Section 3.3 Dividing Polynomials

Before Starting this Section, Review

  1. 1 Multiplying polynomials (Section P.3 , page 33)

  2. 2 Evaluating a function (Section 2.4 , page 219)

  3. 3 Factoring trinomials (Section P.4 , page 43)

Objectives

  1. 1 Learn the division algorithm.

  2. 2 Use synthetic division.

  3. 3 Use the Remainder and Factor Theorems.

Greenhouse Effect and Global Warming

The greenhouse effect refers to the fact that Earth is surrounded by an atmosphere that acts like the transparent cover of a greenhouse, allowing sunlight to filter through while trapping heat. It is becoming increasingly clear that we are experiencing global warming. This means that Earth’s atmosphere is becoming warmer near its surface. Scientists who study climate agree that global warming is due largely to the emission of “greenhouse gases” such as carbon dioxide (CO2),(CO2), chlorofluorocarbons (CFCs), methane (CH4),(CH4), ozone (O3),(O3), and nitrous oxide (N2O).(N2O). The temperature of Earth’s surface increased by 1°F1°F over the 20th century. The late 1990s and early 2000s were some of the hottest years ever recorded. Projections of future warming suggest a global temperature increase of between 2.5°F2.5°F and 10.4°F10.4°F by 2100.

Global warming will result in long-term changes in climate, including a rise in average temperatures, unusual rainfalls, and more storms and floods. The sea level may rise so much that people will have to move away from coastal areas. Some regions of the world may become too dry for farming.

The production and consumption of energy constitute one of the major causes of greenhouse gas emissions. In Example 8, we look at the pattern for the consumption of petroleum in the United States.

The Division Algorithm

  1. 1 Learn the division algorithm.

Dividing polynomials is similar to dividing integers. Both division processes are closely connected to multiplication. The fact that 10=5⋅210=5⋅2 is also expressed as 105=2105=2. By comparison, in equation

x3−1=(x−1)(x2+x+1)Factor using difference of cubes.
x3−1=(x−1)(x2+x+1)Factor using difference of cubes.
(1)

dividing both sides by x−1,x−1, we obtain

x3−1x−1=x2+x+1,x≠1.
x3−1x−1=x2+x+1,x≠1.

Next, consider the equation

x3−2=(x−1)(x2+x+1)−1Subtract 1 from both sidesof equation (1).
x3−2=(x−1)(x2+x+1)−1Subtract 1 from both sidesof equation (1).
(2)

Dividing both sides by (x−1),(x−1), we can rewrite equation (2) in the form

x3−2x−1=x2+x+1−1x−1,x≠1.
x3−2x−1=x2+x+1−1x−1,x≠1.

Here we say that when the dividend x3−2x3−2 is divided by the divisor x−1,x−1, the quotient is x2+x+1x2+x+1 and the remainder is −1.−1. This result is an example of the Division Algorithm.

Side Note

It is helpful to compare the division process for integers with that of polynomials. Notice that

2376_1
2376–1

can also be written as

73=2+13
73=2+13

or

7=3⋅2+1
7=3⋅2+1
.

The Division Algorithm says that the dividend is equal to the divisor times the quotient plus the remainder.

The expression F(x)D(x)F(x)D(x) is improper if degree of F(x)≥F(x)≥ degree of D(x). However, the expression R(x)D(x)R(x)D(x) is always proper because by the Division Algorithm, degree of R(x)<degreeR(x)<degree of D(x).

For an improper expression F(x)D(x)F(x)D(x), you may use the long division and the synthetic division processes reviewed in the next few examples.

Practice Problem 1

  1. Find the quotient and remainder when 3x3+4x2+x+73x3+4x2+x+7 is divided by x2+1.x2+1.

Example 2 Using Long Division

Divide x4−13x2+x+35x4−13x2+x+35 by x2−x−6.x2−x−6.

Solution

Because the dividend does not contain an x3x3 term, we use a zero coefficient for the missing term.

x2+x−6←Quotientx2−x−6x4+0x3−13x2+x+35x4−x3−6x2_x3−7x2+x+35x3−x2−6x_−6x2+7x+35−6x2+6x+36_x−1←Remainder
x2+x−6x2−x−6x4+0x3−13x2+x+35x4−x3−6x2−−−−−−−−−−−−−−−−−−−−−x3−7x2+x+35x3−x2−6x−−−−−−−−−−−−−−−−−6x2+7x+35−6x2+6x+36−−−−−−−−−−−−−x−1←−−−−−−−−−−Quotient←−−−−−−−−−−Remainder

The quotient is x2+x−6,x2+x−6, and the remainder is x−1.x−1.

We can write this result in the form

x4−13x2+x+35x2−x−6=x2+x−6+x−1x2−x−6.
x4−13x2+x+35x2−x−6=x2+x−6+x−1x2−x−6.

Practice Problem 2

  1. Divide x4+5x2+2x+6x4+5x2+2x+6 by x2−x+3.x2−x+3.

Synthetic Division

  1. 2 Use synthetic division.

You can decrease the work involved in finding the quotient and remainder when the divisor D(x) is of the form x−ax−a by using the process known as synthetic division. The procedure is best explained by considering an example.

Long Division Synthetic Division

Each circled term in the long division process is the same as the term above it. Furthermore, the boxed terms are terms in the dividend written in a new position. If these two sets of terms are eliminated in the synthetic division process, a more efficient format results.

The essential steps in the process can be carried out as follows:

  1. Bring down 2 from the first line to the third line.
32−53−14↓2
32⏐↓2−53−14

  1. Multiply 2 in the third line by the 3 in the divisor position and place the product, 6, under −5;−5; then add vertically to obtain 1.

  1. Multiply 1 by 3 and place the product 3 under 3; then add vertically to obtain 6.

  1. Finally, multiply 6 by 3 and place the product, 18, under −14;−14; then add to obtain 4.

The first three terms, namely, 2, 1, and 6, are the coefficients of the quotient, 2x2+x+6,2x2+x+6, and the last number, 4, is the remainder. The procedure for synthetic division is given next.

Practice Problem 3

  1. Divide 2x3−7x2+52x3−7x2+5 by x−3.x−3.

Side Note

A quick way to see if you remembered to put 0 coefficients in place of missing terms when dividing a polynomial of degree n by x −aby x −a is to check that there are n+1n+1 numbers in the row next to a.

Example 4 Using Synthetic Division

Use synthetic division to divide 2x4+x3−16x2+182x4+x3−16x2+18 by x+2.x+2.

Solution

Because x−a=x+2=x−(−2),x−a=x+2=x−(−2), we have a=−2.a=−2. Write the coefficients of the dividend in a line, supplying 0 as the coefficient of the missing x-term. Then carry out the steps of synthetic division.

The quotient is 2x3−3x2−10x+20,2x3−3x2−10x+20, and the remainder is −22;−22; so the result is

2x4+x3−16x2+18x+2=2x3−3x2−10x+20+−22x+2=2x3−3x2−10x+20−22x+2.
2x4+x3−16x2+18x+2==2x3−3x2−10x+20+−22x+22x3−3x2−10x+20−22x+2.

Practice Problem 4

  1. Use synthetic division to divide 2x3+x2−18x−72x3+x2−18x−7 by x+3.x+3.

The Remainder and Factor Theorems

  1. 3 Use the Remainder and Factor Theorems.

In Example 4, we discovered that when the polynomial

F(x)=2x4+x3−16x2+18 is divided by x+2,
F(x)=2x4+x3−16x2+18 is divided by x+2,

the remainder is −22,−22, a constant.

Now evaluate F at −2.−2. We have

F(−2)=2(−2)4+(−2)3−16(−2)2+18Replace x with −2 in F(x).=2(16)−8−16(4)+18=−22Simplify.
F(−2)==2(−2)4+(−2)3−16(−2)2+182(16)−8−16(4)+18=−22Replace x with −2 in F(x).Simplify.

We see that F(−2)=−22,F(−2)=−22, the remainder we found when we divided F(x) by x+2.x+2. Is this result a coincidence? No, the following theorem asserts that such a result is always true.

Note that in the statement of the Remainder Theorem, F(x) plays the dual role of representing a polynomial and a function defined by the same polynomial.

The Remainder Theorem is a consequence of the Division Algorithm. In this case, the divisor D(x) is a linear polynomial D(x)=x − aD(x)=x − a, so the remainder R(x) has to be either the zero polynomial (which has no degree) or a polynomial of degree less than the degree of D(x)=x−aD(x)=x−a. It follows that the remainder has to be a constant polynomial R(x)=RR(x)=R, and we write

F(x)=(x−a)Q(x)+R.
F(x)=(x−a)Q(x)+R.

In particular, if we evaluate F at a, we obtain

F(x)=(x−a)Q(x)+RF(a)=(a−a)Q(a)+R=0⋅Q(a)+R=0+R=RSubstitute x=a, andsimplify.
F(x)F(a)==(x−a)Q(x)+R(a−a)Q(a)+R=0⋅Q(a)+R=0+R=RSubstitute x=a, andsimplify.

This proves the Remainder Theorem.

Example 5 Using the Remainder Theorem

Find the remainder when the polynomial

F(x)=2x5−4x3+5x2−7x+2
F(x)=2x5−4x3+5x2−7x+2

is divided by x−1.x−1.

Solution

We could find the remainder by using long division or synthetic division, but a quicker way is to evaluate F(x) when x=1.x=1. By the Remainder Theorem, F(1) is the remainder.

F(1)=2(1)5−4(1)3+5(1)2−7(1)+2Replace x with 1 in F(x).=2−4+5−7+2=−2
F(1)==2(1)5−4(1)3+5(1)2−7(1)+22−4+5−7+2=−2Replace x with 1 in F(x).

The remainder is −2.−2.

Practice Problem 5

  1. Use the Remainder Theorem to find the remainder when

    F(x)=x110−2x57+5 is divided by x −1.
    F(x)=x110−2x57+5 is divided by x −1.

Sometimes the Remainder Theorem is used in the other direction: to evaluate a polynomial at a specific value of x. This use is illustrated in the next example.

Example 6 Using the Remainder Theorem

Let f(x)=x4+3x3−5x2+8x+75.f(x)=x4+3x3−5x2+8x+75. Find f(−3).f(−3).

Solution

One way of solving this problem is to evaluate f(x) when x=−3.x=−3.

f(−3)=(−3)4+3(−3)3−5(−3)2+8(−3)+75Replace x with −3.=6Simplify.
f(−3)==(−3)4+3(−3)3−5(−3)2+8(−3)+756Replace x with −3.Simplify.

Another way is to use synthetic division to find the remainder when the polynomial f(x) is divided by (x+3).(x+3).

The remainder is 6. By the Remainder Theorem, f(−3)=6.f(−3)=6.

Practice Problem 6

  1. Use synthetic division to find f(−2),f(−2), where

    f(x)=x4+10x2+2x−20.
    f(x)=x4+10x2+2x−20.

Note that if we divide any polynomial F(x) by x−a,x−a, we have

F(x)=(x−a)Q(x)+RDivision algorithmF(x)=(x−a)Q(x)+F(a)R=F(a), by the Remainder Theorem
F(x)=(x−a)Q(x)+RF(x)=(x−a)Q(x)+F(a)Division algorithmR=F(a), by the Remainder Theorem

The last equation says that if F(a)=0,F(a)=0, then

F(x)=(x−a)Q(x)
F(x)=(x−a)Q(x)

and (x−a)(x−a) is a factor of F(x). Conversely, if (x−a)(x−a) is a factor of F(x), then dividing F(x) by (x−a)(x−a) gives a remainder of 0. Therefore, by the Remainder Theorem, F(a)=0.F(a)=0. This argument proves the Factor Theorem.

The Factor Theorem shows that if F(x) is a polynomial, then the following problems are equivalent:

  1. Factoring F(x).

  2. Finding the zeros of the function F(x) defined by the polynomial expression.

  3. Solving (or finding the roots of) the polynomial equation F(x)=0.F(x)=0.

Note that if a is a zero of the polynomial function F(x), then F(x)=(x−a)Q(x).F(x)=(x−a)Q(x). Any solution of the equation Q(x)=0Q(x)=0 is also a zero of F(x). Because the degree of Q(x) is less than the degree of F(x), it is simpler to solve the equation Q(x)=0Q(x)=0 to find other possible zeros of F(x). The equation Q(x)=0Q(x)=0 is called the depressed equation of F(x).

The next example illustrates how to use the Factor Theorem and the depressed equation to solve a polynomial equation.

Example 7 Using the Factor Theorem

Given that 2 is a zero of the function f(x)=3x3+2x2−19x+6,f(x)=3x3+2x2−19x+6, solve the polynomial equation 3x3+2x2−19x+6=0.3x3+2x2−19x+6=0.

Solution

Because 2 is a zero of f(x), we have f(2)=0.f(2)=0. The Factor Theorem tells us that (x−2)(x−2) is a factor of f(x). Next, we use synthetic division to divide f(x) by (x−2).(x−2).

We now have the coefficients of the quotient Q(x), with f(x)=(x−2)Q(x).f(x)=(x−2)Q(x).

f(x)=3x3+2x2−19x+6=(x−2)(3x2+8x−3︸↑Quotient)
f(x)=3x3+2x2−19x+6=(x−2)(3x2+8x−3↑Quotient)

Any solution of the depressed equation 3x2+8x−3=03x2+8x−3=0 is a zero of f. Because this equation is of degree 2, any method of solving a quadratic equation may be used to solve it.

3x2+8x−3=0Depressed equation(3x−1)(x+3)=0Factor3x−1=0orx+3=0Zero-product propertyx=13orx=−3Solve each equation.
3x2+8x−3=0(3x−1)(x+3)=0Depressed equationFactor3x−1=0x=13ororx+3=0x=−3Zero-product propertySolve each equation.

The solution set is {−3, 13, 2}.{−3, 13, 2}.

In Example 7 we were able to completely factor the polynomial function f(x)=3x3+2x2−19x+6=(x−2)(3x−1)(x+3)f(x)=3x3+2x2−19x+6=(x−2)(3x−1)(x+3). Applying the techniques from Section 3.2 allows us to sketch the graph of y=f(x)y=f(x) (see Figure 3.18).

Figure 3.18

Practice Problem 7

  1. Solve the equation 3x3−x2−20x−12=03x3−x2−20x−12=0 given that one solution is −2.−2.

Application

Energy is measured in British thermal units (BTU). One BTU is the amount of energy required to raise the temperature of 1 pound of water 1°F1°F at or near 39.2°F;139.2°F;1 million BTU is equivalent to the energy in 8 gallons of gasoline. Fuels such as coal, petroleum, and natural gas are measured in quadrillion BTU (1 quadrillion=1015),(1 quadrillion=1015), abbreviated “quads.”

Example 8 Crude Oil and Petroleum Consumption

The total crude oil and petroleum products consumption C (in billion barrels) in the United States during 1993–2011 is given in the following table. (Data are rounded to the nearest tenth.)

Year Consumption Year Consumption
1993 6.3 2003 7.3
1995 6.5 2005 7.6
1997 6.8 2007 7.6
1999 7.1 2009 6.9
2001 7.2 2011 6.9
Source: www.eia.gov

These data can be modeled by the function

C(t)=−0.0006t3+0.00576t2+0.10284t+6.33488,
C(t)=−0.0006t3+0.00576t2+0.10284t+6.33488,

where t=0t=0 represents 1993.

The model indicates that C(4)=6.8C(4)=6.8 billion barrels were consumed in 1997 (t=4).(t=4). Find another year between 1997 and 2011 when the model estimates consumption of 6.8 billion barrels.

Solution

We are given that

C(4)=6.8C(t)=(t−4)Q(t)+6.8Division Algorithm and Remainder TheoremC(t)−6.8=(t−4)Q(t)Subtract 6.8 from both sides.
C(4)C(t)C(t)−6.8===6.8(t−4)Q(t)+6.8(t−4)Q(t)Division Algorithm and Remainder TheoremSubtract 6.8 from both sides.

Therefore, 4 is zero of

F(t)=C(t)−6.8=−0.0006t3+0.00576t2+0.10284t−0.46512.
F(t)=C(t)−6.8=−0.0006t3+0.00576t2+0.10284t−0.46512.

Finding another year between 1997 and 2011 when the crude oil and petroleum consumption was 6.8 billion barrels requires us to find another zero of F(t) that is between 4 and 18. Because 4 is a zero of F(t), we use synthetic division to find the quotient Q(t). This yields

Q(t)=−0.0006t2+0.00336t+0.11628.
Q(t)=−0.0006t2+0.00336t+0.11628.

We now solve the depressed (quadratic) equation Q(t)=0Q(t)=0 using the quadratic formula.

Q(t)=−0.0006 t2+0.00336 t+0.11628=0t=−0.00336±√(0.00336)2−4(−0.0006)(0.11628)2(−0.0006)Quadratic formulat=−11.4ort=17Use a calculator.
Q(t)tt===−0.0006 t2+0.00336 t+0.11628=0−0.00336±(0.00336)2−4(−0.0006)(0.11628)−−−−−−−−−−−−−−−−−−−−−−−−−−√2(−0.0006)−11.4ort=17Quadratic formulaUse a calculator.

Because t needs to be between 4(=1997−1993)4(=1997−1993) and 18(=2011−1993),18(=2011−1993), this model tells us that in 2010 (t=17),(t=17), the crude oil and petroleum products consumption in the United States was also 6.8 billion barrels.

Practice Problem 8

  1. The value of C(x)=0.23x3−4.255x2+0.345x+41.05C(x)=0.23x3−4.255x2+0.345x+41.05 is 10 when x=3.x=3. Find another positive number x for which C(x)=10.C(x)=10.

Section 3.3 Exercises

Concepts and Vocabulary

  1. In the division x4−2x3+5x2−2x+1x2−2x+3=x2+2+2x−5x2−2x+3,x4−2x3+5x2−2x+1x2−2x+3=x2+2+2x−5x2−2x+3, the dividend is                        , the divisor is                        , the quotient is                        , and the remainder is                        .

  2. If P(x), Q(x), and F(x) are polynomials and F(x)=P(x)⋅Q(x),F(x)=P(x)⋅Q(x), then the factors of F(x) are                         and                        .

  3. The Remainder Theorem states that if a polynomial F(x) is divided by (x−a),(x−a), then the remainder R=_R=−−−−−−−−.

  4. The Factor Theorem states that (x−a)(x−a) is a factor of a polynomial F(x) if and only if _=0.−−−−−−−−=0.

  5. True or False. You cannot use synthetic division to divide a polynomial by x2−1.x2−1.

  6. True or False. When a polynomial of degree n+1n+1 is divided by a polynomial of degree n, the remainder is a constant polynomial.

  7. True or False. If the polynomial F(x) has (x+2)(x+2) as a factor, then F(2)=0F(2)=0.

  8. True or False. If F(x)=(x−5)(x7+13x3−2x+12)+7F(x)=(x−5)(x7+13x3−2x+12)+7, then F(5)=7F(5)=7.

Building Skills

In Exercises 9–16, use long division to find the quotient and the remainder.

  1. 6x2−x−22x+16x2−x−22x+1

  2. 4x3−2x2+x−32x−34x3−2x2+x−32x−3

  3. 3x4−6x2+3x−7x+13x4−6x2+3x−7x+1

  4. x6+5x3+7x+3x2+2x6+5x3+7x+3x2+2

  5. 4x3−4x2−9x+52x2−x−54x3−4x2−9x+52x2−x−5

  6. y5+3y4−6y2+2y−7y2+2y−3y5+3y4−6y2+2y−7y2+2y−3

  7. z4−2z2+1z2−2z+1z4−2z2+1z2−2z+1

  8. 6x4+13x−11x3−10−x23x2−5−x6x4+13x−11x3−10−x23x2−5−x

In Exercises 17–32, use synthetic division to find the quotient and the remainder when the first polynomial is divided by the second polynomial.

  1. x3−x2−7x+2;x−1x3−x2−7x+2;x−1

  2. 2x3−3x2−x+2;x+22x3−3x2−x+2;x+2

  3. x3+4x2−7x−10;x+2x3+4x2−7x−10;x+2

  4. x3+x2−13x+2;x−3x3+x2−13x+2;x−3

  5. x4−3x3+2x2+4x+5;x−2x4−3x3+2x2+4x+5;x−2

  6. x4−5x3−3x2+10;x−1x4−5x3−3x2+10;x−1

  7. 2x3+4x2−3x+1;x−122x3+4x2−3x+1;x−12

  8. 3x3+8x2+x+1;x−133x3+8x2+x+1;x−13

  9. 2x3−5x2+3x+2;x+12

  10. 3x3−2x2+8x+2;x+13

  11. x5+x4−7x3+2x2+x−1;x−1

  12. 2x5+4x4−3x3−7x2+3x−2;x+2

  13. x5+1;x+1

  14. x5+1;x−1

  15. x6+2x4−x3+5;x+1

  16. 3x5;x−1

In Exercises 33–36, use synthetic division and the Remainder Theorem to find each function value. Check your answer by evaluating the function at the given x-value.

  1. f(x)=x3+3x2+1

    1. f(1)

    2. f(−1)

    3. f(12)

    4. f(10)

  2. g(x)=2x3−3x2+1

    1. g(−2)

    2. g(−1)

    3. g(−12)

    4. g(7)

  3. h(x)=x4+5x3−3x2−20

    1. h(1)

    2. h(−1)

    3. h(−2)

    4. h(2)

  4. f(x)=x4+0.5x3−0.3x2−20

    1. f(0.1)

    2. f(0.5)

    3. f(1.7)

    4. f(−2.3)

In Exercises 37–44, use the Factor Theorem to show that the linear polynomial is a factor of the second polynomial. Check your answer by synthetic division.

  1. x−1;2x3+3x2−6x+1

  2. x−3;3x3−9x2−4x+12

  3. x+1;5x4+8x3+x2+2x+4

  4. x+3;3x4+9x3−4x2−9x+9

  5. x−2;x4+x3−x2−x−18

  6. x+3;x5+3x4+x2+8x+15

  7. x+2;x6−x5−7x4+x3+8x2+5x+2

  8. x−2;2x6−5x5+4x4+x3−7x2−7x+2

In Exercises 45–48, find the value of k for which the first polynomial is a factor of the second polynomial.

  1. x+1;x3+3x2+x+k [Hint: Let f(x)=x3+3x2+x+k. Then f(−1)=0.]

  2. x−1;−x3+4x2+kx−2

  3. x−2;2x3+kx2−kx−2

  4. x−1;k2−3kx2−2kx+6

In Exercises 49–52, use the Factor Theorem to show that the first polynomial is not a factor of the second polynomial. [Hint: Show that the remainder is not zero.]

  1. x−2;−2x3+4x2−4x+9

  2. x+3;−3x3−9x2+5x+12

  3. x+2;4x4+9x3+3x2+x+4

  4. x−3;3x4−8x3+5x2+7x−3

Applying the Concepts

  1. Geometry. The area of a rectangle is (2x4−2x3+5x2−x+2) square centimeters. Its length is (x2−x+2) cm. Find its width.

  2. Geometry. The volume of a rectangular solid is (x4+3x3+x+3) cubic inches. Its length and width are (x+3) and (x+1) inches, respectively. Find its height.

  3. Cell phone sales. The weekly revenue from one type of cell phone sold at a mall outlet is a quadratic function, R, of its price, x, in dollars. The weekly revenue is $3000 if the phone is priced at $40 or $60 (fewer phones are sold at $60) and is $2400 if the phone is priced at $30.

    1. Find an equation for R(x)−3000.

    2. Find an equation for R(x).

    3. Find the maximum weekly revenue and the price that generates this revenue.

  4. Ticket sales. The evening revenue from a theater is a quadratic function, R, of its ticket price, x, in dollars. The evening revenue is $4725 if the tickets are priced at $8 or $12 (fewer tickets are sold at $12) and is $4125 if the tickets are priced at $6.

    1. Find an equation for R(x)−4725.

    2. Find an equation for R(x).

    3. Find the maximum weekly revenue and the price that generates this revenue.

  5. Total energy consumption. The total energy consumption (in quads) in the United States from 1991–2011 can be approximated by the function

    C(t)=−0.0006t3−0.0613t2+2.0829t+82.904,

    where t=0 represents 1991. The model indicates that 97.6 quads of energy were consumed in 2002. Find another year after 2002 when the model estimates 97.6 quads of total energy consumption in the United States. Round your answer to the nearest year. (The model was modified from the data obtained from www.eia.gov.)

  6. Lottery sales. In the lottery game called Florida Lotto, players select 6 numbers out of 53. Various prizes are offered for matching three through six numbers drawn by the lottery. Florida Lotto sales (in millions of dollars) from 1998 through 2011 can be approximated by the function

    s(t)=−0.1779t3−2.8292t2+42.0240t+698.6831,

    where t=0 represents 1998. The model indicated that $737.7 million in Florida Lotto sales occurred in 1999. Find another year after 1999 when the model indicates that $737.7 million in Florida Lotto sales occurred. (The model was slightly modified from the data obtained from www.floridafiscalportal.state.fl.us.)

  7. Marine Corps. The total number of U.S. Marine Corps, M (in thousands), during 1990–2010 can be modeled by the function

    M(t)=−0.0027t3+0.3681t2−5.8645t+195.2782,

    where t=0 represents 1990, t=1 represents 1991, and so on. The model estimates the total number of U.S. Marine Corps in 1992 as 185,000. Find another year between 1992 and 2010 when the model indicates that U.S. Marine Corps were approximately 185,000. Round your answer to the nearest year. (The model was generated using slightly modified data from www.census.gov.)

  8. Unemployment. U.S. unemployment, U (in percent), from 2002–2012 can be modeled by the function

    U(t)=−0.0374t3+0.5934t2−2.0553t+6.7478,

    where t=0 represents 2002, t=1 represents 2003, and so on. The model indicates a 7.7% level of unemployment in 2008 (t=6). Find another year after 2008 with about the same unemployment level. Round your answer to the nearest year. (The model was slightly modified from data obtained from www.multipl.com.)

Beyond the Basics

  1. Show that 12 is a root of multiplicity 2 of the equation

    4x3+8x2−11x+3=0.

  2. Show that −23 is a root of multiplicity 2 of the equation

    9x3+3x2−8x−4=0.

  3. Assuming that n is a positive integer, find the values of n for which each of the following is true.

    1. x+a is a factor of xn+an.

    2. x+a is a factor of xn−an.

    3. x−a is a factor of xn+an.

    4. x−a is a factor of xn−an.

  4. Use Exercise 63 to prove the following:

    1. 711−211 is divisible by 5.

    2. (19)20−(10)20 is divisible by 261.

  5. Synthetic division is a process of dividing a polynomial F(x) by x−a. Modify the process to use synthetic division so that it finds the remainder when the first polynomial is divided by the second.

    1. 2x3+3x2+6x−2;2x−1

      [Hint: 2x−1=2(x−12).]

    2. 2x3−x2−4x+1;2x+3

  6. Let g(x)=3x3−5cx2−3c2x+3c3. Use synthetic division to find each function value.

    1. g(−c)

    2. g(2c)

In Exercises 67–71, let f(x)=ax3+bx2+cx+d,a≠0 and g(x)=kx4+lx3+mx2+nx+p,k≠0. [Hint: Use the end behavior of polynomials and the Intermediate Value Theorem.]

  1. Show that f has exactly one real zero or three real zeros counting multiplicities.

  2. Show that if kp<0, then g has exactly two real zeros or exactly four real zeros counting multiplicities.

  3. Suppose dp<0 and kp<0. Show that the graphs of f and g must intersect.

  4. Give an example to show that the results of Exercise 69 may not hold if dp<0 and kp>0.

  5. Give an example of a fourth-degree polynomial that has no real zeros.

Critical Thinking / Discussion / Writing

    1. Use the fact that f(x)2x−6=12 (f(x)x−3) to explain how to use synthetic division by x−3 to find the quotient and remainder when f(x) is divided by 2x−6.

    2. Repeat part (a), replacing 2x−6 with ax+b, a≠0.

Getting Ready for the Next Section

In Exercises 73–76, use synthetic division to find each quotient and remainder.

  1. x4−x3+2x2+x−3 is divided by x−1.

  2. x3+x2−x+2 is divided by x+2.

  3. −x6+5x5+2x2−10x+3 is divided by x−5.

  4. x4+3x3+3x2+2x+7 is divided by x+1.

In Exercises 77–80, find all of the factors of each integer.

  1. 11

  2. 18

  3. 51

  4. 72

In Exercises 81–84, identify the degree, leading coefficient, and constant term for each polynomial.

  1. x2+2x−1

  2. −x3+x+7

  3. −7x10+3x3−2x+4

  4. 13x9+2x15+3x−11

In Exercises 85 and 86, identify the multiplicity of each zero.

  1. f(x)=(x−1)(x+2)2

  2. g(x)=x2(2−x)3

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