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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. 3x25x3
  2. 3x2(5x3)
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  3x2(5x3)=15x5.
Example A.8.2.
Simplify each expression if possible.
  1. 6t48t4
  2. 6t4(8t4)
Solution.
  1. This expression is a difference of like terms, so we can combine their coefficients to get  6t48t4=14t4
  2. This expression is a product, and the powers have the same base, so we can apply the first law of exponents to get  6t4(8t4)=48t8
Checkpoint A.8.3.
Simplify each expression if possible.
  1. 2a29a3+a2
  2. 2a2(9a3+a2)
Answer.
  1. 3a29a3
  2. 18a5+2a4
Checkpoint A.8.4.
Simplify each expression if possible.
  1. 74a3+2a3
  2. 74a2(2a3)
Answer.
  1. 72a3
  2. 78a5

Subsubsection A.8.1.2 Use formulas

There are several useful formulas for simplifying polynomials.
Example A.8.5.
If  a=5t4,  find a3 and 3a2.
Solution.
We substitute 5t4 for a to find
a3=(5t4)3=53(t4)3=125t12Apply the third law of exponents.3a2=3(5t4)2=352(t4)2=75t8
Example A.8.6.
If a=2y and b=3z2, find b3 and 3a2b.
Solution.
We substitute 2y for a and 3z2 for b to find
b3=(3z2)3=(3)3(z2)3=27z63a2b=3(2y)2(3z2)=3(4y2)(3z2)=36y2z2
Checkpoint A.8.7.
If  a=4x3  and  b=3h, find a3 and ab2.
Answer.
64x9;  36x3h2
Checkpoint A.8.8.
If x=6p2 and y=mq2, find y3 and x2y.
Answer.
m3q6;  36mp4q2

Subsubsection A.8.1.3 Square binomials

Sometimes it is easier to use formulas to square binomials.
Special Products of Binomials.
(a+b)2=(a+b)(a+b)=a2+2ab+b2(ab)2=(ab)(ab)=a22ab+b2(a+b)(ab)=a2b2
Example A.8.9.
Use the identity  (a+b)2=a2+2ab+b2  to expand  (3h2+4k3)2.
Solution.
We substitute 3h2 for a and 4k3 for b into the identity.
(3h2+4k3)2=(3h2)2+2(3h2)(4k3)+(4k3)2=9h4+24h2k3+16k6
Example A.8.10.
Use the identity  (ab)2=a22ab+b2  to expand  (2xy25)2.
Solution.
We substitute 2xy2 for a and 5 for b into the identity.
(2xy25)2=(2xy2)22(2xy2)(5)+52=4x2y420xy2+25
Checkpoint A.8.11.
Expand  (8w43w3)2
Answer.
64w848w7+9w6
Checkpoint A.8.12.
Expand  (a3b+9ab3)2
Answer.
a6b2+18a4b4+81a2b6

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. 4x24x
  2. 4x21
  3. 4x24x+2
Solution.
  1. We factor out a common factor of 4x to get 4x(x1).
  2. This is a difference of two squares that factors as (2x1)(2x+1).
  3. This is the square of a binomial, (2x1)2.
Example A.8.14.
Factor.
  1. 27a23
  2. 27a31
  3. 81a3a
Solution.
  1. We first factor out 3 to find 3(9a21), then factor the difference of two squares to get 3(3a1)(3a+1).
  2. This is a differece of two cubes, which factors as (3a1)(9a2+3a+1).
  3. We first factor out a to get a(81a21), then factor the difference of two squares to get a(9a1)(9a+1) .
Checkpoint A.8.15.
Factor completely 2x3+16y3
Answer.
2(x+2y)(x22xy+4y2)
Checkpoint A.8.16.
Factor completely 4x2y36y3
Answer.
4y(x3y)(x+3y)
Checkpoint A.8.17.
Factor completely 2b36b236b
Answer.
2b(b6)(b+3)
Checkpoint A.8.18.
Factor completely 9b4+9b2
Answer.
9b2(b2+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.
Example A.8.19.
Which of these is the opposite of m2p ?
  1. m2+p
  2. mp2
  3. pm2
Solution.
The opposite of m2p is (m2p)=m2+p, or pm2.
Example A.8.20.
Which of these pairs of binomials are opposites?
  1. 3c5 and 5+3c
  2. 53c and 35c
  3. 5c3 and 35c
Solution.
The opposite of 5c3 is (5c3)=5c+3, or 35c, so (c) is correct.
Checkpoint A.8.21.
Find the opposite of the binomial  2x+1
Answer.
2x1
Checkpoint A.8.22.
Find the opposite of the binomial  b2b
Answer.
bb2
Checkpoint A.8.23.
Find the opposite of the binomial  4n+8
Answer.
4n8
Checkpoint A.8.24.
Find the opposite of the binomial  3z22
Answer.
3z2+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, so the equation of the line is x=15. Because Δx=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, so the equation of the line is y=40. Because Δy=0 between any two points on the line, its slope is 0.
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;  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;  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  12(PQ)
Solution.
12(PQ)=1P2Q=P2Q
Example A.8.32.
Multiply  4(bc)
Solution.
4(bc)=41(bc)=4b1c=4bc
Checkpoint A.8.33.
Multiply  32a(a26)
Answer.
a9
Checkpoint A.8.34.
Multiply  8(m4x2)
Answer.
2mx2

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  tw÷3t2w
Solution.
tww3t2=tww3tt=13t
Example A.8.36.
Divide  a2÷b1b
Solution.
a21bb=a(1b)2b=aab2b
Checkpoint A.8.37.
Divide  34÷cw2t22
Answer.
3t264cw2
Checkpoint A.8.38.
Divide  nn1÷n2p+1
Answer.
p+1n2n

Subsubsection A.8.3.3 Use factoring

Example A.8.39.
Multiply  4y214y2y22y4y+2
Solution.
We factor each numerator and denominator, and look for common factors.
4y214y2y22y4y+2=(2y1)(2y+1)(2y)(2+y)y(1)(y2)2(2y+1)Divide out common factors.=y(2y1)2(y+2)Note:  y2=(2y)
Example A.8.40.
Divide  6ab2a+b÷(4a2b)
Solution.
We multiply the first fraction by the reciprocal of the second fraction.
6ab2a+b÷(4a2b)=23ab2a+b122aabDivide out common factors.=32a(2a+b)
Checkpoint A.8.41.
Multiply  3xy4xy6y22x3y12x
Answer.
18
Checkpoint A.8.42.
Multiply  9x2252x2x216x10
Answer.
(3x+5)(x+1)4
Checkpoint A.8.43.
Divide  (x29)÷x26x+93x
Answer.
3x(x+3)x3
Checkpoint A.8.44.
Divide  x21x+3÷x2x2x2+5x+6
Answer.
(x1)(x+2)x2

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  25+310
Solution.
We factor each denominator and line up any common factors vertically. We use one factor from each column in the LCD.
15=3      510=            5      2
The LCD is 352, or 30.
Example A.8.46.
Find the LCD for the fractions  56a29a2
Solution.
We factor each denominator and line up any common factors vertically. We use one factor from each column in the LCD.
6a=2      3      a9a2=            3      3      a      a
The LCd is 233aa, or 18a2.
Checkpoint A.8.47.
Find the LCD for the fractions.
  1. 7816
  2. 5275+1324
Answer.
  1. 24
  2. 600
Checkpoint A.8.48.
Find the LCD for the fractions.
  1. 34a2b2+710ab3
  2. 38t435t2
Answer.
  1. 20a2b3
  2. 40t4

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:
 3s=?2s2
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 2s3, so we need to multiply the old denominator by 2s. This is the building factor. We multiply top and bottom of the old fraction by the building factor:
3s2s2s=62s2
Example A.8.50.
Write an equivalent fraction with the new denominator:
 2t+1=?t2+t
Solution.
The new denominator factors as t(t+1), so the building factor is t. We multiply top and bottom of the old fraction by t to obtain:
2t+1tt=2tt2+t
Checkpoint A.8.51.
Write an equivalent fraction with the new denominator.
  1. 1n=?n2n
  2. 4a1=?a21
Answer.
  1. n1n2n
  2. 4a+4a21
Checkpoint A.8.52.
Write an equivalent fraction with the new denominator.
  1. bb2=?(b+3)(b2)
  2. x1x2+2x=?(x2x)(x+2)
Answer.
  1. b2+3b(b+3)(b2)
  2. x22x+1(x2x)(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  2xw3xw
Solution.
These are like fractions, so we need only combine their numerators.
2xw3xw=2x3xw=xw
Example A.8.54.
Subtract  3aa+2a
Solution.
These are like fractions, so we need only combine their numerators. Be careful to subtract both terms of the second numerator.
3aa+2a=3(a+2)a=1aa
Example A.8.55.
Add  2+1x
Solution.
We write 2 as a fraction, 21, and build it to the LCD, x.
21xx+1x=2xx+1x=2x+1x
Example A.8.56.
Subtract  a2b3b2
Solution.
We build each fraction to the LCD, 2b2.
a2b3b2=a2bbb3b222=ab2b262b2=ab62b2
Checkpoint A.8.57.
Subtract  4ab2cb2
Answer.
4acb2
Checkpoint A.8.58.
Subtract  p+22qp12q
Answer.
32q
Checkpoint A.8.59.
Add  N+2N
Answer.
N2+2N
Checkpoint A.8.60.
Add  2xyy3x
Answer.
6+y23xy

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. 14xy2
  2. a3b2
  3. 2a1bc2
  4. xy2+x2y
Solution.
We use the fact that  an=1an, and consequently that  1an=an.
  1. x4y2
  2. 1a3b2
  3. 2c2ab
  4. xy2+1x2y
Example A.8.62.
Simplify where possible using the laws of exponents.
  1. 3x3x5
  2. 4a48a8
  3. (2bc3)2
  4. 3x42x3
Solution.
  1. Add the exponents:  3x3x5=3x2
  2. Subtract the exponents:  4a48a8=12a4(8)=a42
  3. Raise each factor to the power 2. Multiply exponents:
    (2bc3)2=22b2(c3)2=c64b2
  4. We cannot add or subtract powers with different exponents.
    3x42x3=3x42x3
Checkpoint A.8.63.
Simplify where possible. Write your answer without negative exponents.
(2x3y4)(34x2y2)
Answer.
3x2y2
Checkpoint A.8.64.
Simplify where possible. Write your answer without negative exponents.
 ab3(3ab)2
Answer.
9a3b
Checkpoint A.8.65.
Simplify where possible. Write your answer without negative exponents.
2x2(2x)2
Answer.
8x414x2
Checkpoint A.8.66.
Simplify where possible. Write your answer without negative exponents.
2x2(2x)2
Answer.
12

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+38
Solution.
We write the whole number with the same denominator as the fraction. Thus,
2+38=168+38=16+38=198
Example A.8.68.
Write an improper fraction for the difference  315
Solution.
We write the whole number with the same denominator as the fraction. Thus,
315=15515=1515=145
Checkpoint A.8.69.
Write an improper fraction for each sum or difference.
  1. 1+34
  2. 2310
Answer.
  1. 74
  2. 1710
Checkpoint A.8.70.
Write an improper fraction for each sum or difference.
  1. 123100
  2. 3+325
Answer.
  1. 77100
  2. 7825

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÷15=351115
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.
15×35+11=525+11=536
Note the pattern: divisor × quotient + remainder = starting number
Example A.8.72.
Check that the division is correct:
(3n2+n6)÷(n+2)=3n5+4n+2
Solution.
The answer tells us that n+2 divides into 3n2+n6 to give a quotient of 3n5, with a remainder of 4. If we multiply n+2 by 3n5, and then add 4, we should get 3n2+n6 back again.
(n+2)(3n5)+4=(3n2+n10)+4=3n2n6
Checkpoint A.8.73.
Check the division.
25÷4=614
Answer.
4(6)+1=25
Checkpoint A.8.74.
Check the division.
1331÷28=471528
Answer.
28(47)+15=1331
Checkpoint A.8.75.
Check the division.
(n2+3n+6)÷(n+1)=n+2+4n+1
Answer.
(n+1)(n+2)+4=n2+3n+6
Checkpoint A.8.76.
Check the division.
(2x3+7x2+9x+40)÷(2x3)=x2+5x+12+402x3
Answer.
(2x3)(x2+5x+12)+40=2x3+7x2+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. 2x22x=3
  2. (2x1)2=3
  3. 2x2x=3
Solution.
  1. Because  2x22x3  does not factor, we use the quadratic formula.
    x=2±(2)24(2)(3)2(2)=2±284=1±72
  2. We use extraction of roots.
    2x1=±3x=1±32
  3. We write the equation in standard form and factor the left side.
    2x2x3=0(2x3)(x+1)=02x3=0    x+1=0x=32    x=1
Checkpoint A.8.78.
Solve each equation by the easiest method.
  1. 3x2+10x=8
  2. x2+6x+9=8
  3. 81x218x+1=0
  4. 9x2+18x=27
Answer.
  1. x=4, 23
  2. x=2±22
  3. x=19, 19
  4. 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  2x3+9x28x+36=0
Solution.
We graph the equation  y=2x3+9x28x36  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. 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  x2+2x+3=152x
Solution.
We graph the equations  y1=x2+2x+3  and  y2=152x  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,  y1=y2  when  x=6  or  x=2 , so  x=6  and  x=2  are the solutions.
Checkpoint A.8.81.
Use a graph to solve the equation  2x3+7x27x12=0
Answer.
x=4, 1, 32
Checkpoint A.8.82.
Use a graph to solve the equation  24x+4=11+2xx2
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  8x+2+xx3
  2. Solve  8x+2=xx3
  3. Solve  8x+2+1=xx3
  4. Simplify  8x+1xx3+2x
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  3x+3=1x+3
Answer.
3(x+3)+3x(x+3)=x
Checkpoint A.8.85.
Write the first step for the problem.
Combine  3x+31x+3
Answer.
3(x+3)x(x+3)+3x(x+3)x(x+3)xx(x+3)
Checkpoint A.8.86.
Write the first step for the problem.
Simplify  3x+131x+3
Answer.
3(x+3)x(x+3)3x(x+3)x
Checkpoint A.8.87.
Write the first step for the problem.
Solve  3x=1x+3
Answer.
3(x+3)=x
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