Factor simple trinomials of the form x² + bx + c

Ontario Grade 10 Mathematics

Quadratic Relations

Breaking a quadratic expression into two binomial factors

In Grade 9 you learned to expand expressions like

(x + 3)(x + 5)

to get

x^2 + 8x + 15

. Factoring reverses that process: you start with

x^2 + 8x + 15

and find the two binomials that multiply together to produce it. This skill is central to the quadratic unit in MFM2P. It lets you rewrite a quadratic expression in a form that reveals useful information, such as the values of

x

that make the expression equal zero. This lesson focuses on trinomials — three-term expressions — that follow the pattern

x^2 + bx + c

, where

b

and

c

are integers.

What you will learn

Explain what it means to factor a trinomial of the form x² + bx + c.
Find two integers whose product equals c and whose sum equals b.
Write a trinomial as a product of two binomials and verify the result by expanding.
Factor trinomials where b or c is negative, choosing the correct signs for each factor.

Grade 9 Bridge: What Expanding Tells Us

Before factoring, it helps to watch expansion in slow motion so you can see the pattern it leaves behind.

Expand

(x + p)(x + q)

step by step using the distributive property. Multiply each term in the first binomial by each term in the second:

x \cdot x = x^2

, then

x \cdot q = qx

, then

p \cdot x = px

, and finally

p \cdot q = pq

. Collecting the two middle terms gives

x^2 + (p + q)x + pq

.

This shows a key fact: the coefficient of the middle term,

b

, equals

p + q

, and the constant at the end,

c

, equals

p \times q

. Factoring exploits exactly this relationship — you search for two numbers that add to

b

and multiply to

c

, then write them as the constants inside the two binomials.

(x+p)(x+q) = x^2 + (p+q)x + pq

Expanding $(x + p)(x + q)$ always produces $x^2 + (p+q)x + pq$ .
The middle coefficient $b = p + q$ and the constant term $c = p \times q$ .
Factoring is the reverse of expanding: you find $p$ and $q$ given $b$ and $c$ .

The Factor-Pair Method

To factor

x^2 + bx + c

, you need two integers

p

and

q

such that

p + q = b

and

p \times q = c

. A reliable way to find them is to list every pair of integers that multiply to

c

, then check which pair adds to

b

. This is called the factor-pair method.

Start by writing down factor pairs of

c

— that is, pairs of integers whose product is

c

. Remember to include pairs with negative integers when

c

is negative. Once you have your list, scan each pair for the one whose sum equals

b

. That pair gives you

p

and

q

, and the factored form is

(x + p)(x + q)

.

The order of the two binomials does not matter:

(x + p)(x + q)

and

(x + q)(x + p)

are identical. Always verify your answer by expanding it back out and confirming you recover the original trinomial.

In summary, you need two integers where their product equals

c

and their sum equals

b

.

List all integer pairs whose product is $c$ .
Identify the pair whose sum is $b$ .
Write the factored form as $(x + p)(x + q)$ .
Always verify by expanding your answer.

Working with Negative Values of b and c

The sign of

c

tells you whether

p

and

q

have the same sign or opposite signs. If

c

is positive, both integers are positive or both are negative (their product is positive either way). If

c

is negative, one integer is positive and the other is negative (opposite signs always give a negative product).

Once you know the signs, the sign of

b

narrows things down further. If

c

is positive and

b

is positive, both

p

and

q

are positive. If

c

is positive and

b

is negative, both

p

and

q

are negative. If

c

is negative, the integer with the larger absolute value takes the same sign as

b

.

For example, to factor

x^2 - 7x + 12

, you need two numbers that multiply to

12

and add to

-7

. Since

c = 12 > 0

and

b = -7 < 0

, both numbers must be negative. The pair

-3

and

-4

satisfies both conditions:

(-3) \times (-4) = 12

and

(-3) + (-4) = -7

. The factored form is

(x - 3)(x - 4)

.

If $c > 0$ : both integers share the same sign; $b$ tells you which sign.
If $c < 0$ : the integers have opposite signs; the larger absolute value takes the sign of $b$ .
Use these sign rules to shorten your search through factor pairs.

When a Trinomial Cannot Be Factored Over the Integers

Not every trinomial of the form

x^2 + bx + c

can be written as

(x + p)(x + q)

with integer values of

p

and

q

. If you list every factor pair of

c

and none of them sums to

b

, the trinomial is called prime over the integers — it cannot be factored using integers.

For example,

x^2 + 3x + 5

requires two integers that multiply to

5

and add to

3

. The only factor pairs of

5

are

(1, 5)

and

(-1, -5)

. Neither pair sums to

3

, so this trinomial is prime over the integers. In MFM2P, you only need to factor trinomials that do factor neatly — but knowing that factoring can fail is important so you do not force a wrong answer.

If no integer pair satisfies both conditions, the trinomial is prime over the integers.
Exhaust all factor pairs before concluding a trinomial cannot be factored.
In this course, exam trinomials will always factor over the integers.

Checking Your Work by Expanding

The fastest way to confirm your factoring is correct is to expand your answer and match it to the original trinomial. Use the distributive property (sometimes called FOIL for two binomials): multiply the First terms, the Outer terms, the Inner terms, and the Last terms, then collect like terms.

If your expansion matches the original expression exactly — same coefficients, same signs — your factoring is correct. If there is any mismatch, recheck your factor pair. A sign error is the most common source of mistakes, so pay close attention to positive and negative signs during verification.

Building the habit of checking every factoring answer by expanding will save marks on tests and help you catch errors before they become bigger problems later in the unit.

Always expand your factored form and compare it to the original trinomial.
A mismatch in signs or coefficients means a factor pair was chosen incorrectly.
Checking takes only seconds and prevents common sign errors.

Sign Rules for Choosing Factor Pairs

c is …	b is …	Both integers are …	Quick example
Positive	Positive	Both positive	x^2+7x+12 \to (x+3)(x+4)
Positive	Negative	Both negative	x^2-7x+12 \to (x-3)(x-4)
Negative	Positive	Opposite signs; larger absolute value is positive	x^2+2x-15 \to (x+5)(x-3)
Negative	Negative	Opposite signs; larger absolute value is negative	x^2-2x-15 \to (x-5)(x+3)

Worked example

Example 1: Positive b and Positive c

Factor

x^2 + 9x + 20

.

Identify b and c
Read off the values of $b$ and $c$ directly from the trinomial $x^2 + 9x + 20$ . Here $b = 9$ and $c = 20$ . You need two integers $p$ and $q$ such that $p + q = 9$ and $p \times q = 20$ .
$b = 9, \quad c = 20$
Use sign rules
Since $c = 20$ is positive and $b = 9$ is positive, both integers must be positive. This limits your search to positive factor pairs only.
List factor pairs of 20
Write every pair of positive integers whose product is $20$ , and note each sum so you can compare quickly.
$1 \times 20 = 20, \quad 2 \times 10 = 20, \quad 4 \times 5 = 20$
Identify the correct pair
Check the sums: $1 + 20 = 21$ , $2 + 10 = 12$ , and $4 + 5 = 9$ . The pair $4$ and $5$ gives a sum of $9$ , which matches $b$ .
$p = 4, \quad q = 5$
Write the factored form
Place $p = 4$ and $q = 5$ as the constants inside the two binomials.
$(x + 4)(x + 5)$
Verify by expanding
Expand $(x+4)(x+5)$ to confirm: $x \cdot x = x^2$ , $x \cdot 5 = 5x$ , $4 \cdot x = 4x$ , and $4 \cdot 5 = 20$ . Collect like terms to get $x^2 + 9x + 20$ . This matches the original, so the factoring is correct.
$(x+4)(x+5) = x^2 + 9x + 20 \checkmark$

Answer:

(x + 4)(x + 5)

Check: Expanding

(x+4)(x+5)

gives

x^2 + 5x + 4x + 20 = x^2 + 9x + 20

, which matches the original trinomial.

Worked example

Example 2: Negative b and Negative c

Factor

x^2 - 3x - 18

.

Identify b and c
From the trinomial $x^2 - 3x - 18$ , read $b = -3$ and $c = -18$ . You need two integers $p$ and $q$ with $p + q = -3$ and $p \times q = -18$ .
$b = -3, \quad c = -18$
Use sign rules
Since $c = -18$ is negative, the two integers must have opposite signs — one positive and one negative. Since $b = -3$ is negative, the integer with the larger absolute value must be negative.
List factor pairs of 18 with opposite signs
List pairs of integers with opposite signs whose product is $-18$ , writing the larger absolute value as negative. Compute each sum so you can compare to $b$ .
$1 \times (-18), \quad 2 \times (-9), \quad 3 \times (-6), \quad 6 \times (-3), \quad 9 \times (-2)$
Identify the correct pair
Compute the sums: $1 + (-18) = -17$ , $2 + (-9) = -7$ , $3 + (-6) = -3$ , $6 + (-3) = 3$ , $9 + (-2) = 7$ . The pair $3$ and $-6$ gives a sum of $-3$ , matching $b$ .
$p = 3, \quad q = -6$
Write the factored form
Place $p = 3$ and $q = -6$ as the constants inside the two binomials.
$(x + 3)(x - 6)$
Verify by expanding
Expand $(x+3)(x-6)$ : $x \cdot x = x^2$ , $x \cdot (-6) = -6x$ , $3 \cdot x = 3x$ , and $3 \cdot (-6) = -18$ . Collecting like terms: $x^2 + (-6x + 3x) - 18 = x^2 - 3x - 18$ . This matches the original trinomial.
$(x+3)(x-6) = x^2 - 3x - 18 \checkmark$

Answer:

(x + 3)(x - 6)

Check: Expanding

(x+3)(x-6)

gives

x^2 - 6x + 3x - 18 = x^2 - 3x - 18

, which matches the original trinomial.

Common mistakes and how to avoid them

Choosing a factor pair that multiplies to c but forgetting to check that it also adds to b.

Correction: Both conditions must hold simultaneously: the product must equal c AND the sum must equal b. Always verify both before writing the factored form.

Assigning the wrong signs to p and q, for example writing

(x-4)(x+5)

instead of

(x+4)(x+5)

.

Correction: Use the sign rules table: let c and b guide the signs before listing pairs. Always verify by expanding.

Skipping the verification step and submitting an answer without checking.

Correction: Always expand your factored form and confirm it matches the original trinomial exactly before moving on.

Stopping the factor-pair search too early without trying all pairs.

Correction: List all factor pairs of c systematically — from smallest to largest — so no pair is skipped.

Thinking that if

c

is negative, both integers must be negative.

Correction: A negative product requires one positive and one negative integer. Only a positive product allows both integers to share the same sign.

Lesson summary

Factoring reverses expansion: you rewrite $x^2 + bx + c$ as $(x+p)(x+q)$ where $p + q = b$ and $p \times q = c$ .
List all integer factor pairs of $c$ , then identify the pair whose sum equals $b$ .
Use sign rules to narrow your search: the sign of $c$ determines whether the integers share a sign or have opposite signs, and the sign of $b$ identifies which sign dominates.
Not every trinomial factors over the integers; if no pair satisfies both conditions, the trinomial is prime over the integers.
Always verify your answer by expanding the factored form and confirming it equals the original trinomial.

Check your understanding

Question 1

Which pair of integers should be used to factor

x^2 + 11x + 24

?

3 and 8
4 and 6
2 and 12
1 and 24

Show answer and explanation

3 and 8

You need two integers that multiply to 24 and add to 11. Check:

3 \times 8 = 24

and

3 + 8 = 11

. The other pairs do not add to 11:

4 + 6 = 10

,

2 + 12 = 14

,

1 + 24 = 25

.

Question 2

What is the fully factored form of

x^2 - 10x + 16

?

$(x - 2)(x - 8)$
$(x + 2)(x - 8)$
$(x - 4)(x - 4)$
$(x - 2)(x + 8)$

Show answer and explanation

(x - 2)(x - 8)

You need two integers that multiply to 16 and add to

-10

. Since

c = 16 > 0

and

b = -10 < 0

, both integers are negative. The pair

-2

and

-8

satisfies:

(-2)\times(-8) = 16

and

(-2)+(-8) = -10

. So the factored form is

(x-2)(x-8)

.

Question 3

Which trinomial is prime over the integers (cannot be factored using integers)?

$x^2 + 4x + 6$
$x^2 + 5x + 6$
$x^2 - 5x + 6$
$x^2 + x - 6$

Show answer and explanation

x^2 + 4x + 6

For

x^2+4x+6

, the factor pairs of 6 are

(1,6)

summing to 7 and

(2,3)

summing to 5 — neither sums to 4. The other trinomials all factor:

x^2+5x+6=(x+2)(x+3)

,

x^2-5x+6=(x-2)(x-3)

, and

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

.

Question 4

A student factors

x^2 - x - 12

as

(x + 3)(x - 4)

. Is this correct?

Yes, because $3 \times (-4) = -12$ and $3 + (-4) = -1$ .
No; the correct factoring is $(x - 3)(x + 4)$ .
No; the correct factoring is $(x - 3)(x - 4)$ .
No; the correct factoring is $(x + 6)(x - 2)$ .

Show answer and explanation

Yes, because

3 \times (-4) = -12

and

3 + (-4) = -1

.

Check:

3 \times (-4) = -12 = c

and

3 + (-4) = -1 = b

. Expanding

(x+3)(x-4)

gives

x^2 - 4x + 3x - 12 = x^2 - x - 12

, which matches. The student's answer is correct.

Key terms

Trinomial: A polynomial with exactly three terms, such as $x^2 + bx + c$ .
Factor (verb): To rewrite an expression as a product of two or more simpler expressions.
Binomial: A polynomial with exactly two terms, such as $x + 4$ .
Factor pair: A pair of integers whose product equals a given number, used when searching for the constants $p$ and $q$ .
Coefficient: The numerical part of a term; in $9x$ , the coefficient is 9.
Constant term: The term in a polynomial that has no variable; in $x^2 + 5x + 6$ , the constant term is 6.
Prime over the integers: A polynomial that cannot be factored into binomials with integer coefficients.
Verify: To check an answer — in factoring, this means expanding the factored form and confirming it equals the original expression.

Continue through MFM2P

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