A1.3 Factor quadratic expressions using an appropriate strategy

Ontario Grade 11 Mathematics

Quadratic Functions

Ontario Grade 11 Mathematics — A1.3

Factoring rewrites an expression as a product. For example, writing

12

as

3\times4

is factoring a number. In this lesson, you will use that idea with quadratic expressions. Factoring is useful because it shows how an expression is built from simpler factors. Your goal is to choose a strategy that fits the expression, carry it out carefully, and check the result by expanding.

What you will learn

Recognize a quadratic expression and explain what it means to factor it.
Choose a suitable strategy for a quadratic expression.
Factor common forms and check a factorization by expanding.

1. Review: factors, products, and quadratic expressions

A factor is a quantity multiplied by another quantity. Since

3(x+2)

is a product, its factors are

3

and

(x+2)

. To expand it, multiply each term inside the brackets by

3

:

3(x+2)=3x+6

. Factoring reverses this process.

A quadratic expression has a variable raised to the power of

2

, with no higher power. It can be written in the form

ax^2+bx+c

, where

a

,

b

, and

c

are numbers and

a\ne0

. For instance,

2x^2+7x+3

is quadratic. An expression such as

x^3+2x

is not quadratic because it includes

x^3

.

The greatest common factor, or GCF, is the largest factor shared by every term. For

6x+9

, the GCF is

3

, so

6x+9=3(2x+3)

. Before using another factoring strategy, look for a GCF. This can make the remaining expression simpler.

ax^2+bx+c

Factoring writes a sum or difference as a product.
Expanding can be used to check a factorization.
Look for a GCF before applying another strategy.

2. Match the expression to a strategy

Start by checking how many terms the expression has, whether the first term is a square, and whether all terms share a factor. A strategy is appropriate when the expression has the pattern needed for that strategy. Do not force a pattern that is not present.

For a three-term expression

x^2+bx+c

, look for two numbers whose product is

c

and whose sum is

b

. Those numbers become the constant parts of two binomial factors. A binomial is an expression with two terms, such as

(x+3)

. For example, the numbers

2

and

5

multiply to

10

and add to

7

, so

x^2+7x+10=(x+2)(x+5)

.

When the coefficient of

x^2

is not

1

, a useful approach is to split the middle term and factor by grouping. Grouping means placing the four terms into two pairs, factoring each pair, and then taking out the common bracket. You will see this in the worked example.

For a two-term expression, check for a difference of squares. This pattern is a square minus another square:

A^2-B^2=(A-B)(A+B)

. For example,

x^2-25=(x-5)(x+5)

. A sum of squares does not match this pattern.

Another useful three-term pattern is a perfect-square trinomial. It comes from squaring a binomial:

(A+B)^2=A^2+2AB+B^2

or

(A-B)^2=A^2-2AB+B^2

. Check that the first and last terms are squares, then check that the middle term is twice their square roots multiplied together, with the correct sign.

A^2-B^2=(A-B)(A+B)

For $x^2+bx+c$ , use a product-and-sum pair.
For a non-unit leading coefficient, splitting the middle term can lead to grouping.
Recognize difference-of-squares and perfect-square patterns only when their conditions fit.

3. A visual plan for choosing

The table gives a short decision guide. It does not replace checking the factors. It helps you decide what to try first based on the expression's visible structure.

If no pattern works immediately, return to the GCF check and consider splitting the middle term for a three-term quadratic. Some expressions cannot be factored using integer factors. In that case, do not claim a product of integer binomials unless expanding it reproduces the original expression.

The number of terms and the coefficients offer clues.
Always expand your proposed factors to verify them.

4. Worked example and application

In the worked example, the leading coefficient is not

1

, and there is no common factor in all three terms. Splitting the middle term creates four terms that can be grouped into pairs. The common bracket in those pairs then gives the factorization.

The numbers used to split the middle term must multiply to $ac$ and add to $b$ .
Grouping is successful when both pairs produce the same bracket.

Strategy clues

What you notice	Strategy to try	Example
Every term shares a factor	Factor out the GCF first	$8x^2+12x=4x(2x+3)$
Three terms; leading coefficient is 1	Find two numbers with product $c$ and sum $b$	$x^2+7x+10=(x+2)(x+5)$
Three terms; leading coefficient is not 1	Split the middle term, then group	$6x^2+x-2=(3x+2)(2x-1)$
Two squares with subtraction	Use difference of squares	$x^2-25=(x-5)(x+5)$
Three terms fit a squared binomial	Use a perfect-square pattern	$x^2+6x+9=(x+3)^2$

Worked example

Factor a quadratic by splitting and grouping

Factor

6x^2+x-2

.

Check the expression
There are three terms, and the GCF is $1$ . Since the coefficient of $x^2$ is $6$ , use splitting and grouping rather than the simpler product-and-sum rule for a leading coefficient of $1$ .
Find a pair
Multiply the leading coefficient, $6$ , by the constant term, $-2$ . Find two integers with product $-12$ and sum $1$ . The pair $4$ and $-3$ works.
$4(-3)=-12,\quad 4+(-3)=1$
Split the middle term
Replace $x$ with $4x-3x$ . Their coefficients add to $1$ , so the expression's value is unchanged.
$6x^2+4x-3x-2$
Group and factor each pair
Group the first two terms and the last two terms. Factor the GCF from each pair. Both groups contain the bracket $(3x+2)$ .
$2x(3x+2)-(3x+2)$
Take out the common bracket
Since $(3x+2)$ is a factor of both terms, factor it out. The remaining factors are $2x$ and $-1$ .
$(3x+2)(2x-1)$

Answer:

6x^2+x-2=(3x+2)(2x-1)

Check: Expand to verify:

(3x+2)(2x-1)=6x^2-3x+4x-2=6x^2+x-2

. The result matches the original expression.

Common mistakes and how to avoid them

Stopping after splitting the middle term.

Correction: Splitting creates an equivalent four-term expression, not the final factors. Continue by grouping and taking out the common bracket.

Choosing two numbers with the right product but the wrong sum.

Correction: For

x^2+bx+c

, the pair must multiply to

c

and add to

b

. For splitting when the leading coefficient is not

1

, the pair must multiply to

ac

and add to

b

.

Using a difference-of-squares pattern on a sum of squares.

Correction: The pattern requires subtraction:

A^2-B^2

. Check the sign between the squared terms.

Changing the sign when factoring a negative term from a pair.

Correction: Divide every term in the pair by the factor you take out. For example,

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

.

Assuming a proposed factorization is correct because it looks familiar.

Correction: Expand the factors. The product must match every term of the original expression.

Lesson summary

Factoring rewrites a quadratic expression as a product.
Check for a GCF before trying another strategy.
Use the expression's structure to choose a suitable strategy.
For three terms with a leading coefficient other than $1$ , splitting and grouping can work.
Expand the factors to check your answer.

Check your understanding

Question 1

Which pair of numbers can factor

x^2+9x+20

?

4 and 5
2 and 10
-4 and -5
1 and 20

Show answer and explanation

4 and 5

The pair must multiply to

20

and add to

9

. The numbers

4

and

5

meet both conditions, so the expression factors as

(x+4)(x+5)

.

Question 2

What is the complete factorization of

x^2-36

?

$(x-6)(x+6)$
$(x-36)(x+1)$
$(x-6)^2$
It cannot be factored.

Show answer and explanation

(x-6)(x+6)

x^2-36

is a difference of squares because

36=6^2

. The factors are

(x-6)(x+6)

.

Question 3

Which pair is used to split the middle term when factoring

3x^2+8x+4

by grouping?

2 and 6
1 and 4
2 and 2
-2 and -6

Show answer and explanation

2 and 6

Here,

ac=3(4)=12

and

b=8

. The numbers

2

and

6

multiply to

12

and add to

8

. Splitting gives

3x^2+2x+6x+4

, which groups to

(3x+2)(x+2)

.

Key terms

Factor: A quantity that is multiplied by another quantity.
Quadratic expression: An expression whose highest power of the variable is $2$ .
Greatest common factor (GCF): The greatest factor shared by every term in an expression.
Binomial: An algebraic expression with two terms.
Expand: Multiply factors to rewrite a product as a sum or difference.

Continue through MCF3M

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About this lesson

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Mathematics (MCF3M), expectation A1.3. It is a study resource, not an official curriculum publication.

Official curriculum reference

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A1.3 · Factor quadratic expressions using an appropriate strategy

Ontario Grade 11 Mathematics

Ontario Grade 11 Mathematics — A1.3

What you will learn

1. Review: factors, products, and quadratic expressions

2. Match the expression to a strategy

3. A visual plan for choosing

4. Worked example and application

Strategy clues

Factor a quadratic by splitting and grouping

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through MCF3M

About this lesson