MCF3M Functions & Applications · Unit 2 — The Algebra of Quadratic Expressions
Lesson 2.4
Factoring Quadratic Expressions: ax² + bx + c
Use the ac method, split the middle term, and factor by grouping. Take out a GCF first when needed, then expand to check.
Estimated time: 30–45 minutes
Learning goals
- factor quadratic expressions with a leading coefficient greater than 1
- identify a, b and c and find two integers with product ac and sum b
- split the middle term and factor by grouping
- use guess-and-check as an alternative strategy
- take out a GCF first and keep it in the final answer
- check factors by expanding
Step 1 of 4
Warm Up
Success criteria
Step 2 of 4
Learn
Why does the leading coefficient matter?
For x² + bx + c, the constants in (x + p)(x + q) add to b and multiply to c. For 2x² + 7x + 3, one bracket needs 2x. Now the middle terms come from 2x times one constant, plus x times the other.
For example, (2x + 1)(x + 3) gives 6x + x = 7x. The constants 1 and 3 add to 4, not 7. So simply finding constants with sum 7 and product 3 will not work.
Split bx into two terms whose coefficients multiply to ac and add to b. Then factor by grouping.
Splitting changes how the expression looks, not its value.
The ac method
First check for a GCF. Next, identify a, b and c, including their signs. Multiply a × c. Find two integers with product ac and sum b.
Rewrite bx using those two numbers. Group the four terms into two pairs. Factor each pair, then take out the shared binomial. Finally, expand to check.
Why ac? In (px + q)(rx + s), the middle coefficients are ps and qr. Their sum is b, and their product is (pr)(qs) = ac. Splitting puts these middle products back into the expression.
Key term
Factoring by grouping
Make two pairs of terms, factor each pair, then take out the binomial both pairs share.
2x(x + 4) + (x + 4) = (2x + 1)(x + 4)
Worked example
Example A — Factor 2x² + 9x + 4.
Identify
No GCF greater than 1. a = 2, b = 9, c = 4.
ac = 2 × 4 = 8
Find the pair
8 and 1 multiply to 8 and add to 9.
8 × 1 = 8; 8 + 1 = 9
Split
Replace 9x with 8x + x.
2x² + 8x + x + 4
Group
Take 2x from the first pair and 1 from the second.
2x(x + 4) + 1(x + 4)
Factor
Both pairs share x + 4. Collect the outside terms 2x and 1.
(2x + 1)(x + 4)
Check
Multiply all four pairs and collect the middle terms.
2x² + 8x + x + 4 = 2x² + 9x + 4
Answer: (2x + 1)(x + 4)
Algebra workspace · Split the Middle Term
Product ac and sum b both need to match.
Predict first
For 3x² + 10x + 3, should the splitting numbers multiply to c or ac?
Make a prediction to open the workspace.
Worked example
Example B — Factor 3x² − 11x + 6.
Identify
a = 3, b = −11, c = 6. No common factor greater than 1.
ac = 18
Find the pair
Positive product, negative sum: use two negative numbers.
(−9)(−2) = 18; −9 − 2 = −11
Split
Replace −11x with −9x − 2x.
3x² − 9x − 2x + 6
Group
Factor 3x from the first pair and −2 from the second so the brackets match.
3x(x − 3) − 2(x − 3)
Factor
Take out the common bracket x − 3.
(3x − 2)(x − 3)
Check
Expand and collect like terms.
3x² − 9x − 2x + 6 = 3x² − 11x + 6
Answer: (3x − 2)(x − 3)
Keep the signs
Worked example
Example C — Factor completely 4x² + 20x + 24.
GCF first
Each coefficient is divisible by 4. Keep that 4 outside.
4(x² + 5x + 6)
Factor inside
The inside leading coefficient is 1. Use 2 and 3: product 6, sum 5.
4(x + 2)(x + 3)
Keep the GCF
Without the 4, your answer would be only one quarter of the original expression.
Check
Expand the binomials, then multiply every term by 4.
4(x² + 5x + 6) = 4x² + 20x + 24
Answer: 4(x + 2)(x + 3)
Another strategy: guess-and-check
For 2x² + 9x + 4, start with (2x + ?)(x + ?). The constants must multiply to 4. Try (2x + 4)(x + 1): its middle terms add to 6x, so it is wrong. Try (2x + 1)(x + 4): 8x + x = 9x, so it works.
This is a systematic search, not a random guess. Test factor pairs and their order. The ac method is usually easier when there are many pairs. If no integer pair has product ac and sum b, there is no factorization into integer binomials.
Real-world idea
A rectangle gives the factors meaning
Imagine a rectangular school display. Its area is A(x) = 2x² + 9x + 4 square metres, where x is a positive length in metres. Factoring finds possible side lengths: 2x + 1 and x + 4. The input x sets the lengths; the output A(x) is the area. Each binomial factor represents one dimension.
Four smaller rectangles
Split the display along the 2x and 1 parts of one side, and the x and 4 parts of the other. The four areas are 2x², 8x, x and 4. Adding them gives 2x² + 9x + 4; multiplying the full dimensions gives (2x + 1)(x + 4). This is an area model of the distributive property.
Worked example
Check the display's dimensions when x = 2 metres.
Find the dimensions
Substitute x = 2 into each binomial.
2x + 1 = 5; x + 4 = 6
Multiply
A rectangle's area is length times width.
A = 5 × 6 = 30 m²
Check the model
Substitute the same input into the expanded area expression.
A(2) = 2(2²) + 9(2) + 4 = 8 + 18 + 4 = 30 m²
Answer: 5 m by 6 m; area 30 m²
Step 3 of 4
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Step 4 of 4
Mastery Check
Lesson 2.4 mastery check
6 questions · no hints during the check. Answer them all, then check your score — you can retry as many times as you like.