Mastering Determining the Number of Zeros in a Polynomial Function

To justify this theorem, the graph of a polynomial function will be examined for non-negative values of

x .

Consider an arbitrary polynomial function

p (x)

of degree

n .

p (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0}

Here,

a_{n}

is the leading coefficient and

a_{0}

the constant term. Consider the case where both

a_{n}

and

a_{0}

are positive. Because

a_{0} > 0,

the graph intersects the

y -

axis above the

x -

axis. Also, since

a_{n} > 0,

the end behavior of the graph for positive values of

x

is up.

In this case, the number of sign changes must be even. This is because every time a positive sign changes to a negative sign, it must return to positive again, making the number of sign changes a multiple of $2 .$ The number of times the graph crosses the $x -$ axis must also be even, since the graph begins and ends above the $x - axis .$

Without sign changes, a polynomial function only increases or decreases as

x

increases. Therefore, each change in direction of the graph is related to a sign change in the polynomial. To explore this idea, consider the following polynomial function.

f (x) = x^{3} - 8 x^{2} + 24

It has already been said that only non-negative values of

x

are considered, so only non-negative values of

x

will be considered for the graph of

f (x) .

The

y -

intercept of the graph is

24 .

Then, the graph goes down from

24

for a while before going up. This happens because in the first interval, the term

- 8 x^{2}

directs the graph before the leading term

x^{3}

takes direction. For the values of this first interval,

x^{3}

is less than

8 x^{2}

and, therefore,

x

is less than

8 .

x^{3} < 8 x^{2} \Rightarrow x < 8

This value can be substituted into the function

f (x)

to find its corresponding output.

f (x) = x^{3} - 8 x^{2} + 24

Substitute

$x = 8$

f (8) = (8)^{3} - 8 (8)^{2} + 24

MultBasePow

$a \cdot a^{m} = a^{1 + m}$

f (8) = 8^{3} - 8^{3} + 24

SubTerm

Subtract term

f (8) = 24

When

x = 8,

the value of

f

returns to the original value of

24

before continuously going up as

x

increases. In the interval that goes from

0

8,

the term

- 8 x^{2}

drives the graph down. Then the value of

x^{3}

catches up and directs the graph upwards. Now consider a variation of the function.

f_{2} (x) = 8 x^{3} - x^{2} + 24

The same will be done to find the values where

x^{2}

is greater than

8 x^{3} .

8 x^{3} < x^{2}

DivEqn

$LHS / x^{2} = RHS / x^{2}$

8 x < 1

DivEqn

$LHS / 8 = RHS / 8$

x < \frac{1}{8}

This time the interval is significantly shorter than in the previous case. This can also be seen in the graph.

Here, the effect of the term $- x^{2}$ is negligible compared to the effect of the term $8 x^{3} .$ This indicates that a change in direction of the graph of a function is caused by a sign change, but a sign change does not always result on a change in direction of the graph. Consider the graph of the initial polynomial function again.

The function starts above the $x -$ axis and has $4$ significant direction changes. Therefore, there are at least $4$ sign changes in the equation. Also, each time the graph moves in a different direction, it crosses the $x -$ axis, which suggests that the number of zeros can be the same as the number of sign changes. Consider now a different graph.

Here, the number of zeros is less than the number of times the graph changes direction. Since sign changes indicate changes in direction, it is possible to have fewer zeros than sign changes. Also, since there cannot be more zeros without changing direction, there cannot be more zeros than sign changes. Two conclusions can be made from this reasoning.

The number of sign changes and the number of zeros are both even.
The number of zeros has to be less than or equal to the number of sign changes.

This is equivalent to what is written in the rule. The other cases can be examined on a similar way. Also, negative zeros are found by considering $p (- x)$ because this is a reflection in the $y -$ axis. Note that this is an informal justification and cannot be taken as a formal proof.

Positive Real Zeros	Negative Real Zeros	Imaginary Zeros	Total Zeros
$2$	$2$	$0$	$4$
$2$	$0$	$2$	$4$
$0$	$2$	$2$	$4$
$0$	$0$	$4$	$4$

Positive Real Zeros

Negative Real Zeros

Imaginary Zeros

Total Zeros

2

2

0

4

2

0

2

4

0

2

2

4

0

0

4

4

Positive Real Zeros	Negative Real Zeros	Imaginary Zeros	Total Zeros
$2$	$2$	$0$	$4$
$2$	$0$	$2$	$4$
$0$	$2$	$2$	$4$
$0$	$0$	$4$	$4$

Positive Real Zeros

Negative Real Zeros

Imaginary Zeros

Total Zeros

2

2

0

4

2

0

2

4

0

2

2

4

0

0

4

4

Real Positive Zeros	Real Negative Zeros	Imaginary Zeros	Total Zeros
$4$	$1$	$0$	$5$
$2$	$1$	$2$	$5$
$0$	$1$	$4$	$5$

Real Positive Zeros

Real Negative Zeros

Imaginary Zeros

Total Zeros

4

1

0

5

2

1

2

5

0

1

4

5

Possible Number of Positive Real Zeros
$4$	$4 - 2 = 2$	$4 - 4 = 0$

Possible Number of Positive Real Zeros

4

4 - 2 = 2

4 - 4 = 0

Real Positive Zeros	Real Negative Zeros	Imaginary Zeros	Total Zeros
$4$	$1$	$0$	$5$
$2$	$1$	$2$	$5$
$0$	$1$	$4$	$5$

Real Positive Zeros

Real Negative Zeros

Imaginary Zeros

Total Zeros

4

1

0

5

2

1

2

5

0

1

4

5

Real Positive Zeros	Real Negative Zeros	Imaginary Zeros	Total Zeros
$3$	$2$	$0$	$5$
$3$	$0$	$2$	$5$
$1$	$2$	$2$	$5$
$1$	$0$	$4$	$5$

Real Positive Zeros

Real Negative Zeros

Imaginary Zeros

Total Zeros

3

2

0

5

3

0

2

5

1

2

2

5

1

0

4

5

Zero Type	Number
Positive Real	$2$
Negative Real	$1$
Complex	$2$

Zero Type

Number

Positive Real

2

Negative Real

1

Complex

2

Positive Real	Negative Real	Complex	Total
$3$	$0$	$2$	$5$
$1$	$0$	$4$	$5$

Positive Real

Negative Real

Complex

Total

3

0

2

5

1

0

4

5

Positive Real	Negative Real	Complex	Total
$3$	$0$	$0$	$3$
$1$	$0$	$2$	$3$

Positive Real

Negative Real

Complex

Total

3

0

0

3

1

0

2

3

{{ article.displayTitle }}

Catch-Up and Review

How Many Zeros Does a Polynomial Have?

The Number of Zeros of a Polynomial Function

Fundamental Theorem of Algebra

Corollary

Proof

Degree $1$

Degree $2$

Degree $3$

Conclusion

Example

How Many Solutions Does the Polynomial Equation Have?

Descartes' Rule of Signs

Proof

Example

Finding the Number of Zeros of a Polynomial Function

Answer

Hint

Solution

Real Positive, Real Negative, and Non-real Zeros

Determining the Characteristics of Polynomial Functions

Hint

Solution

Writing a Polynomial Function with Real Positive Zeros

Answer

Hint

Solution

Finding the Zeros of a Function Given its Graph

Hint

Solution

Solving the Challenge

	{{ 'ml-lesson-number-slides' \| message : article.intro.bblockCount }}
	{{ 'ml-lesson-number-exercises' \| message : article.intro.exerciseCount }}
	{{ 'ml-lesson-time-estimation' \| message }}

{{ article.displayTitle }}

Catch-Up and Review

Corollary

Proof

Degree 1

Degree 2

Degree 3

Conclusion

Example

Proof

Example

Answer

Hint

Solution

Hint

Solution

Answer

Hint

Solution

Hint

Solution

Degree $1$

Degree $2$

Degree $3$