C3.7 Solve polynomial and rational applications

Ontario Grade 12 Mathematics

Polynomial and Rational Functions

Turn a situation into an equation, solve it, and check what the answers mean.

Applications connect algebra to quantities such as area, dimensions, time, and rates. The main work is to translate the situation into an equation, solve that equation, and decide which solutions make sense. A solution that works algebraically may still be impossible in the context, so checking is part of solving.

What you will learn

Choose a variable and represent the quantities in an application.
Build and solve polynomial or rational equations from a situation.
Check algebraic solutions against the original context and state the answer with appropriate units.

1. Review the tools: variables, equations, and restrictions

A variable is a letter used to represent an unknown quantity. An equation says that two expressions have the same value. To solve an application, first decide what the variable represents. Then express other quantities using that variable.

A polynomial is an expression made from numbers and variables with non-negative whole-number exponents, combined by addition, subtraction, or multiplication. For example, an expression such as

x^2+5x

is a polynomial. A polynomial equation can often be rearranged into the form

P(x)=0

and solved by factoring, when it factors simply.

A rational expression is a fraction whose numerator and denominator are polynomials. A rational equation contains one or more rational expressions. Its denominator cannot be zero, so identify values that make a denominator zero before solving. These values are excluded even if later algebra appears to produce them.

Useful prerequisites include expanding brackets, factoring simple quadratics, and solving an equation by doing the same operation to both sides. When multiplying a rational equation by a common denominator, keep its restrictions in mind. The multiplication can produce a polynomial equation, but the original restrictions still apply.

P(x)=0

State what the variable represents and include its units.
List values that make an original denominator zero.
After solving, check each candidate in the original situation.

2. Translate a situation into a model

A model is an equation that represents the relationships described in a situation. Begin with known information and the relationship it gives. For a rectangle, area equals length times width. If one dimension is unknown, the other can sometimes be written in terms of that variable.

Some applications lead directly to polynomial equations. For instance, multiplying two dimensions that each depend on

x

can produce a quadratic expression. Other applications involve a quotient, such as distance divided by time, or a dimension expressed as a fixed area divided by another dimension. These can lead to rational equations.

Make a small table or list when several quantities change together. Write each quantity in terms of the variable before forming the final equation. This makes it easier to see whether the relationship is additive, multiplicative, or a quotient.

Use the context to set sensible restrictions. A length must be positive. A time cannot be negative. A denominator representing a dimension cannot equal zero. These restrictions are not optional details; they help determine which algebraic answer is meaningful.

A=lw

Use the relationship named in the situation, not a guessed equation.
Keep units attached to quantities in your written explanation.
Use context restrictions to reject impossible answers.

3. Solve and interpret the equation

Once the equation is formed, simplify it carefully. If it is rational, state any excluded values, then multiply both sides by a common denominator to clear the fractions. This usually leaves a polynomial equation that can be solved with familiar algebra.

When factoring a quadratic, look for two numbers whose product and sum match its coefficients. Set each factor equal to zero to find candidate solutions. A candidate is not automatically the final answer: it must satisfy the original equation and the situation's restrictions.

Substitution is a reliable check. Put a candidate into the original relationship, not only the rearranged equation. Confirm that both sides agree and that quantities such as lengths remain positive. If two candidates both work, explain what each represents. If only one fits the stated conditions, report only that one.

Finish with a sentence that answers the question. Include units where relevant. A bare number or variable value may not tell the reader what was found.

a(x)b(x)=0 \Longrightarrow a(x)=0\ or\ b(x)=0

Clear denominators only after recording restrictions.
Treat solutions of the rearranged equation as candidates until checked.
State the result in the language and units of the application.

4. Guided application: a rectangle with fixed area

Suppose a rectangular garden has an area of

48\text{ m}^2

and a perimeter of

28\text{ m}

. Find its dimensions. This problem uses both forms of modelling: the fixed-area relationship can express one dimension as a fraction, while the perimeter condition leads to a polynomial equation.

Let the width be

x

metres. The width must be positive. Since area is length times width, the length is

48/x

metres, so

x

cannot be zero. The perimeter is twice the sum of length and width. Set that expression equal to

28

.

The rational equation becomes a quadratic after multiplying by

x

. Factor the quadratic and check each solution. Both positive dimensions satisfy the given conditions, so both describe the same rectangle with its length and width named in either order.

2\left(x+\frac{48}{x}\right)=28

Area determines the length as a quotient when the width is chosen.
Perimeter supplies the second relationship needed to solve.
Both positive dimension pairs are valid; they differ only in which dimension is called width.

How the relationships build the model

Quantity or condition	Expression or equation
Width	$x$ metres
Length from the area	$48/x$ metres
Perimeter condition	$2(x+48/x)=28$
Polynomial equation after clearing the denominator	$x^2-14x+48=0$

Worked example

Find the dimensions from area and perimeter

A rectangular garden has an area of

48\text{ m}^2

and a perimeter of

28\text{ m}

. Determine its dimensions.

Choose a variable
Let the width be $x$ metres. Since a width is a length, require $x>0$ . The area relationship gives the length as $48/x$ metres, and this also confirms that $x$ cannot be zero.
$\text{length}=\frac{48}{x},\quad x>0$
Form the perimeter equation
A rectangle has two widths and two lengths, so its perimeter is twice the sum of one width and one length. Set this equal to the given perimeter.
$2\left(x+\frac{48}{x}\right)=28$
Clear the denominator
Multiply both sides by $x$ . This is allowed because the context already requires $x>0$ , so $x$ is not zero. Rearrange the resulting equation so one side is zero.
$2x^2-28x+96=0$
Factor and solve
Divide by $2$ , then factor the quadratic. A product is zero when at least one factor is zero, giving two candidate widths.
$x^2-14x+48=0\quad\Longrightarrow\quad(x-6)(x-8)=0$
Check in the situation
The candidate widths are $6$ m and $8$ m. Their corresponding lengths from the area relationship are $8$ m and $6$ m. Each pair has area $48\text{ m}^2$ and perimeter $28$ m, and both dimensions are positive.
$6\cdot8=48,\quad 2(6+8)=28$

Answer: The garden is

6

m by

8

m.

Check: The area is

6\cdot8=48\text{ m}^2

, and the perimeter is

2(6+8)=28\text{ m}

.

Common mistakes and how to avoid them

Using

48x

for the length after choosing width

x

.

Correction: Area is length times width, so solve

x\cdot\text{length}=48

. The length is

48/x

, not

48x

.

Multiplying a rational equation by its denominator without checking whether that denominator can be zero.

Correction: Record the excluded value first. In an application, also use restrictions such as positive lengths.

Reporting every algebraic candidate without checking it.

Correction: Substitute candidates into the original relationship and check whether they fit the context and its units.

Giving a numerical answer without identifying the quantity.

Correction: State what the number measures and include the appropriate unit.

Lesson summary

Choose a variable and express related quantities in terms of it.
Use the situation's relationships to form a polynomial or rational equation.
For rational equations, note denominator restrictions before clearing fractions.
Solve, check candidates in the original context, and state the result with units.

Check your understanding

Question 1

A rectangle has area

35\text{ m}^2

and width

x

metres. Which expression gives its length?

$35x$
$35/x$
$x/35$
$35-x$

Show answer and explanation

35/x

Area is length times width. Solving

x\cdot\text{length}=35

gives length

35/x

.

Question 2

For the equation

\frac{20}{x}=5

, which value must be excluded before solving?

$x=0$
$x=4$
$x=5$
$x=20$

Show answer and explanation

x=0

The denominator cannot be zero, so

x=0

is excluded. Solving gives

x=4

, which is allowed.

Question 3

A candidate solution gives a negative width for a garden. What should you do?

Accept it because it solves the equation.
Reject it because a width in this context must be positive.
Change the negative sign to positive without checking.
Ignore the units and report the number.

Show answer and explanation

Reject it because a width in this context must be positive.

An algebraic candidate must also fit the context. A garden width cannot be negative.

Key terms

Variable: A letter that represents an unknown or changing quantity.
Polynomial: An expression made from constants and variables with non-negative whole-number exponents, combined using addition, subtraction, or multiplication.
Rational expression: A fraction in which the numerator and denominator are polynomials, with denominator values of zero excluded.
Candidate solution: A value found by solving an equation that still needs to be checked in the original situation.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 12 Mathematics (MHF4U), expectation C3.7. 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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C3.7 · Solve polynomial and rational applications

Ontario Grade 12 Mathematics

Turn a situation into an equation, solve it, and check what the answers mean.

What you will learn

1. Review the tools: variables, equations, and restrictions

2. Translate a situation into a model

3. Solve and interpret the equation

4. Guided application: a rectangle with fixed area

How the relationships build the model

Find the dimensions from area and perimeter

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through MHF4U

About this lesson and its review