The Real Zeros of a Polynomial Function

REMAINDER THEOREMLet f be a polynomial function. If f (x) is divided by x – c, then the remainder is f (c).

Let’s look at an example to see how this theorem is useful.

1232 23 xxxxf

-2 2 -3 2 -1 using synthetic division let’s divide by x + 2

2 -7 16 -33

-4 14 -32

the remainderFind f(-2)

3312223222 23 f

So the remainder we get in synthetic division is the same as the answer we’d get if we put -2 in the function.

The root of x + 2 = 0 is x = -2

161835343 23 f

FACTOR THEOREM

Let f be a polynomial function. Then x – c is a factor of f (x) if and only if f (c) = 0

If and only if means this will be true either way:

1. If f(c) = 0, then x - c is a factor of f(x)

2. If x - c is a factor of f(x) then f(c) = 0.

12 -51 153-3 -4 5 0 8

-4 17 -51 161

?854 offactor a 3 Is 23 xxx

Try synthetic division and see if the remainder is 0

NO it’s not a factor. In fact, f(-3) = 161

We could have computed f(-3) at first to determine this. Not = 0 so not a factor

Opposite sign goes

here

Our goal in this section is to learn how we can factor higher degree polynomials. For example we want to factor:

23 234 xxxxxfWe could randomly try some factors and use synthetic division and know by the factor theorem that if the remainder is 0 then we have a factor. We might be trying things all day and not hit a factor so in this section we’ll learn some techniques to help us narrow down the things to try.

The first of these is called Descartes Rule of Signs named after a French mathematician that worked in the 1600’s.

Rene Descartes 1596 - 1650

Descartes’ Rule of Signs

Let f denote a polynomial function written in standard form.

The number of positive real zeros of f either equals the number of sign changes of f (x) or else equals that number less an even integer.

The number of negative real zeros of f either equals the number of sign changes of f (-x) or else equals that number less an even integer.

23 234 xxxxxfstarts Pos. changes Neg. changes Pos.

1 2

There are 2 sign changes so this means there could be 2 or 0 positive real zeros to the polynomial.

Descartes’Rule of Signs

Let f denote a polynomial function written in standard form.

The number of positive real zeros of f either equals the number of sign changes of f (x) or else equals that number less an even integer.

The number of negative real zeros of f either equals the number of sign changes of f (-x) or else equals that number less an even integer.

23 234 xxxxxf

simplify f(-x)There are 2 sign changes so this means there could be 2 or 0 negative real zeros to the polynomial.

23 234 xxxxxf

23 234 xxxxxf

starts Pos. changes Neg. changes Pos.1 2

243 45 xxxf

Use Descartes’ Rule of Signs to determine how many positive and how many negative real zeros the polynomial may have.

starts Neg. changes Pos.

There is one sign change so there is one positive real zero.

1

243243 4545 xxxxxf

starts Pos. Never changes

There are no negative real zeros.

Counting multiplicities and complex (imaginary) zeros, the total number of zeros will be the same as the degree of the polynomial.

Descartes rule says one positive and no negative real zeros so there must be 4 complex zeros for a total of 5. We’ll learn more complex zeros in Section 4.7.

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Back to our original polynomial we want to factor:

We’d need to try a lot of positive or negative numbers until we found one that had 0 remainder. To help we have:

The Rational Zeros Theorem

What this tells us is that we can get a list of the POSSIBLE rational zeros that might work by taking factors of the constant divided by factors of the leading coefficient.

1, 2

1

1

Factors of the constant

Factors of the leading coefficient

Both positives and negatives would work for factors

23 234 xxxxxf

1, 2

1

So a list of possible things to try would be any number from the top divided by any from the bottom with a + or - on it. In this case that just leaves us with 1 or 2

23 234 xxxxxf Let’s try 1

1 2 -1 -21 1 1 -3 -1 2

1 2 -1 -2 0

YES! It is a zero since the remainder is 0

Since 1 is a zero, we can write the factor x - 1, and use the quotient to write the polynomial factored.

221 23 xxxx We found a positive real zero so Descartes Rule tells us there is another one

1, 2

1

We could try 2, the other positive possible. IMPORTANT: Just because 1 worked doesn’t mean it won’t work again since it could have a multiplicity. Let’s try 1 again,

but we try it on the factored version for the remaining factor (once you have it partly factored use that to keep going---don't start over with the original).

1 3 2 1 1 2 -1 -2

1 3 2 0 YES! the remainder is 0

221 23 xxxxxf

Once you can get it down to 3 numbers here, you can put the variables back in and factor or use the quadratic formula, we are done with trial and error.

12232 xxxx

Let’s take our polynomial then and write all of the factors we found:

23 234 xxxxxf

121 2 xxx

In this factored form we can find intercepts and left and right hand behavior and graph the polynomial

Left & right hand behavior

Touches at 1 crosses at -1 and -2.

Plot intercepts

“Rough” graph

There ended up being two positive real zeros, 1 and 1 and two negative real zeros, -2, and -1.

Let’s try another one from start to finish using the theorems and rules to help us.

92729132 234 xxxxxfUsing the rational zeros theorem let's find factors of the constant over factors of the leading coefficient to know what numbers to try.

1, 3, 9

1, 2

factors of constant

factors of leading coefficient

So possible rational zeros are all possible combinations of numbers on top with numbers on bottom:

2

9,9,

2

3,3,

2

1,1

92729132 234 xxxxxf

2

9,9,

2

3,3,

2

1,1 Let’s see if Descartes

Rule helps us narrow down the choices.

starts Pos. Stays positive

No sign changes in f(x) so no positive real zeros---we just ruled out half the choices to try so that helps!

2

9,9,

2

3,3,

2

1,1

92729132 234 xxxxxfstarts Pos. changes Neg. changes Pos. Changes Neg. Changes Pos.

1 2 3 4

4 sign changes so 4 or 2 or 0 negative real zeros.

92729132 234 xxxxxf

Let’s try -1 again

2

9,9,

2

3,3,

2

1,1

Yes! We found a zero. Let’s work with reduced polynomial then.

-2 -11 -18 -9-1 2 13 29 27 9

2 11 18 9 0

-2 - 9 -9 -1 2 11 18 9

2 9 9 0

Yes! We found another one. We are done with trial and error since we can put variables back in and solve the remaining quadratic equation. 332992 2 xxxx

So remaining zeros found by setting these factors = 0 are -3/2 and -3. Notice these were in our list of choices.

Let’s try -1

Yes! We found a zero. Let’s work with reduced polynomial then.

92729132 234 xxxxxf

So our polynomial factored is:

3321 2 xxxxf

Intermediate Value

Theorem

Let’s solve the equation: 05332 23 xxxTo do that let’s consider the function

5332 23 xxxxfIf we find the zeros of the function, we would be solving the equation above since we want to know where the function = 0

By Descartes Rule: There is one sign change in f(x) so there is one positive real zero.

5332 23 xxxxfThere are 2 sign changes in f(-x) so there are 2 or 0 negative real zeros.

Using the rational zeros theorem, the possible rational zeros are:

1, 5

1, 2 2

5,

2

1,5,1

5332 23 xxxxf2

5,

2

1,5,1

Let’s try 1 2 -1 -4

1 2 -3 -3 -5

2 -1 -4 -9

1 is not a zero and f(1) = -9

Let’s try 5 10 35 160

5 2 -3 -3 -5

2 7 32 155

5 is not a zero and f(5) = 155

On the next screen we’ll plot these points and the y intercept on the graph and think about what we can tell about this graph and its zeros.

5332 23 xxxxf2

5,

2

1,5,1

155 is a lot higher than this but that gives us an idea it’s up high

f(1) = -9

f(5) = 155

To join these points in a smooth, continuous curve, you would have to cross the x axis somewhere between 1 and 5. This is the Intermediate Value Theorem in action. We can see that since Descartes Rule told us there was 1 positive real zero, that is must be between 1 and 5 so you wouldn’t try 1/2, but you'd try 5/2 instead.

f(0) = -5

f(1) = -9

f(5) = 155

Intermediate Value TheoremLet f denote a polynomial function. If a < b and if f(a) and f(b) are of opposite sign, then there is at least one zero of f between a and b.

In our illustration, a = 1 and b =5

In our illustration, f(a) = -9 and f(b) = 155 which are opposite signs

So if we find function values for 2 different x’s and one is positive and the other negative, there must be a zero of the function between these two x values

We use this theorem to approximate zeros when they are irrational numbers. The function below has a zero between -1 and 0. We’ll use the Intermediate Value Theorem to approximate the zero to one decimal place.

28 234 xxxxf

First let’s verify that there is a zero between -1 and 0. If we find f(-1) and f(0) and they are of opposite signs, we’ll know there is a zero between them by the Intermediate Value Theorem.

-1 -7 8 -8 -1 1 8 -1 0 2

1 7 -8 8 -6

So f(-1) = -6 and f(0) = 2These are opposite signs.

f(-1) = -6

f(0) = 2

28 234 xxxxf

-1

The graph must cross the x-axis somewhere between -1 and 0

Let’s try half way between at x = - 0.5

f(-0.5) = 0.8125

-0.5 -3.75 2.375 -1.1875

-0.5 1 8 -1 0 2

1 7.5 -4.75 2.375 0.8125

Does the sign change occur between f(-1) and f(-0.5) or between f(-0.5) and f(0)?

neg

pos

pos

So let’s try something between -1 and - 0.5. Let’s try - 0.7. Do this with synthetic division or direct substitution.

f(-0.7) = -0.9939Sign change

f(-0.6) = 0.0416

- 0.6 is the closest to zero so this is the zero approximated to one decimal place

Notice that the sign change is between - 0.7 and - 0.6