Q10 Factor quadratics using common factors, trinomials, and differences of squares

Ontario Grade 10 Mathematics

Quadratic Relations

Common Factors, Trinomials, and Differences of Squares

Factoring is the reverse of expanding. When you expanded in Grade 9, you multiplied brackets together to get a longer expression. Factoring starts with the longer expression and finds the brackets that produced it. This skill is essential in MPM2D because it lets you solve quadratic equations, find the x-intercepts of parabolas, and simplify algebraic fractions. This lesson walks through three factoring strategies in order of difficulty: pulling out a common factor, factoring a trinomial, and recognizing a difference of squares. Work through each section carefully, check the examples by expanding your answer, and then try the quick-check questions at the end.

What you will learn

Factor a polynomial by pulling out the greatest common factor (GCF).
Factor a trinomial of the form $x^2 + bx + c$ by finding two numbers that multiply to $c$ and add to $b$ .
Factor a trinomial of the form $ax^2 + bx + c$ where $a \neq 1$ using decomposition.
Recognize and factor a difference of squares $a^2 - b^2$ .
Choose the correct factoring strategy for a given quadratic expression.

Bridge from Grade 9: What Factoring Means

In Grade 9 (MTH1W) you learned to expand expressions like

3(x + 4)

to get

3x + 12

. Factoring reverses that process: you start with

3x + 12

and find

3(x + 4)

. The two numbers or expressions you multiply together are called factors.

A quadratic expression is a polynomial whose highest power is

x^2

, such as

x^2 + 5x + 6

or

4x^2 - 9

. Factoring a quadratic means rewriting it as a product of two or more simpler expressions. You should always check your answer by expanding the brackets — if you get the original expression back, the factoring is correct.

Before factoring any expression, always ask: is there a greatest common factor (GCF) I can pull out first? Removing the GCF makes every subsequent step easier.

Factoring is the reverse of expanding.
A quadratic has a highest power of $x^2$ .
Always look for a GCF before trying any other strategy.
Check every factored answer by expanding it.

Strategy 1: Greatest Common Factor (GCF)

The GCF of a polynomial is the largest number and the highest power of any variable that divides evenly into every term. To factor by GCF, divide each term by the GCF and write the result inside a single set of brackets.

For example, consider

6x^2 + 9x

. The GCF of the coefficients 6 and 9 is 3. The variable

x

appears in both terms (as

x^2

and

x^1

), so the lowest power is

x^1 = x

. The GCF is therefore

3x

. Dividing each term gives

6x^2 \div 3x = 2x

and

9x \div 3x = 3

, so

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

.

Sometimes the GCF is a number only, and sometimes it is a variable only. Either way, the process is the same: identify the GCF, divide every term by it, and place the results inside brackets. If a quadratic has a GCF, always remove it first, even if more factoring is needed afterward.

ax^2 + ax = ax(x + 1)

Find the largest number and lowest variable power shared by all terms.
Divide every term by the GCF and write the result in brackets.
The GCF sits in front of the bracket.
After removing the GCF, check whether the bracket can be factored further.

Strategy 2: Factoring Trinomials

A trinomial is a polynomial with exactly three terms. The most common quadratic trinomial looks like

x^2 + bx + c

, where

b

is the coefficient of

x

and

c

is the constant term. To factor it, you need two integers

p

and

q

such that

p \times q = c

and

p + q = b

. The factored form is then

(x + p)(x + q)

.

For example, to factor

x^2 + 7x + 12

, list pairs of integers that multiply to 12:

(1, 12),\, (2, 6),\, (3, 4)

and their negatives. The pair that also adds to 7 is

(3, 4)

, so

x^2 + 7x + 12 = (x + 3)(x + 4)

. Expand to check:

(x+3)(x+4) = x^2 + 4x + 3x + 12 = x^2 + 7x + 12

. ✓

When the leading coefficient is not 1, for example

2x^2 + 7x + 3

, the process is called decomposition. Multiply the leading coefficient by the constant:

2 \times 3 = 6

. Find two numbers that multiply to 6 and add to 7: those are 1 and 6. Rewrite the middle term using these numbers:

2x^2 + x + 6x + 3

. Group in pairs and factor each group:

x(2x + 1) + 3(2x + 1)

. Both groups share the factor

(2x + 1)

, giving

(x + 3)(2x + 1)

.

Pay close attention to signs. If

c

is positive,

p

and

q

have the same sign (both positive or both negative, matching the sign of

b

). If

c

is negative,

p

and

q

have opposite signs.

x^2 + bx + c = (x + p)(x + q)

For $x^2 + bx + c$ , find $p$ and $q$ where $pq = c$ and $p + q = b$ .
The factored form is $(x + p)(x + q)$ .
For $ax^2 + bx + c$ with $a \neq 1$ , use decomposition: multiply $a \times c$ , split the middle term, then group and factor.
Watch signs carefully: same signs when $c > 0$ , opposite signs when $c < 0$ .

Strategy 3: Difference of Squares

A difference of squares is an expression of the form

A^2 - B^2

, where one perfect square is subtracted from another. It factors neatly into

(A + B)(A - B)

. You can verify this pattern by expanding:

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

. The middle terms cancel, which is exactly why this pattern works.

To use this strategy, you need to recognize perfect squares. Common perfect square numbers include 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100. Variable expressions like

x^2

,

4x^2

, and

9y^2

are also perfect squares because their square roots are integers or simple expressions.

For example,

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

. Another example:

4x^2 - 49 = (2x)^2 - 7^2 = (2x + 7)(2x - 7)

. Two conditions must both be true: the expression must have exactly two terms, and there must be a subtraction (not addition) between them. A sum of squares, such as

x^2 + 25

, does not factor over the integers.

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

The pattern is $A^2 - B^2 = (A + B)(A - B)$ .
Both terms must be perfect squares and separated by a minus sign.
Identify $A$ and $B$ by taking the square root of each term.
A sum of squares ( $A^2 + B^2$ ) does not factor using this method.

Choosing the Right Strategy

When you see a quadratic to factor, work through a quick mental checklist. First, is there a GCF? If yes, remove it — always. Second, how many terms remain after removing the GCF? Two terms that fit

A^2 - B^2

call for the difference-of-squares pattern. Three terms call for trinomial factoring (find

p

and

q

, or use decomposition if the leading coefficient is not 1).

Sometimes you need more than one strategy in sequence. For instance,

2x^2 - 50

has a GCF of 2, giving

2(x^2 - 25)

, and then

x^2 - 25

is a difference of squares, giving

2(x + 5)(x - 5)

. Always check whether the expression inside the brackets can be factored further before declaring you are done.

Not every quadratic factors over the integers. If you cannot find integer values of

p

and

q

for a trinomial, or if a two-term expression is not a difference of squares, the expression may be prime (non-factorable). At the Grade 10 level, exam questions will always lead to integer factors.

Always remove the GCF first.
Two terms and a minus sign → try difference of squares.
Three terms → try trinomial factoring or decomposition.
Apply more than one strategy in sequence when needed.
Check by expanding your final answer.

Factoring Strategy Selector

What you see after removing the GCF	Number of terms	Strategy to use	Factored form pattern
$ax^2 + bx$ (no constant)	2	GCF only	$x(ax + b)$
$A^2 - B^2$ (two perfect squares, minus sign)	2	Difference of squares	$(A+B)(A-B)$
$x^2 + bx + c$ (leading coefficient is 1)	3	Find $p$ and $q$	$(x+p)(x+q)$
$ax^2 + bx + c$ , $a \neq 1$	3	Decomposition	$(px+r)(qx+s)$

Worked example

Factor $3x^2 - 12x - 36$ completely

Factor

3x^2 - 12x - 36

completely.

Look for a GCF
Check all three coefficients: 3, 12, and 36. The GCF of these numbers is 3. There is no variable factor common to all terms because the last term, $-36$ , has no $x$ . Divide every term by 3.
$3x^2 - 12x - 36 = 3(x^2 - 4x - 12)$
Identify the trinomial inside the brackets
Inside the brackets you now have $x^2 - 4x - 12$ . This is a trinomial with leading coefficient 1, so look for two integers $p$ and $q$ where $p \times q = -12$ and $p + q = -4$ .
$p × q = -12, p + q = -4$
List factor pairs of $-12$ and find the right pair
Because the product is negative, $p$ and $q$ have opposite signs. Pairs that multiply to $-12$ : $(1, -12),\, (-1, 12),\, (2, -6),\, (-2, 6),\, (3, -4),\, (-3, 4)$ . The pair that adds to $-4$ is $(2, -6)$ because $2 + (-6) = -4$ .
$p = 2, q = -6$
Write the factored trinomial
Replace $x^2 - 4x - 12$ with $(x + 2)(x - 6)$ . Do not forget to keep the GCF of 3 in front.
$3(x + 2)(x - 6)$
Check by expanding
Expand $(x + 2)(x - 6)$ first: $x^2 - 6x + 2x - 12 = x^2 - 4x - 12$ . Then multiply by 3: $3x^2 - 12x - 36$ . This matches the original expression, so the answer is correct.
$3(x^2 - 4x - 12) = 3x^2 - 12x - 36\ ✓$

Answer:

3(x + 2)(x - 6)

Check: Expanding

3(x+2)(x-6)

gives

3(x^2 - 4x - 12) = 3x^2 - 12x - 36

, which matches the original expression.

Worked example

Factor $5x^2 - 45$ completely

Factor

5x^2 - 45

completely.

Look for a GCF
The coefficients are 5 and 45. The GCF is 5. There is no common variable factor because the second term has no $x$ . Divide both terms by 5.
$5x^2 - 45 = 5(x^2 - 9)$
Examine the bracket for further factoring
Inside the bracket you have $x^2 - 9$ . This is a two-term expression with a minus sign. Check whether both terms are perfect squares: $x^2 = (x)^2$ and $9 = (3)^2$ . Both conditions are met, so this is a difference of squares.
$x^2 - 9 = x^2 - 3^2$
Apply the difference-of-squares pattern
Use $A^2 - B^2 = (A + B)(A - B)$ with $A = x$ and $B = 3$ . Replace $x^2 - 9$ with $(x + 3)(x - 3)$ and keep the factor of 5 outside.
$5(x + 3)(x - 3)$
Check by expanding
Expand $(x + 3)(x - 3)$ : the outer and inner terms cancel, leaving $x^2 - 9$ . Multiply by 5 to get $5x^2 - 45$ . This matches the original expression.
$5(x^2 - 9) = 5x^2 - 45\ ✓$

Answer:

5(x + 3)(x - 3)

Check: Expanding

5(x+3)(x-3)

gives

5(x^2 - 9) = 5x^2 - 45

, which matches the original expression.

Common mistakes and how to avoid them

Forgetting to factor out the GCF first, which makes trinomial factoring much harder than it needs to be.

Correction: Before trying any other strategy, always check for a common factor across all terms and remove it.

Thinking

x^2 + 25

is a difference of squares and writing

(x + 5)(x - 5)

.

Correction: Difference of squares requires subtraction.

x^2 + 25

does not factor over the integers. Only

x^2 - 25

fits the pattern.

Finding

p

and

q

that add to

b

but forgetting to check that they also multiply to

c

.

Correction: Both conditions must be satisfied:

p + q = b

AND

p \times q = c

. Always verify both before writing the factored form.

Writing

(x - 5)^2

instead of

(x + 5)(x - 5)

for a difference of squares.

Correction: A difference of squares produces two different brackets: one with addition and one with subtraction. It is not a perfect square trinomial.

Stopping after one factoring step when the expression inside the brackets can still be factored further.

Correction: After each factoring step, look inside every bracket. Keep factoring until no bracket can be simplified further.

Lesson summary

Always remove the GCF first — it simplifies every step that follows.
A difference of squares $A^2 - B^2$ factors into $(A+B)(A-B)$ ; both terms must be perfect squares separated by a minus sign.
For a trinomial $x^2 + bx + c$ , find two integers $p$ and $q$ where $pq = c$ and $p + q = b$ , then write $(x+p)(x+q)$ .
For a trinomial $ax^2 + bx + c$ with $a \neq 1$ , use decomposition: multiply $a \times c$ , split the middle term, group, and factor.
Use the strategy selector: two terms with minus → difference of squares; three terms → trinomial or decomposition.
Always verify your answer by expanding the brackets and confirming you get the original expression back.

Check your understanding

Question 1

Which of the following is the fully factored form of

x^2 - 3x - 10

?

$(x - 5)(x + 2)$
$(x + 5)(x - 2)$
$(x - 5)(x - 2)$
$(x + 5)(x + 2)$

Show answer and explanation

(x - 5)(x + 2)

You need two integers that multiply to

-10

and add to

-3

. The pair

-5

and

2

works:

(-5)(2) = -10

and

(-5) + 2 = -3

. So the factored form is

(x - 5)(x + 2)

.

Question 2

What is the fully factored form of

4x^2 - 36

?

$(2x - 6)(2x + 6)$
$(4x - 9)(x + 4)$
$4(x + 3)(x - 3)$
$2(2x^2 - 18)$

Show answer and explanation

4(x + 3)(x - 3)

First, the GCF of

4x^2

and

36

is 4, giving

4(x^2 - 9)

. Then

x^2 - 9 = (x+3)(x-3)

by the difference of squares. The fully factored form is

4(x+3)(x-3)

. Option A is not fully factored because

(2x-6)

and

(2x+6)

each still contain a factor of 2.

Question 3

For the trinomial

x^2 + 2x - 15

, which pair of integers

p

and

q

satisfies both

pq = -15

and

p + q = 2

?

$p = 3,\ q = -5$
$p = 5,\ q = -3$
$p = -3,\ q = -5$
$p = 15,\ q = -1$

Show answer and explanation

p = 5,\ q = -3

Check each option:

5 \times (-3) = -15

✓ and

5 + (-3) = 2

✓. Option A gives

3 + (-5) = -2

, not

2

. Option C gives a product of

15

, not

-15

. Option D gives

15 + (-1) = 14

, not

2

.

Question 4

Which expression is a difference of squares and can be factored using the pattern

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

?

$x^2 + 16$
$x^2 - 7$
$9x^2 - 25$
$4x^2 + 49$

Show answer and explanation

9x^2 - 25

9x^2 - 25 = (3x)^2 - 5^2

, so

A = 3x

and

B = 5

, giving

(3x+5)(3x-5)

. The other options either have a plus sign (so they are sums, not differences) or contain a term like

x^2 - 7

where 7 is not a perfect square.

Key terms

Factor (verb): To rewrite an expression as a product of two or more simpler expressions.
Quadratic expression: A polynomial whose highest power of the variable is 2, for example $x^2 + 5x + 6$ .
Greatest Common Factor (GCF): The largest number and highest variable power that divides evenly into every term of a polynomial.
Trinomial: A polynomial with exactly three terms, such as $x^2 + 7x + 12$ .
Difference of squares: An expression of the form $A^2 - B^2$ , which factors as $(A+B)(A-B)$ .
Perfect square: A number or expression that is the square of an integer or a simple expression, such as 9 (since $3^2 = 9$ ) or $4x^2$ (since $(2x)^2 = 4x^2$ ).
Decomposition: A trinomial-factoring method for $ax^2 + bx + c$ where $a \neq 1$ : multiply $a \times c$ , split the middle term, group in pairs, and factor each group.
Prime (polynomial): A polynomial that cannot be factored further over the integers.

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About this lesson

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 10 Mathematics (MPM2D), study topic Q10. It is a study resource, not an official curriculum publication.

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Q10 · Factor quadratics using common factors, trinomials, and differences of squares

Ontario Grade 10 Mathematics

Common Factors, Trinomials, and Differences of Squares

What you will learn

Bridge from Grade 9: What Factoring Means

Strategy 1: Greatest Common Factor (GCF)

Strategy 2: Factoring Trinomials

Strategy 3: Difference of Squares

Choosing the Right Strategy

Factoring Strategy Selector

Factor $3x^2 - 12x - 36$ completely

Factor $5x^2 - 45$ completely

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Question 4

Key terms

Continue through MPM2D

About this lesson