C1.8 Determine a polynomial family from zeros and a point

Ontario Grade 12 Mathematics

Polynomial and Rational Functions

Use zeros to build factors, then use a point to find the scale factor

A polynomial’s zeros show where its graph meets the

x

-axis. Each zero gives a factor in the polynomial. But zeros alone usually do not determine the exact polynomial: multiplying by different non-zero constants keeps the same zeros. A point on the graph can identify the needed constant. In this lesson, you will move from the given zeros to a polynomial family, then use one point to select a specific polynomial.

What you will learn

Connect a polynomial’s zeros to its linear factors.
Write a family of polynomials that has the specified zeros.
Use a point on the graph to determine the unknown constant.

1. Prerequisite bridge: zeros and factors

A zero is an

x

-value that makes a function’s output equal to zero. If

f(3)=0

, then

3

is a zero of

f

. On a graph, the point

(3,0)

is an

x

-intercept.

A factor is an expression that is multiplied by other expressions. For example, when

x=3

makes a factor equal zero, that factor can be written as

(x-3)

. The sign inside the factor is opposite the zero: a zero of

-4

gives the factor

(x+4)

.

This link works because substituting the zero into its factor gives zero. A product with a zero factor has value zero. So if a polynomial has the factor

(x-3)

, then

x=3

is one of its zeros.

x=r\;\Longleftrightarrow\;x-r=0

A zero $r$ corresponds to a factor $(x-r)$ .
A polynomial with a factor $(x-r)$ has zero $r$ .

2. From zeros to a polynomial family

A polynomial family is a set of polynomials with a shared feature. Here, the shared feature is a specified set of zeros. The zeros determine factors, but they do not usually determine the number multiplying those factors.

That number is called the leading constant in this form. We write it as

a

. For a polynomial with distinct zeros

r_1

and

r_2

, a family with those zeros can be written as

f(x)=a(x-r_1)(x-r_2)

, where

a

is non-zero. Changing

a

changes the vertical scale of the graph, but the listed zeros remain the same.

A repeated zero is a zero that occurs more than once. If a zero

r

has multiplicity

m

, its factor appears

m

times, or as a power:

(x-r)^m

. Multiplicity tells how many copies of the factor belong in the polynomial. Use the stated multiplicities when they are provided.

A point is written as

(x,y)

, where

x

is the input and

y

is the output. If the point lies on the polynomial’s graph, then its coordinates satisfy

y=f(x)

. Substituting those coordinates into the family gives an equation for

a

. Solving that equation selects the polynomial that passes through the point.

f(x)=a\prod_{i=1}^{n}(x-r_i)^{m_i}

Build one factor $(x-r)$ for each zero $r$ , including repeats.
Keep an unknown non-zero constant when zeros are given but the polynomial is not fully specified.
Substitute a known point into the family to solve for the constant.

3. A factor table and a useful check

The table shows how each zero becomes a factor. It also shows how multiplicity changes that factor. The product of the listed factors gives the variable part of the family; the constant

a

is still needed.

After finding a value for

a

, check two things. First, the polynomial should equal zero at every given zero. Second, substituting the coordinates of the given point should produce its stated

y

-value. These checks catch sign errors and missed repeated factors.

The factor’s sign is chosen so that it becomes zero at the stated zero.
A repeated zero requires a repeated factor, represented by a power.

4. Guided example: use a point to find the polynomial

Suppose a polynomial has zeros

-2

and

1

, with

1

having multiplicity

2

. Its graph passes through

(2,-32)

. The factors from the zeros are

(x+2)

and

(x-1)^2

. Begin with an unknown non-zero constant, then use the point to determine it.

The point supplies both an input and an output. Substitute

x=2

into the family and set the result equal to

-32

. Once the constant is found, write the specific polynomial and check the zeros and the point.

f(x)=a(x+2)(x-1)^2

The zero $-2$ gives $(x+2)$ .
The zero $1$ with multiplicity $2$ gives $(x-1)^2$ .
The point determines the constant multiplying the factors.

5. Applying the method

For a new question, first list every zero and its multiplicity. Translate each zero into a factor, then include an unknown non-zero constant. Next, substitute the coordinates of the given point. Solve for the constant and state the resulting polynomial.

If the point’s input makes one of the factors zero, the family would give an output of zero there. A point with a non-zero output could not lie on that graph. This is a useful consistency check before solving.

The point must be given as a point on the polynomial’s graph. If the information is inconsistent, no polynomial in the stated family can satisfy it. Otherwise, the point fixes the constant and identifies one polynomial from the family.

y=f(x)

Translate zeros into factors before using the point.
Use the full point, not just its $x$ -coordinate.
Check the final polynomial against both the zeros and the point.

Turning zeros into factors

Zero	Multiplicity	Factor
$-2$	$1$	$(x+2)$
$1$	$2$	$(x-1)^2$

Worked example

Find the polynomial from two zeros and a point

A polynomial has zeros

-2

and

1

, where

1

has multiplicity

2

. Its graph passes through

(2,-32)

. Determine the polynomial.

Build the family
The zero $-2$ gives the factor $(x+2)$ . The zero $1$ gives $(x-1)$ , and its multiplicity of $2$ means this factor appears twice. Include an unknown non-zero constant $a$ because the zeros do not fix the polynomial’s scale.
$f(x)=a(x+2)(x-1)^2$
Substitute the point
Since $(2,-32)$ lies on the graph, the output at input $2$ must be $-32$ . Substitute $x=2$ and $f(2)=-32$ into the family.
$-32=a(2+2)(2-1)^2$
Solve for the constant
The factors at $x=2$ multiply to $4$ . Divide both sides by $4$ to find the value of $a$ .
$a=\frac{-32}{4}=-8$
Write and check the polynomial
Replace $a$ with $-8$ . At $x=-2$ and $x=1$ , a factor is zero. At $x=2$ , the value is $-8(4)(1)=-32$ , as required.
$f(x)=-8(x+2)(x-1)^2$

Answer: The polynomial is

f(x)=-8(x+2)(x-1)^2

.

Check: Substituting

x=-2

or

x=1

gives

f(x)=0

. Substituting

x=2

gives

f(2)=-32

, so the zeros and point are both satisfied.

Common mistakes and how to avoid them

Writing

(x-r)

with the same sign as the zero.

Correction: Use

(x-r)

for zero

r

. For example, zero

-2

gives

(x+2)

.

Leaving out a repeated factor.

Correction: Use the given multiplicity as the factor’s exponent. A zero of

1

with multiplicity

2

gives

(x-1)^2

.

Assuming the zeros alone determine the exact polynomial.

Correction: Keep a non-zero constant

a

in the family. Use the given point to determine its value.

Substituting only the point’s input and forgetting its output.

Correction: Use both coordinates: substitute the input into

f(x)

and set the result equal to the point’s

y

-coordinate.

Lesson summary

Each zero $r$ gives a factor $(x-r)$ .
A zero with multiplicity $m$ gives the factor $(x-r)^m$ .
Include an unknown non-zero constant to represent the family.
Substitute the coordinates of a point on the graph to find the constant.
Check that the final polynomial has the stated zeros and passes through the point.

Check your understanding

Question 1

A polynomial has zeros

-3

and

2

, both with multiplicity

1

. Which expression gives its family before a point is used?

$f(x)=a(x-3)(x+2)$
$f(x)=a(x+3)(x-2)$
$f(x)=(x+3)(x-2)$
$f(x)=a(x+3)^2(x-2)$

Show answer and explanation

f(x)=a(x+3)(x-2)

Zero

-3

gives

(x+3)

, and zero

2

gives

(x-2)

. The unknown non-zero constant

a

remains because zeros alone do not determine the scale.

Question 2

A polynomial family is

f(x)=a(x-1)(x+2)^2

. The graph passes through

(0,8)

. What is

a

?

$1$
$2$
$-1$
$4$

Show answer and explanation

1

Substitute

x=0

and

f(0)=8

:

8=a(-1)(2)^2=-4a

. This gives

a=-2

. The correct option is therefore the one showing

-2

.

Key terms

Zero: An input value that makes a function’s output equal to zero.
Factor: An expression multiplied by other expressions to form a product.
Multiplicity: The number of times a zero’s factor occurs in a polynomial.
Polynomial family: A set of polynomials that share a feature, such as the same specified zeros.

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About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 12 Mathematics (MHF4U), expectation C1.8. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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C1.8 · Determine a polynomial family from zeros and a point

Ontario Grade 12 Mathematics

Use zeros to build factors, then use a point to find the scale factor

What you will learn

1. Prerequisite bridge: zeros and factors

2. From zeros to a polynomial family

3. A factor table and a useful check

4. Guided example: use a point to find the polynomial

5. Applying the method

Turning zeros into factors

Find the polynomial from two zeros and a point

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Key terms

Continue through MHF4U

About this lesson and its review