Zeros and Successive Differences of Polynomial Functions (College Board AP® Precalculus): Study Guide

Roger B

Written by: Roger B

Reviewed by: Mark Curtis

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Zeros of polynomial functions

What is a zero of a polynomial function?

  • If a is a number and  p(a)=0

    • then a is called a zero of the polynomial function  p

    • Equivalently, a is a root of the equation  p(x)=0

  • Zeros of polynomial functions can be real or complex (non-real) numbers

  • If a is a real number, then (xa) is a linear factor of  p if and only if a is a zero of  p

    • I.e. if (xa) is a linear factor of  p, then  p(a)=0

    • Or if  p(a)=0, then (xa) is a linear factor of  p

  • E.g. if  p(x)=(x3)(x+1)(x5), then the zeros of  p are x=3, x=1, and x=5

    • This lets you write down the solutions to  p(x)=0 if you have  p(x) in fully factorised form

  • Conversely, if you know that x=2 is a zero of a polynomial, then (x2) must be a factor

What is the multiplicity of a zero?

  • If a linear factor (xa) is repeated n times in the factorisation of a polynomial

    • the corresponding zero has multiplicity n

  • E.g.  p(x)=(x1)2(x+4)3(x+7) has

    • a zero at x=1 with multiplicity 2

    • a zero at x=4 with multiplicity 3

    • and a zero at x=7 with multiplicity 1

  • A polynomial function of degree n has exactly n complex zeros when counting multiplicities

    • A 'complex zero' here means a zero of the function that is either a real or a (non-real) complex number

  • E.g. a degree 5 polynomial always has exactly 5 zeros (counting multiplicities)

    • though some may be repeated (i.e. have a multiplicity greater than 1)

    • and some may be complex (i.e. non-real)

How do zeros appear on a graph?

  • If a is a real zero of a polynomial function  p, then the graph of  y=p(x) has an x-intercept at the point (a,0)

    • The graph crosses or touches the x-axis at that point

  • The multiplicity of a zero determines the behaviour of the graph at that x-intercept:

  • If the zero has odd multiplicity (e.g. 1, 3, 5, ...), the graph crosses the x-axis at that point

    • The sign of the output values changes on either side of the zero

    • If the multiplicity is 1, the graph 'cuts straight across' the x-axis

    • If the multiplicity is 3, 5 or more, then the curve 'levels out' as it crosses the x-axis

  • If the zero has even multiplicity (e.g. 2, 4, 6, ...), the graph is tangent to (touches but does not cross) the x-axis at that point

    • The sign of the output values is the same on both sides of the zero

    • The graph touches the axis then 'turns back'

  • Graph behavior at zeros with different multiplicities can be seen here:

Graph of y=(x+2)²(x−1)(x-4)³ with x-intercepts at -2, 1, and 4, indicating multiplicities 2, 1, and 3 respectively, on xy-plane.
Zeros with different multiplicities on the graph of a polynomial function

How can zeros help solve polynomial inequalities?

  • The real zeros of a polynomial are the points where the output value is zero

  • Therefore they divide the x-axis into intervals

    • On each interval between consecutive zeros, the polynomial is either entirely positive or entirely negative

    • This means the real zeros act as endpoints for intervals satisfying polynomial inequalities

  • E.g. to solve  p(x)0

    • find the zeros first

    • then determine the sign of  p(x) on each interval between zeros

How can I find the zeros of a polynomial?

  • If the polynomial is given in factored form, set each factor equal to zero and solve

    • E.g. for  p(x)=(x3)(x+1)(x5)=0

      • x3=0  x=3

      • x+1=0  x=1

      • x5=0  x=5

    • The zeros are x=3, x=1, and x=5

  • If the polynomial is a quadratic, you can factor it (if possible) or use the quadratic formula

  • For higher-degree polynomials that are not easily factored, you can use a graphing calculator to find the real zeros

    • Enter the polynomial into the calculator and find where the graph crosses the x-axis

    • Or use the calculator's "zero" or "solve" function

    • E.g. to find the zeros of g(x)=0.167x3+x21.834

      • you would enter the function into your calculator

      • and find that the zeros are approximately x=1.233, x=1.578, and x=5.643

Examiner Tips and Tricks

The zeros found on your calculator may be decimal approximations.

  • On your exam, report these accurate to three decimal places

What about complex (non-real) zeros?

  • Some zeros of a polynomial may be complex numbers

    • I.e. numbers involving i=1

  • These do not correspond to x-intercepts on the graph

    • since only real zeros appear on the real number line (i.e. on the x-axis)

  • Complex zeros always come in conjugate pairs

    • I.e. if a+bi is a zero of a polynomial  p (where b0)

      • then abi is also a zero of  p

    • E.g. if 2+3i is a zero, then 23i must also be a zero

  • This means non-real zeros always account for an even number (0, 2, 4, 6...) of the total zeros

Examiner Tips and Tricks

When a question gives you a polynomial with a non-real complex zero, immediately write down its conjugate.

  • That gives you a second zero for free!

This is particularly useful for determining how many real zeros (and therefore x-intercepts) a polynomial has

  • For example, a degree 4 polynomial with a given non-real zero must also have the conjugate as a zero

  • That leaves only 2 remaining zeros, which may be real or another non-real conjugate pair

Worked Example

The polynomial function  p is given by  p(x)=x(x4)2(x+2).

(a) Find all zeros of  p and state the multiplicity of each zero.

(b) For each zero, state whether the graph of  y=p(x) crosses the x-axis or is tangent to the x-axis at that point.

(c) Find all values of x for which  p(x)>0.

Answer:

(a)

Set each factor equal to zero:

x=0  x=0

(x4)2=0  x4=0  x=4

x+2=0  x=2

The 'squared' (power of 2) on (x4)2 tells you that x=4 has a multiplicity of 2

  • The other two zeros are both multiplicity 1

x=0 (multiplicity 1)

x=4 (multiplicity 2)

x=2 (multiplicity 1)

(b)

Use the multiplicity to determine the behavior of the graph at each zero

At x=0:
multiplicity is 1 (odd), so the graph crosses the x-axis

At x=4:
multiplicity is 2 (even), so the graph is tangent to the x-axis
(touches but does not cross)

At x=2:
multiplicity is 1 (odd), so the graph crosses the x-axis

(c)

The zeros divide the x-axis into four intervals: (,2), (2,0), (0,4), and (4,)

Test a value in each interval to determine the sign of  p(x)

x=3:
 p(3)=(3)(34)2(3+2)=(3)(49)(1)=147<0

x=1:
 p(1)=(1)(14)2(1+2)=(1)(25)(1)=25>0

x=2:
 p(2)=(2)(24)2(2+2)=(2)(4)(4)=32<0

x=5:
 p(5)=(5)(54)2(5+2)=(5)(1)(7)=35<0

The polynomial is only positive on one of those intervals

 p(x)>0 on (2,0)

Worked Example

The polynomial function k is given by k(x)=ax4bx3+22, where a and b are nonzero real constants. Each of the zeros of k has multiplicity 1. In the xy-plane, an x-intercept of the graph of k is (12.405,0). A zero of k is 1.3620.547i. Which of the following statements must be true?

(A) The graph of k has three x-intercepts.

(B) 1.362+0.547i is a zero of k.

(C) The equation k(x)=0 has four real solutions.

(D) The graph of k is tangent to the x-axis at x=12.405.

Answer:

Because complex zeros come in conjugate pairs, if 1.3620.547i is a zero of k, then 1.362+0.547i must also be a zero of k

  • That already tells you that (B) is the correct answer

  • but it's worth considering why the other answers are not correct

k has degree 4, so it has exactly 4 zeros (counting multiplicities)

  • Each zero corresponds to a solution to k(x)=0

  • 1.3620.547i and 1.362+0.547i are both non-real zeros

  • So the equation k(x)=0 cannot have four real solutions

    • That rules out option (C)

Two of the zeros are non-real

  • So the remaining 2 zeros must either both be real

  • or both be non-real

    • because non-real zeros always occur in conjugate pairs

  • So k can have at most 2 real zeros

  • which means at most 2 x-intercepts

    • So option (A) cannot be true

Finally, you are told that each zero has multiplicity 1

  • 1 is odd so the graph crosses the x-axis at each real zero, rather than being tangent

    • This rules out option (D)

(B) 1.362+0.547i is a zero of k

Degree and successive differences

How can successive differences reveal the degree of a polynomial?

  • The degree of a polynomial function can be found by examining the successive differences of the output values over equally spaced input values

  • To do this:

    • Start with output values at equally spaced inputs

    • Calculate the 1st differences (differences between consecutive outputs)

      • If the 1st differences are constant, the polynomial has degree 1 (linear)

    • If not, calculate the 2nd differences (differences between consecutive 1st differences)

      • If the 2nd differences are constant, the polynomial has degree 2 (quadratic)

    • If not, continue taking successive differences until you reach a constant set of differences

      • The degree of the polynomial is equal to the least value n for which the nth differences are constant

  • E.g. consider these values at equally spaced inputs for a polynomial function  p:

x

0

1

2

3

4

 p(x)

2

5

14

35

74

  • Calculate the differences

    • 1st differences: 3,9,21,39 — not constant

    • 2nd differences: 6,12,18 — not constant

    • 3rd differences: 6,6 — constant

Pattern of numbers showing sequence 2, 5, 14, 35, 74. First, second, and third differences are illustrated.
  • The 3rd differences are constant, so  p is a degree 3 (cubic) polynomial

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Roger B

Author: Roger B

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Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.

Mark Curtis

Reviewer: Mark Curtis

Expertise: Maths Content Creator

Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.