Page 1

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lyn mial Functi lyn mial Functi nsns

00 00 00 00

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In this presentation, we will learn all the theorems we need to know about finding the zeros of a polynomial function.

1. Fundamental Theorem of Algebra

2. Factor Theorem

3. Linear Factorization Theorem

4. Remainder Theorem

5. Conjugates Pairs Theorem

6. Rational Zeros Theorem

7. Descartes’ Rule of Signs

8. Boundedness (a.k.a. Bounds on Zeros) Theorem

9. Intermediate Value Theorem

Table of Contents

Page 3Fundamental Theorem of Algebra

Fundamental Theorem of Algebra (FTA):Every polynomial function of degree n (n 1) has at least one complex zero. In fact, there are exactly n complex zeros (not necessarily distinct).

Ex. 1: f(x) = x3 – 2x2 – 9x + 18 has 3 complex zeros since the degree is 3.

However, FTA doesn’t tell us what the 3 complex zeros are nor whether they are distinct or not. To find these zeros, we either need to factor f(x) or use synthetic division. In this example, f(x) can be factored by grouping:

f(x) = x3 – 2x2 – 9x + 18 = x2(x – 2) – 9(x – 2) = (x2 – 9)(x – 2) = (x + 3)(x – 3)(x – 2) which gives us –3, 3, and 2 as the 3 (complex) zeros.

Ex. 2: f(x) = x4 – 1 has 4 complex zeros since the degree is 4.

To find the 4 complex zeros, we can factor f(x) as follows:

f(x) = x4 – 1 = (x2 – 1)(x2 + 1) = (x – 1)(x + 1)(x – i)(x + i). So the 4 complex zeros are 1, –1, i and –i.

Since 1, –1, i and –i are zeros of f(x), we have f(1) = 0, f(–1) = 0, f(i) = 0 and f(–i) = 0, and vice versa. This leads to the Factor Theorem, as we will see in the next slide.

Note: A complex number is a number can be written in a + bi form where a, b are real and i = . A real number is also considered a complex number. By letting b = 0, we have, for example, 3 = 3 + 0i.

1

Note: In the previous context, you might have heard we said that x2 + 1 is not factorable. That means x2 + 1 is not factorable over the real numbers. It’s, however, factorable over the complex numbers.

Page 4Factor Theorem

Factor Theorem:If f(c) = 0, then x – c is a factor of f(x) and vice versa.

Ex. 1: f(x) = x4 – 1

Since f(1) = 0, f(–1) = 0, f(i) = 0 and f(–i) = 0, we have x – 1, x + 1, x – i, and x + i as factors of f(x). And vice versa, since f(x) = x4 – 1 = (x – 1)(x + 1)(x – i)(x + i) f(1) = 0, f(–1) = 0, f(i) = 0 and f(–i) = 0.

Note: If f(x) is a polynomial function and f(c) = 0 for some complex number c, the following statements are equivalent:

1. c is a zero of f(x).2. c is a root of f(x).3. c is a solution of the equation f(x) = 0.4. If c is real, c is an x-intercept of the graph of y = f(x). 5. x – c is a factor of f(x).

Example: f(x) = x4 – 1 We know f(1) = 0, f(–1) = 0, f(i) = 0 and f(–i) = 0, so, 1, –1, i, and –i are the 4 complex zeros(or roots) of f(x). However, only 1 and –1 are the x-intercepts of the graph of y = f(x).

Page 5Linear Factorization Theorem

Linear Factorization Theorem:Every polynomial function f of degree n (n 1) can be factored into n linear factors (not necessarily distinct) of the form f(x) = an(x – r1)(x – r2)...(x – rn).

This theorem is an extension of the factor theorem. The distinction is: with the factor theorem, we only need to know one complex zero of a polynomial function f(x), whereas, with the linear factorization theorem, we need to know all complex zeros of f(x). For example, if we only know that –1 is a zero of f(x) = x4 – 1, we can say x + 1 is a factor of f(x). However, in general, we can’t factor f(x) into its linear factorization form unless we know all the zeros. On the other hand, since we do know all 4 zeros of f(x) = x4 – 1: 1, –1, i, –i, we can factor f(x) as f(x) = (x – 1)(x + 1)(x – i)(x + i).

Recall that we got the factored form of f(x) = x4 – 1 first, and it’s because of the factored form we know all its zeros. However, for many polynomials, it’s the other way around. We must know all the zeros of a polynomial first, before we can write it in linear factorization form.

Ex. 1: Write f(x) = 2x3 + 7x – 9 in linear factorization form.

Notice that f(x) above is a cubic polynomial where factor by grouping is not feasible (we don’t have 4 terms). Before we show you how to do the above problem, let’s revisit f(x) = x4 – 1.

f(x) = x4 – 1 = (x2 – 1)(x2 + 1) = (x – 1)(x + 1)(x – i)(x + i)

These two factors are not linear factors, since the degree of each factor is 2 (they are called quadratic factors).

These four factors are linear factors, since the degree of each factor is 1. So we say this is the linear factorization form of f(x).

Page 6Implication of Factor Theorem and Linear Factorization Theorem

What do the factor theorem and linear factorization theorem imply?

Two things:1. If we are given all the linear factors or the linear factorization form of a polynomial

function f(x), we know all the complex zeros of f(x).

2. If we are given all the complex zeros of a polynomial function f(x), we know all the linear factors of f(x), but not necessary its linear factorization form.

Ex. 1: Given that f(x) = 2(x – 1)(x + 1)(x – i)(x + i). What are the complex zeros of f(x)?Answer: 1, –1, i, –i.

Ex. 2: Given that f(x) has 1, –1, i, –i as all the complex zeros, what is f(x)?Answer: Many answers are possible: f(x) = x4 – 1, f(x) = 2(x4 – 1), f(x) = 3(x4 – 1), etc.

Let’s go back to this problem: Write f(x) = 2x3 + 7x – 9 in linear factorization form.

Hint: What is an easy zero you can spot? Answer: 1 since 2(1)3 + 7(1) – 9 = 0.

How do we find the linear factorization form? Answer: Find all zeros and take advantage that 1 is a zero.1 2 0 7 9

2 2 9

2 2 9 0

2

2

2 2 9 0

2 2 4(2)(9)

2(2)

2 68 2 2 17 1 17

4 4 2 2

x x

x

ix i

Even though the zeros are 1, and .However, the linear factorization form is not f(x) = . It is, rather,

f(x) = since the leading coefficient is 2.

1712 2 i 171

2 2 i

17 171 12 2 2 2( 1)( ( ))( ( ))x x i x i

17 171 12 2 2 22( 1)( ( ))( ( ))x x i x i

Page 7Remainder Theorem

Remainder Theorem:If a polynomial function f(x) is divided by x – c, the reminder is f(c).

Ex. 1: What is the remainder when x4 – 3x2 + 2x – 2 is divided by x – 2?

Answer: Let f(x) = x4 – 3x2 + 2x – 2. By the remainder theorem, when f(x) is divided by x – 2, the remainder is f(2). So f(2) = (2)4 – 3(2)2 + 2(2) – 2 = 16 – 12 + 4 – 2 = 6. That is, the remainder is 6.

Are there other ways to verify it? Answer: Yes, by long division or by synthetic division. xxx

x

x

xx

xx

xx

xx

xx

xxxxx42

6

84

24

2

2

42

32

2

22302

23

2

2

23

23

34

234

By synthetic division: 2 1 0 3 2 2

2 4 2 8

1 2 1 4 6

We can see that there are 3 ways to answer questions like the one above: 1. By long division (not suggested, since it’s too time

consuming).2. By synthetic division.3. By remainder theorem.

When do we use synthetic division? When do we use remainder theorem?

By Long Division:

remainder

remainder

Page 8

When do we use synthetic division? When do we use remainder theorem?

Answer: Most of the time, it doesn’t matter. However, sometimes, one way is better than the other.

Ex. 1: What is the remainder when 17x3 +52x2 + 23x + 62 is divided by x + 3? By synthetic division: By remainder theorem:

f(–3) = 17(–3)3 + 52(–3)2 + 23(–3) + 62= 17(–27) + 52(9) – 69 + 62= –459 + 468 – 69 + 62= 9 – 7 = 2

3 17 52 23 62

51 3 60

17 1 20 2

In the case above, synthetic division is better than remainder theorem. In general, synthetic division is better when f(x) (i.e., the dividend) has rather large coefficients.

Ex. 2: What is the remainder when 2x10 + 3x3 – 4 is divided by x + 1? By synthetic division: By remainder theorem:

f(–1) = 2(–1)10 + 3(–1)3 – 4= 2(1) + 3(–1) – 4= 2 – 3 – 4 = –5

1 2 0 0 0 0 0 0 3 0 0 4

2 2 2 2 2 2 2 1 1 1

2 2 2 2 2 2 2 1 1 1 5

In the case above, remainder theorem is better than synthetic division. In general, remainder theorem is better when f(x) (i.e., the dividend) has rather large degree but not so many terms, and the divisor is either x + 1 or x – 1.

Remainder Theorem (cont’d)

Page 9

What does the remainder theorem tell?

Remainder Theorem (cont’d) and Conjugates Pairs Theorem

Answer: 1. It tells the remainder when a polynomial f(x) is divided by a divisor of the form x – c. That

is, to get the remainder, plug c into f(x) and evaluate f(c). 2. It tells whether x – c is a factor of f(x) or not. That is, if f(c) = 0, then x – c is a factor of

f(x). Otherwise, it’s not.

Ex. 1: Is x – 1 a factor of f(x) = 2x10 + 3x3 – 5? Solution: Since f(1) = 2(1)10 + 3(1)3 – 5 = 2 + 3 – 5 = 0, so x – 1 is a factor of f(x).

Ex. 2: Is x + 1 a factor of f(x) = 2x10 + 3x3 – 5? Solution: Since f(–1) = 2(–1)10 + 3(–1)3 – 5 = 2 – 3 – 5 = –6 0, so x + 1 is not a factor of f(x).

Conjugates Pairs Theorem (CPT):If a polynomial function f(x) has real coefficients only and a + bi is a zero, then so is a – bi. [That is, imaginary zeros must come in conjugates pairs.]

If 2 – i is a zero of a polynomial function f(x) with real coefficients, what must be another zero? 2 + i

If 3i is a zero of f(x) with real coefficients, what must be another zero? –3i

Page 10Conjugates Pairs Theorem (cont’d)

Ex. 1: Given that f(x) = x4 – 6x3 + 9x2 + 6x – 10 has a complex zero 3 + i, find the other complex zeros and write f(x) in its linear factored form.Solution: Again, take advantage that 3 + i is zero and its implication of 3 – i being also a zero (and yes, we can use imaginary zeros in the synthetic divisions too.)

The CPT also applies to irrational zeros of the form (not mentioned in the book):

If f(x) has rational coefficients only and is a zero, then so is , where a, b and c are rational numbers. [That is, irrational zeros of this form occur in conjugates pairs.]

If is a zero of a polynomial function f(x) with rational coefficients, what must be another zero? ______

If is a zero of f(x) with rational coefficients, what must be another zero? ____

Conjugates Pairs Theorem (CPT):If a polynomial function f(x) has real coefficients only and a + bi is a zero, then so is a – bi. [That is, imaginary zeros must occur in conjugates pairs.]

a b c

a b c a b c

2 3 5

4 6

Note: A rational number is a real number that can be expressed in a/b form where a, b are integers (b 0). That is, a rational number is simply a fraction. If a real number can’t be expressed in a/b form, it’s irrational

2 3 5

4 6

3 1 6 9 6 10

3 10 3 10

1 3 1 3 0

i

i i

i i

3 1 3 1 3

3 0 3

1 0 1 0

i i i

i i

2

2

1 0

1

1

x

x

x

(3 + i)(–3 + i) = (i + 3)(i – 3) = i2 – 9 = –1 – 9 = –10

(3 + i)(3 – i) = 9 – i2 = 9 – (–1) = 10

The other 3 zeros are 3 – i, 1 and –1, and the factored form is

f(x) = (x – (3 + i))(x – (3 – i))(x – 1)(x + 1)

Page 11

Note:

Conjugates Pairs Theorem (cont’d)

The author of our textbook mentions CPT for imaginary zeros only and he fails to mention CPT for irrational zeros. It is not because he doesn’t know it (we bet everything he does). It is also not the issue that imaginary zeros require f(x) to have real coefficients only whereas irrational zeros require f(x) to have the rational coefficients only. (After all, all polynomial functions we’ve seen so far have integer coefficients only).

Conjugates Pairs Theorem (CPT):If a polynomial function f(x) has real coefficients only and a + bi is a zero, then so is a – bi. [That is, imaginary zeros must come in conjugates pairs.]

If f(x) has rational coefficients only and is a zero, then so is , where a, b and c are rational numbers. [That is, irrational zeros of this form come in conjugates pairs.]

a b c a b c

3 23

3

3

2 0

2

2

x

x

x

It is because, unlike imaginary zeros which always occur in conjugate pairs, irrational zeros do not. For example, it’s not difficult to see that is a zero of f(x) = x3 – 2. However, it doesn’t imply that will be also a zero. In fact, it isn’t: .

3 233 3( 2) ( 2) 2 2 2 4 0f

So why not—after all, is irrational and is its conjugate? It’s because is not of the form _______. However, if a zero is of the form , then its conjugate will be also a zero.

Ex. 1: Given f(x) = x3 – 7x2 + 7x – 1 has as a zero. Find the other two zeros without doing extensive algebraic work.Solution: By CPT, is also a zero, and by inspection, 1 is the remaining zero since 13 – 7(1)2 + 7(1) – 1 = 0.

3 2 3 2 3 2a b c

3 2 2

a b c

3 2 2

Page 12Rational Zeros Theorem

Rational Zeros Theorem (RZT):The possible rational zeros of a polynomial function f of degree n 1 with integer coefficients are of the form p/q, where p is a factor of a0, the constant term, and q is a

factor of an, the leading coefficient.

f(x) = an xn + an–1xn–1 + ··· + a1x + a0

leading coefficient constant term

Yes! When you apply the RZT, you can ignore these terms.

Ex. 1: f(x) = x3 – 2x2 – 9x + 18

factors of 18 1, 2, 3, 6, 9, 181, 2, 3, 6, 9, 18

factors of 1 1

p

q

12

factors of 2 1, 21, , 2

factors of 2 1, 2

p

q

factors of 6 1, 2, 3, 61, 2, 3, 6

factors of 1 1

p

q

Ex. 2: f(x) = x3 + 6x2 + 11x + 6

Ex. 3: f(x) = 2x3 + 7x2 – 2

Page 13Rational Zeros Theorem (cont’d)

Rational Zeros Theorem (RZT):The possible rational zeros of a polynomial function f of degree n 1 are of the form p/q, where p is a factor of a0, the constant term, and q is a factor of an, the leading coefficient.

What does the RZT tell? What it doesn’t tell?It tells what numbers we need to try in the synthetic division. However, it doesn’t tell which ones work unless we carry out the synthetic division. Also, the theorem only says possible rational zeros, which means they might or might not be zeros of the polynomial—it’s not a guarantee. In the worst case scenario, it can be the case that none of the listed potential rational zeros is actually a zero.

Ex. 1: f(x) = x3 – 12Though we can list ±1, ±2, ±3, ±4, ±6, and ±12 as the potential rational zeros. However, f(x) has two imaginary zeros and one real zero and unfortunately, this real zero is , which is irrational.

Why p/q (i.e., factors of the constant term over factors of the leading coefficient)?

Let’s use Ex. 3: f(x) = 2x3 + 7x2 – 2

If the number we try is p/q: say ½ If the number we try is NOT p/q: say –312 2 7 0 2

1 4 2

2 8 4 0

If we use a number from p/q, there is a chance it will make this number 0. In this case, it is 0!

3 2 7 0 2

6 3 9

2 1 3 7

Since this number must be a multiple of 3,but this number isn’t. There is no way theycan up to 0!

121, , 2

p

q

Note:3

3

3

12 0

12

12

x

x

x

3 12

Page 14Descartes’ Rule of Signs

Descartes’ Rule of Signs (DRS):1. The number of positive real zeros of f is either the number of sign changes of the

coefficients (from one to the next) of f(x) or that number minus an even number.2. The number of negative real zeros of f is either the number of sign changes of the

coefficients (from one to the next) of f(–x) or that number minus an even number.

What does the DRS tell?1. It tells the possible number of positive real zeros,2. It tells the possible number of negative real zeros, and hence,3. It tells the possible number of imaginary zeros.

Number of

Pos. Real Zeros

Neg. Real Zeros

Imaginary Zeros

3 2 0

3 0 2

1 2 2

1 0 4

Ex. 1: f(x) = x5 + 2x4 – 3x3 + 4x2 + 5x – 6

Solution: Since f(x) has 3 sign changes, by DRS, f(x) has 3 or 1 positive real zeros. f(–x) = (–x)5 + 2(–x)4 – 3(–x)3 + 4(–x)2 + 5(–x) – 6

= –x5 + 2x4 – 3(–x3) + 4x2 – 5x – 6 = –x5 + 2x4 + 3x3 + 4x2 – 5x – 6

Since f(–x) has 2 sign changes, by DRS, f(x) has 2 or 0 negative real zeros.

Why minus an even number?From the above example, even there are 3 sign changes in f(x) and 2 sign changes in f(–x), we can’t say there are exactly 3 pos. and 2 neg.real zeros. Otherwise we are ruling out imaginary zeros and they do exist sometimes. Furthermore, since imaginary zeros must occur in conjugate pairs, i.e. there must be an even number of imaginary zeros, this is why we have to reduce the number of sign changes by an even number.

Page 15Descartes’ Rule of Signs (cont’d)

What the DRS doesn’t tell?It doesn’t tell exactly how many of each kind (positive real, negative real, and imaginary). It also doesn’t tell what the zeros are, unless we use numbers from the RZT and try them in synthetic division.

On the other hand, it is most handy when there are no sign changes. For example, f(x) = x3 + 6x2 + 11x + 6 has no positive real zeros since there are no sign changes! So use negative numbers for synthetic divisions.

A shortcut for f(–x):A shortcut for evaluating and simplifying f(–x) (without replacing x by –x) is:1. Keep the sign of the coefficient if it is a coefficient of x to an even power.2. Change the sign of the coefficient if it is a coefficient of x to an odd power.

Ex. 1: Given f(x) = x3 + 6x2 + 11x + 6, what is f(–x)?Solution: f(–x) = –x3 + 6x2 – 11x + 6

A misconception:A misconception for some students is: when they are trying a negative number in synthetic division, they think they should use the coefficients from the f(–x). However, that isn’t the case and should never be the case. We still use the coefficients of f(x) when we are trying negative numbers in synthetic division. We only use the coefficients of f(–x) for determining the possible number of negative real zeros, not for finding the actual zeros.

Note (the long way): f(–x) = (–x)3 + 6(–x)2 + 11(–x) + 6

= –x3 + 6x2 – 11x + 6

Note:

Using the coefficients of f(x), we see that –1 is a zero of f(x). However, using coefficients from f(–x), we will miss the fact that –1 is a zero:

–1 1 6 11 6 –1 –5 –6 1 5 6 0

–1 –1 6 –11 6 1 –7 –18 –1 7 –18 –12

Page 16Intermediate Value Theorem

Intermediate Value Theorem: If a < b and f(a) and f(b) are of opposite signs, then there is at least one real zero of f

between a and b.

Notes: 1. In either case 1 or case 2, there can be more than one real zero between a and b. (See right.)2. If both f(a) and f(b) are positive (or both are negative), then

there might be or might not be a zero between a and b. We just can’t tell. (See right.)

Ex. 1: The table on the right shows some function values of f(x) = x3 + 4x2 – 3x – 4. By the Intermediate Value Theorem, we can conclude that f(x) has a zero between which two pairs of consecutive integers: between –1 and 0; and between 1 and 2.

a bx

y

a bx

y

a bx

y

Why? Well, since f is a polynomial function, it is continuous everywhere (i.e., there is no gaps or holes). It must cross the x-axis if this is f(a) and f(b) are of opposite signs.

Case 1: f(a) is negative and f(b) is positive. Case 2: f(a) is positive and f(b) is negative.

a bx

y

There must be a real zero between a and b(a, f(a))

(b, f(b))

a bx

y

There must be a real zero between a and b

(a, f(a))

(b, f(b))

x f(x)

–2 10

–1 2

0 –4

1 –2

2 14

Page 17Intermediate Value Theorem (cont’d)

Intermediate Value Theorem: If a < b and f(a) and f(b) are of opposite signs, then there is at least one real zero of f between a and b.

Ex. 1: Given that polynomial function f(x) has f(2) = –3, f(3) = 1, and f(4) = 5, we can conclude f(x) must have a zero between the following pairs of integers except:a) 2 and 3 b) 2 and 4 c) 1 and 4 d) 3 and 4

Ex. 2: Given that polynomial function f(x) has f(2) = –3, f(3) = 1, and f(4) = –5, which of the following is the correct conclusion about the number of zeros of f(x) between 2 and 4?a) There are no zeros since f(2) and f(4) are both negative.b) There must be exactly 1 zero.c) There must be exactly 2 zeros.d) There must be at least 2 zeros.

BoundsAn upper bound for a set S is a number that is the largest element of the set; a lower bound is a number that is the smallest element of the set.

If S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, an upper bound can be 11 and a lower bound can be 0. However, upper bounds and lower bounds are not unique. Other upper bounds for S include 23, 45, 678 and 10 (where 10 is called the least upper bound). Other lower bounds include ½, –34, –567 and 1 (where 1 is called the greatest lower bound).

Page 18Boundedness Theorem

Boundedness Theorem (BT):Suppose f(x) (with a positive leading coefficient) is divided by x – c using synthetic division,1. if c > 0 and the numbers in the last row of the synthetic division are either positive or

zero, then c is an upper bound for the real zeros of f, 2. if c < 0 and the numbers in the last row are alternately positive and negative (zero

entries can be treated as positive or negative to your advantage), then c is a lower bound.

2 8 7 6 3

16 18 24

8 9 12 27

1 8 7 6 3

8 1 5

8 1 5 2

1 8 7 6 3

8 15 9

8 15 9 6

2 8 7 6 3

16 46 80

8 23 40 77

Ex 1: Given f(x) = 8x3 + 7x2 – 6x – 3 and the following synthetic divisions

From the above synthetic divisions and by the Boundedness Theorem, give the lower bound(s) of f(x): –2 and the upper bound(s) of f(x): 1, 2

Note:This theorem is in many other textbooks, but it’s not in ours. A comparable theorem in our textbook is called the Bounds on Zeros Theorem. We find that to be difficult to explain and also not easy to memorize. Hence, we will not explain it here.

If a positive number is an upper bound for the real zeros of a polynomial function, it means we don’t need to try any numbers higher than it. Similarly, if a negative number is a lower bound, it means we don’t need to try any numbers lower than it. Also, in other context, an upper bound doesn’t need to be positive nor a lower bound be negative. However, in the context of finding upper and lower bounds of real zeros of a polynomial function using the BT, an upper bound must be positive and a lower bound be negative.

Page 19Boundedness Theorem (cont’d)

Boundedness Theorem (BT):Suppose f(x) (with a positive leading coefficient) is divided by x – c using synthetic division,1. if c > 0 and the numbers in the last row of the synthetic division are either positive or zero, then

c is an upper bound for the real zeros of f, 2. if c < 0 and the numbers in the last row are alternately positive and negative (zero entries can be

treated as positive or negative to your advantage), then c is a lower bound.

Ex. 1: Let f(x), g(x) and h(x) be polynomial functions all with a positive leading coefficient. The test for whether 3 is an upper bound and –2 a lower bound of the zeros of these functions are given in the synthetic divisions below (where the symbol (+) means a positive number and (–) means negative):

1 2 3 4

5 6 7

3

( ) ( ) ( ) ( )

F F F F

F F F

1 2 3 4

5 6 7

3

( ) ( ) 0 ( )

G G G G

G G G

1 2 3 4

5 6 7

3

( ) ( ) ( ) ( )

H H H H

H H H

By the BT, we can conclude that 3 is an upper bound for the zeros of function(s): f(x) and g(x)

1 2 3 4

8 9 10

2

( ) ( ) ( ) ( )

F F F F

F F F

1 2 3 4

8 9 10

2

( ) 0 ( ) ( )

G G G G

G G G

1 2 3 4

8 9 10

2

( ) 0 ( ) ( )

H H H H

H H H

By the BT, we can conclude that 2 is a lower bound for the zeros of function(s): f(x) and g(x)

Note:We can only apply the theorem for polynomial functions with a positive leading coefficient. However, if a polynomial function has a negative leading coefficient, we can multiply the polynomial by –1, so that it has a positive coefficient, and hence the theorem is applicable. Multiplying a polynomial by –1 will affect the graph of the polynomial (flipping the graph over the x-axis), but will not affect its real zeros (i.e., the x-intercepts).

-1.5 -1 -0.5 0.5 1 1.5

-15

-10

-5

5

10

15 y = f(x) = 8x3 + 7x2 – 6x – 3

y = –f(x) = –8x3 – 7x2 + 6x + 3

We just consider this 0 as (–).

If we consider this 0 as (+), there’s no change from first to second; if we consider this 0 as (–), there’s no change from second to third.

Page 20Boundedness Theorem (cont’d)

24,12,8,6,4,32,1 q

p

10 1 0 3 6 28 24

10 100 ( ) ( ) ( )

1 10 97 ( ) ( ) ( )

In the context of finding zeros, the numbers we use should be from the RZT. However, in the context of finding an upper bound for the zeros, it doesn’t need to be from the RZT. For example, we can pick 5 and 20 (notice they are not from the RZT) and determine which is an upper bound for the zeros.

Ex. 1: f(x) = x5 – 3x3 + 6x2 – 28x + 24

By RZT:

)()()(397201

)()()(40020

2428630120

5 1 0 3 6 28 24

5 25 ( ) ( ) ( )

1 5 22 ( ) ( ) ( )

3 1 0 3 6 28 24

3 9 18 36 24

1 3 6 12 8 0

“Good” Bounds vs. “Bad” BoundsAs we can see on the right, both 5 and 20 are upper bounds of the real zeros of f(x). However, 5 is a much better upper bound than 20 since knowing 5 is an upper bound, we don’t need to try 6, 8, 12 and 24. Knowing 20 is an upper bound, we just know we don’t need to try 24.

Similarly, for lower bounds, both –10 and –3 are lower bounds of the real zeros of f(x). However, –3 is a much better lower bound than –10 since knowing –3 is a lower bound, we don’t need to try –4, –6, –8, –12 and –24. Knowing –10 is a lower bound, we just know we don’t need to try –12 and –24.

Notice –3 is probably the best case scenario, not only it’s a lower bound for the zeros, it’s also a zero!

Page 21Let’s Find the Zeros! . . . And Final Comments

24,12,8,6,4,32,1 q

pEx. 1: f(x) = x5 – 3x3 + 6x2 – 28x + 24

By RZT:

To find all the complex zeros of a polynomial function (especially when the degree ≥ 3 and factoring doesn’t seem feasible), we must use synthetic division. If a number we try in the synthetic division is a zero (regardless it’s by careful analysis or simply by luck), it reduces the degree of the polynomial by 1. If the degree is still ≥ 3, we can use the last row of the synthetic division (by omitting the last 0) as the new coefficients to find the remaining zeros and we can also apply theorems such as the RZT to the new coefficients.

The goal is to reduce the degree to 2 so that we can treat it as a quadratic equation and solve it either by factoring or by quadratic formula. For example, the above polynomial is degree 5, we need to find 3 zeros (by using synthetic division) first, and for the last 2 zeros, we can always use quadratic formula if it’s not factorable.

3 1 0 3 6 28 24

3 9 18 36 24

1 3 6 12 8 0

1, 2, 4, 8p

q Re-apply RZT:

1 1 3 6 12 8

1 2 4 8

1 2 4 8 0

2 1 2 4 8

2 0 8

1 0 4 0

2

2

4 0

4

4

2

x

x

x

x i

Zeros: –3, 1, 2, ±2i