2} \end{equation*}
Solution.
The answer tells us that \(n+2\) divides into \(3n^2+n-6\) to give a quotient of \(3n-5\text{,}\) with a remainder of 4. If we multiply \(n+2\) by \(3n-5\text{,}\) and then add 4, we should get \(3n^2+n-6\) back again.
\begin{equation*} (n+2)(3n-5)+4 = (3n^2+n-10)+4 = 3n^2-n-6 \end{equation*}
Checkpoint A.8.73.
Check the division.
\begin{equation*} 25 \div 4 = 6\dfrac{1}{4} \end{equation*}
Answer.
\(4(6)+1=25\)
Checkpoint A.8.74.
Check the division.
\begin{equation*} 1331 \div 28 = 47\dfrac{15}{28} \end{equation*}
Answer.
\(28(47)+15=1331\)
Checkpoint A.8.75.
Check the division.
\begin{equation*} (n^2+3n+6) \div (n+1) = n+2 +\dfrac{4}{n+1} \end{equation*}
Answer.
\((n+1)(n+2)+4 = n^2+3n+6\)
Checkpoint A.8.76.
Check the division.
\begin{equation*} (2x^3+7x^2+9x+40) \div (2x-3) = x^2+5x+12 + \dfrac{40}{2x-3} \end{equation*}
Answer.
\((2x-3)(x^2+5x+12)+40 = 2x^3+7x^2+9x+40\)

Subsection A.8.5 Equations with Fractions

Subsubsection A.8.5.1 Solve quadratic equations

Once we have cleared the fractions from an equation, we may have a quadratic equation to solve. We can choose the easiest method to solve: factoring, extracting roots, or the quadratic formula.
Example A.8.77.
Solve each quadratic equation by the easiest method.
  1. \(\displaystyle 2x^2-2x=3\)
  2. \(\displaystyle (2x-1)^2=3\)
  3. \(\displaystyle 2x^2-x=3\)
Solution.
  1. Because \(~2x^2-2x-3~\) does not factor, we use the quadratic formula.
    \begin{equation*} x = \dfrac{2 \pm \sqrt{(-2)^2-4(2)(-3)}}{2(2)} = \dfrac{2\pm \sqrt{28}}{4} = \dfrac{1 \pm \sqrt{7}}{2} \end{equation*}
  2. We use extraction of roots.
    \begin{align*} 2x-1 \amp = \pm \sqrt{3}\\ x \amp = \dfrac{1 \pm \sqrt{3}}{2} \end{align*}
  3. We write the equation in standard form and factor the left side.
    \begin{align*} 2x^2-x-3 \amp = 0\\ (2x-3)(x+1) \amp = 0\\ 2x-3=0~~~~x+1 \amp = 0\\ x = \dfrac{3}{2}~~~~x \amp = -1 \end{align*}
Checkpoint A.8.78.
Solve each equation by the easiest method.
  1. \(\displaystyle 3x^2+10x=8\)
  2. \(\displaystyle x^2+6x+9=8\)
  3. \(\displaystyle 81x^2-18x+1=0\)
  4. \(\displaystyle 9x^2+18x=27\)
Answer.
  1. \(\displaystyle x=-4,~\dfrac{2}{3}\)
  2. \(\displaystyle x=-2\pm2\sqrt{2}\)
  3. \(\displaystyle x=\dfrac{1}{9},~\dfrac{1}{9}\)
  4. \(\displaystyle x=-3,~1\)

Subsubsection A.8.5.2 Solve equations graphically

If we can’t solve an equation algebraically, we may be able use a graph to find at least an approximation for the solution.
Example A.8.79.
Use a graph to solve the equation \(~2x^3+9x^2-8x+36=0\)
Solution.
We graph the equation \(~y=2x^3+9x^2-8x-36~\) and look for the points where \(~y=0~\) (the \(x\)-intercepts).
cubic
From the graph, we estimate the solutions at \(~x=-4.5,~x=-2,\) and \(~x=2\text{.}\) By substituting each of these values into the original equation, you can verify that they are indeed solutions.
Example A.8.80.
Use a graph to solve the equation \(~x^2+2x+3 = 15-2x\)
Solution.
We graph the equations \(~y_1=x^2+2x+3~\) and \(~y_2=15-2x~\) and look for points on the two graphs where the coordinates are equal (intersubsection points).
parabola and line
From the graph, we see that the points with \(~x=-6~\) and \(~x=2~\) have the same \(y\)-coordinate on both graphs. In other words, \(~y_1=y_2~\) when \(~x=-6~\) or \(~x=2~\text{,}\) so \(~x=-6~\) and \(~x=2~\) are the solutions.
Checkpoint A.8.81.
Use a graph to solve the equation \(~2x^3+7x^2-7x-12=0\)
Answer.
\(x = -4,~ -1,~ \dfrac{3}{2}\)
Checkpoint A.8.82.
Use a graph to solve the equation \(~\dfrac{24}{x+4}=11+2x-x^2\)
Answer.
\(x=-1,~4\)

Subsubsection A.8.5.3 Choose the correct technique

Example A.8.83.
Choose the appropriate technique for each problem.
  1. Cross-multiply
  2. Multiply each term by the LCD
  3. Multiply top and bottom by the LCD
  4. Find building factors
  1. Combine \(~\dfrac{8}{x+2} + \dfrac{x}{x-3}\)
  2. Solve \(~\dfrac{8}{x+2} = \dfrac{x}{x-3}\)
  3. Solve \(~\dfrac{8}{x+2} + 1 = \dfrac{x}{x-3}\)
  4. Simplify \(~\dfrac{\dfrac{8}{x} + 1}{\dfrac{x}{x-3}+\dfrac{2}{x}}\)
Solution.
  1. To add fractions, we find an LCD and build each fraction, so choice IV is correct.
  2. To solve a proportion, we can cross-multiply, so choice I is correct.
  3. To clear fractions from an equation, we multiply by the LCD, so choice II is correct.
  4. To simplify a complex fraction, we apply the fundamental pricnicple of fractions, so choice III is correct.
Checkpoint A.8.84.
Write the first step for the problem.
Solve \(~\dfrac{3}{x} + 3 = \dfrac{1}{x+3}\)
Answer.
\(3(x+3)+3x(x+3)=x\)
Checkpoint A.8.85.
Write the first step for the problem.
Combine \(~\dfrac{3}{x} + 3 - \dfrac{1}{x+3}\)
Answer.
\(\dfrac{3(x+3)}{x(x+3)} + \dfrac{3x(x+3)}{x(x+3)} - \dfrac{x}{x(x+3)}\)
Checkpoint A.8.86.
Write the first step for the problem.
Simplify \(~\dfrac{\dfrac{3}{x} + 1} {3- \dfrac{1}{x+3}}\)
Answer.
\(\dfrac{3(x+3)-x(x+3)}{3x(x+3)-x}\)
Checkpoint A.8.87.
Write the first step for the problem.
Solve \(~\dfrac{3}{x} = \dfrac{1}{x+3}\)
Answer.
\(3(x+3)=x\)
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