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C1.7 · Build polynomial equations from given conditions

Learn to build polynomial equations from given conditions through clear examples and targeted practice.

Ontario Grade 12 Mathematics

Polynomial and Rational Functions

Turn information about roots, points, and graph features into a polynomial model.

A polynomial equation can describe a graph or a set of values. In this lesson, you will build a polynomial equation from information such as its zeros and a point on its graph. First, recall that a zero is an input value that makes a function equal to zero. For example, if f(3)=0f(3)=0, then 33 is a zero of ff. A polynomial may also be written in factored form, which makes its zeros visible. The main idea is to translate each condition into part of an equation, then use all the conditions to complete the model.

What you will learn

1. Prerequisite bridge: factors, zeros, and degree

A polynomial is an expression made from constants and variables raised to whole-number powers, combined using addition, subtraction, and multiplication. For example, 2x3−x+42x^3-x+4 is a polynomial.
A factor is an expression that is multiplied by another expression. If a polynomial has a factor x−2x-2, then substituting x=2x=2 makes that factor equal to zero. The whole product is therefore zero, so 22 is a zero of the polynomial.
The degree of a polynomial is the greatest exponent of the variable after the expression is simplified. For a polynomial written as a product of linear factors, the degree is the number of those factors, counting repeats. A repeated factor indicates a repeated zero. For example, (x+1)2(x+1)^2 has the zero −1-1 repeated twice.
x−r=0  ⟺  x=rx-r=0\iff x=r

2. Translate conditions into factors

A condition is information the polynomial must satisfy. A stated zero gives a factor directly: if rr is a zero, include x−rx-r. If a zero is repeated, include the corresponding factor more than once.
Suppose a polynomial has zeros −2-2 and 44, and the zero −2-2 is repeated. The known factors are (x+2)2(x+2)^2 and (x−4)(x-4). Their product is a starting model. If the polynomial is known to have degree 33, these factors already account for the full degree.
Sometimes the polynomial has a larger degree than the factors supplied by the zeros. In that case, more information is needed to determine the remaining coefficient or factor. A point on the graph is useful because its coordinates tell you an input and the corresponding output.
f(x)=a(x−r1)(x−r2)⋯(x−rn)f(x)=a(x-r_1)(x-r_2)\cdots(x-r_n)

3. Use a point to determine the scale

The coefficient aa in a factored model is often unknown. It changes the vertical scale of the graph without changing the listed zeros. To find it, substitute the coordinates of a known point into the model.
A point (p,q)(p,q) on the graph means that the function value at input pp is qq. Substitute x=px=p and set the expression equal to qq. Solve the resulting equation for the unknown coefficient.
This approach works because the point condition and the zero conditions describe the same polynomial. The final equation should satisfy both kinds of information. Check by substituting the zeros and the given point into the completed model.
f(p)=qf(p)=q

Condition-to-model guide

Given conditionHow it enters the modelExample
Zero rrInclude the factor x−rx-rZero 55 gives x−5x-5
Repeated zero rrRepeat its factorZero −2-2 twice gives (x+2)2(x+2)^2
Point (p,q)(p,q)Substitute x=px=p and set f(p)=qf(p)=qPoint (2,7)(2,7) gives f(2)=7f(2)=7
Degree nnCount factors, including repeatsThree linear factors give degree 33

Worked example

Build a cubic from zeros and a point

Build a polynomial equation of degree 33 with zeros −1-1 and 33, where −1-1 is repeated, and whose graph passes through (1,16)(1,16).
  1. Translate the zeros
    The zero −1-1 gives the factor x+1x+1. Since it is repeated, use that factor twice. The zero 33 gives the factor x−3x-3. Together these factors have degree 33, as required.
    f(x)=a(x+1)2(x−3)f(x)=a(x+1)^2(x-3)
  2. Use the point condition
    The graph passes through (1,16)(1,16), so the output must be 1616 when the input is 11. Substitute those values into the factored model.
    16=a(1+1)2(1−3)16=a(1+1)^2(1-3)
  3. Solve for the coefficient
    Evaluate the factors. Their product is −8-8, so the equation becomes 16=−8a16=-8a. Dividing both sides by −8-8 gives the coefficient.
    a=−2a=-2
  4. Write the polynomial equation
    Replace the unknown coefficient with −2-2. The factored form clearly displays the zeros and their repetition.
    f(x)=−2(x+1)2(x−3)f(x)=-2(x+1)^2(x-3)
Answer: The polynomial is f(x)=−2(x+1)2(x−3)f(x)=-2(x+1)^2(x-3).
Check: At x=−1x=-1, the repeated factor is zero; at x=3x=3, the other factor is zero. At x=1x=1, the value is −2(2)2(−2)=16-2(2)^2(-2)=16. The polynomial has degree 33.

Common mistakes and how to avoid them

Using x+rx+r for a zero rr.
Correction: Use x−rx-r. For example, zero −4-4 gives x+4x+4, since substituting −4-4 makes that factor zero.
Forgetting to repeat a factor when a zero is repeated.
Correction: Include the factor the stated number of times. A zero repeated twice gives a squared factor.
Assuming the zeros determine the whole polynomial.
Correction: Zeros determine factors, but an unknown coefficient may remain. Use an additional condition, such as a point, to determine it.
Using the point coordinates in the wrong order.
Correction: For a point (p,q)(p,q), substitute pp for the input and set the result equal to qq.

Lesson summary

Check your understanding

Question 1

A polynomial has degree 33, zeros 22 and −1-1, and the zero 22 is repeated. Which factored model includes these conditions?
  1. a(x−2)2(x+1)a(x-2)^2(x+1)
  2. a(x+2)2(x−1)a(x+2)^2(x-1)
  3. a(x−2)(x+1)a(x-2)(x+1)
  4. a(x−2)2(x−1)a(x-2)^2(x-1)
Show answer and explanation
a(x−2)2(x+1)a(x-2)^2(x+1)
The zero 22 gives x−2x-2 and is repeated, while zero −1-1 gives x+1x+1. There are three linear factors, matching degree 33.

Question 2

A polynomial is given by f(x)=a(x−2)(x+1)f(x)=a(x-2)(x+1) and passes through (0,6)(0,6). What is aa?
  1. −3-3
  2. −1-1
  3. 11
  4. 33
Show answer and explanation
−3-3
Substitute the point condition: 6=a(0−2)(0+1)=−2a6=a(0-2)(0+1)=-2a. Dividing by −2-2 gives a=−3a=-3.

Key terms

Polynomial
An expression formed from constants and variables with whole-number exponents, using addition, subtraction, and multiplication.
Zero
An input value that makes a function's output equal to zero.
Factor
An expression that is multiplied by another expression.
Degree
The greatest exponent of the variable in a polynomial after simplification.
Multiplicity
The number of times a zero's factor appears in the polynomial.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 12 Mathematics (MHF4U), expectation C1.7. It is a study resource, not an official curriculum publication.

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