Polynomials: Modeling, Basic Rules, and the Factor Theorem

LAUNCH

A container manufacturer has just received a large order for metal boxes that must be able to hold 50 cubic inches. He plans to make these boxes out of rectangular pieces of metal 8 inches by 10 inches by cutting out squares from the corners and folding up the sides. He needs to know what size square he should cut out to achieve his goal. How would you use the given information to solve this problem?

We are assuming that, in planning to solve this problem, you immediately employed the use of your algebraic skills. Before we review this problem, we would like to point out that what you have just engaged in is the process of mathematical modeling where you attempted to model the essence of the problem by using mathematical concepts. In this case you most likely used a polynomial to model the problem. In fact, it was a cubic polynomial. Mathematical modeling is big business these days, and consultants are highly sought after to solve problems using such techniques. The general approach is to begin by finding a simple model for the problem, using polynomials if possible, since they are usually easy to work with. If the polynomial model does not fit the situation, you try to use other functions. We will have more to say about this in Chapter 9.

Getting back to our launch problem, let us see how algebra can be used to model the situation.

To help us visualize the situation, we use the helpful problem-solving strategy of drawing a diagram (see Figure 3.1 below). Of course, if you use this in your classroom, then cutting out the squares from an 8"× 10" piece of paper would demonstrate this very clearly.

x x

x x

10 – 2x

10 – 2x 8

10

10 – 2x 8 – 2x 8 – 2x

8 – 2x x

Fold up along the dotted lines

Our box Cut squares from corners

Figure 3.1

Notice that we have let x represent our unknown, the side of the square to be cut out in inches.

Then the dimensions of the box it forms have length: (10− 2x), width: (8 − 2x), and height x.

The volume of the box will therefore be length times width times height, or (10− 2x)(8 − 2x) x.

Since the manufacturer wants the volume of this box to be 50, we want to solve the equation x(8− 2x)(10 − 2x) = 50. We hope that this is the equation you arrived at as well. If we simplify

the expression on the left, we get the equation 80x− 36x²+ 4x³= 50. We notice that the left side of this equation is a polynomial. We are now interested in the solution. If we graph the curve y = 80x− 36x²+ 4x³ and restrict ourselves to x between 0 and 4 which are the only values of x which are physically possible in this problem situation, we get the following picture (Figure 3.2).

5 and 2. (Try to find the solutions using your calculator!)

Before getting into a deep discussion of finding roots of polynomials, we review the definition of a polynomial. This is probably the most misunderstood word in secondary school mathematics.

A polynomial in x, denoted by p(x), consists of one or more terms of the form cxⁿ where c is a constant, and n is a nonnegative integer. For example p(x) = 7 is a polynomial, since this can be written as 7x⁰. (It is also a monomial, but that doesn’t stop it from being a polynomial!) Each of the following are also polynomials: p(x) = 3x + 2, p(x) = 7x²+πx −√

2. But1

x = x⁻¹is not a polynomial since it has a negative exponent and the exponents in polynomials must be nonnegative integers.

Also √

x = x¹² is not a polynomial, since the exponent is fractional. When a value x = c makes p(x) = 0, we say that x = c is a root or a zero of p(x). Thus, zeroes of the polynomial p(x) = x²− 5x + 6 are x = 2 and x = 3, since both p(2) = 0 and p(3) = 0. We can talk about the roots of any function, f (x), regardless of whether or not it is a polynomial. These are simply the numbers that make f (x) = 0.

In the previous chapter we discussed the division algorithm for polynomials which explained that, if we had a polynomial, a(x), of degree n, and we divided it by a polynomial b(x) of smaller degree, then there would be a quotient q(x) and a remainder r (x) such that a(x) = b(x)q(x) + r (x), where the degree of r (x) is smaller than the degree of b(x). We gave some examples to illustrate the method of long division that would be used to find q(x) and r (x).

In this section we concentrate on the specific case when the divisor is a polynomial of the form x− c. As we shall soon see, this is a particularly important case to consider because it is related to finding the roots of polynomials. Thus, in this case, a(x) = (x− c)q(x) + r (x). Since our divisor is of degree one and our remainder must be of degree less than 1, it has to be of degree 0. Thus, it must be a constant. So, in this case we will simply write a(x) = (x− c)q(x) + r. Our first theorem is based on this idea and is a standard one in precalculus courses.

Theorem 3.1 When a polynomial p(x) is divided by x− c, the remainder, r, that you get, is p(c).

Note: It is important that the divisor be written in the form x− c.

Proof. When we divide p(x) by x− c, we get a quotient q(x) and a remainder r and p(x) = (x − c) q(x) + r. Now replace x by c and we get that p(c) = (0)q(c) + r = r. That is, the remainder r , when we divide p(x) by x− c, is p(c). 

Thus, if we divide the polynomial p(x) = x³− 3x²+ 4x− 8 by x − c = x − 1, the remainder will be p(1) or−6 since here, c is 1. If we divided the same polynomial by x + 2, which can be written as x− (−2), the remainder will be p(−2) or −36 since here, c = −2. Let us illustrate this second result in long division form. (A review of long division occurs in Section 2.7.)

Example 3.2 Show, using long division that, when we divide p(x) = x³− 3x²+ 4x− 8 by x + 2, we get a remainder of−36.

Solution. Here is the long division:

x² −5x +14

x + 2 x³ −3x² +4x −8 (Line a)

x³ +2x² (Line b)

−5x² +4x (Line c)

−5x² −10x (Line d) 14x −8 (Line e) 14x +28 (Line f)

−36

A corollary of Theorem 3.1, known as the Factor Theorem, is:

Corollary 3.3 If p(x) is a polynomial and if p(c) = 0, then x− c is a factor of p(x).

Proof. Since p(c) is 0, we have by the previous theorem, that the remainder when p(x) is divided by x− c, that is, r , is zero. Thus, our division algorithm statement, p(x) = (x − c)q(x) + r , now reads p(x) = (x− c)q(x). That is, x − c is a factor of p(x). We are done. To find q(x), the other factor, we simply divide p(x) by (x− c). After all, q(x) is the quotient! 

To illustrate, suppose that p(x) = x²− 4. Since p(2) = 0, x − 2 is a factor of p(x). In a similar manner, since p(−2) = 0, (x − −2), that is, x + 2 is a factor of p(x).

Let us illustrate this with another example.

Example 3.4 Find the roots of the polynomials (a) p(x) = x³− 2x²− 5x + 6

(b) q(x) = x³− 2x²+ 6x + 5 without the use of a calculator.

Solution. (a) By inspection, we see that x = 1 is a root of the first equation, since p(1) = 0. Thus, x− 1 is a factor of p(x). Now divide p(x) by x − 1 using long division and you find that the other factor is x²− x − 6. So p(x) = (x − 1)(x²− x − 6) = (x − 1)(x − 3)(x + 2). It follows that p(x) = 0 when x = 1, x = 3, and x =−2.

(b) Again, x = 1 makes q(x) = 0, and again x− 1 is a factor of q (x) . By long division we see that the other factor of q(x) is x²− x − 5. So q (x) = (x − 1)(x²− x − 5). Now q(x) = 0 when x = 1 or x²− x − 5 = 0, and this latter is zero when x = ^1±√₂²¹ by the quadratic formula.

There are some interesting factoring results that can be obtained by Corollary 3.3. We illustrate some of them.

Example 3.5 It is a common fact taught in secondary school that the expressions xⁿ− bⁿare always divisible by x− b. Show how this follows from our Corollary 3.3. More specifically, show that

xⁿ− bⁿ= (x− b)(xⁿ⁻¹+ xⁿ⁻²b + xⁿ⁻³b²+. . . + xbⁿ⁻²+ bⁿ⁻¹)

Solution. Let p(x) = xⁿ− bⁿ. Since p(b) = bⁿ− bⁿ= 0, by the above Corollary 3.3, x− b is a fac-tor of p(x). We can find the other facfac-tor by long division, or by synthetic division. (See the next section for a relatively complete discussion of synthetic division.) In fact, the other factor is (xⁿ⁻¹+ xⁿ⁻²b + xⁿ⁻³b²+. . . + xbⁿ⁻²+ bⁿ⁻¹.) Thus, p(x) = (x − b)(xⁿ⁻¹+ xⁿ⁻²b + xⁿ⁻³b²+ . . . xbⁿ⁻²+ bⁿ⁻¹), which is what we were trying to prove.

Let us show what this says in two special cases, n = 3 and n = 4. When n = 3, we have

x³− b³= (x− b)

x²+ xb + b²

(3.1) and, when n = 4, we have

x⁴− b⁴= (x− b)(x³+ x²b + xb²+ b³). (3.2)

You should verify that, if you multiply the expressions on the right side of the equality in both equations (3.1) and (3.2), we get the left sides of these equations, respectively.

Student Learning Opportunities

1 Find the remainder when 3x²− 4x + 1 is divided by x − 3.

2 Find the remainder when x⁴⁴+ 3x²³− 2 is divided by x + 1.

3 Find all roots of the following equations by first observing that, in each case, there is a simple number, either 0, 1, or 2 that satisfies each equation.

(a) x²− x + x³− 1 = 0.

(b) x³− 8x + 7 = 0.

(c) x³− 5x²+ 6x = 0.

(d) 2x³− 11x²+ 17x− 6 = 0.

4 The polynomial p(x) has the property that p(2) = p(3) = p(−1) = 0. Find two such polynomi-als. Find a third such polynomial that also makes p(4) = 14.

5 Show that a³+ b³ can be factored into (a + b)(a²− ab + b²). Show that (a⁵+ b⁵) = (a + b) (a⁴− a³b + a²b²− ab³+ b⁴). Generalize to finding factors of aⁿ+ bⁿwhen n is a positive odd integer.

6 Factor each of the following completely: You may need to use the results of the previous problem.

(a) x⁴− 1 (b) y³+ 8

(c) a⁶− b⁶ (d) 8x³+ 27y³

(e) 16x⁶− 81y⁶

7 Find all real solutions of the polynomial equations below by factoring.

(a) x⁴− 2x³+ 3x²= 0 (b) 2x³− x²− 18x + 9 = 0

(c) x⁶− 2x³=−1

8 For which values of m is x− 1 a factor of x³+ m²x²+ 3mx + 1?

9 (C) Show that the polynomial p(x) = x⁵+ b⁵always has a root and use the root to factor p(x).

10 If two factors of the polynomial 2x³− hx + k = 0 are x + 2 and x − 1, what are the values of h and k?

11 Find 4 different factors all in terms of x and y, that when multiplied equal 8^2x− 27^2y. [Hint: First write this as a²− b² and factor. Then each factor will be the sum or difference of two cubes.]

12 What is the sum of the prime factors of 2¹⁶− 1?

13 If the polynomial p1(x) = ax²+ bx + c has roots r and s, show that the polynomial that has roots 1

r and 1

s is p2(x) = cx²+ bx + a.

14 (C) After doing the previous problem, one of your students asks if it is true that if we have a cubic polynomial with roots r , s, and t, then a polynomial that has roots 1

r, 1 s, and 1

t is just the polynomial with the coefficients reversed? How do you respond? Justify your answer.

15 Model the following problem using polynomials: A grain silo consists of a main part which is a cylinder, topped by a hemispherical roof. Suppose the height of the cylindrical portion is to be 50 feet and the volume of the silo, including the hemisphere on top is 20, 000 cubic feet.

What is the radius of the cylindrical portion?(The volume of a cylinder is given by V =πr²h and the volume of a sphere is V = ⁴₃πr³. For more information on volume, see Chapter 4.) 16 The United States Post Office will not accept a box whose girth (distance around) plus length

is more than 108 inches. Suppose that we want to build a container with a square base whose volume is as large as possible and whose girth is precisely the maximum 108 inches. Model this situation by letting x be the length of the side of the square base, and expressing the volume in terms of x. Then use your calculator to estimate the dimensions of the box.

17 If p is an odd number greater than 1, show that ( p− 1)

p−1 2



− 1 is divisible by p − 2. [Hint:

Let p = 2n + 1.]

18 Show that x−a is a factor of x²(a− b) + a²(b− x) + b²(x− a). [Hint: Call the given expression p(x).]

19 Show that x− c is a factor of (x − b)³+ (b− c)³+ (c− x)³.

20 When the polynomial p(x) = 2x³+ ax²+ b is divided by x− 1, the remainder is 1, but when it is divided by x + 1, the remainder is−1. If possible, find the values of a and b that will make this true, or prove that it is impossible.