MCF3M Functions & Applications · Unit 3 — Quadratic Functions: Models and Forms
Lesson 3.4
Solving Quadratic Equations by Factoring
Connect factors to graph roots, solve with the zero-product property, keep every solution, and interpret break-even points.
Estimated time: 45 minutes
Learning goals
- rearrange an equation with zero on one side
- factor completely using appropriate strategies
- apply the zero-product property
- solve each variable-containing factor
- check in the original equation
- recognize repeated roots
- avoid losing roots through division
- solve contextual problems
Step 1 of 4
Warm Up
Success criteria
Step 2 of 4
Learn
See it — the same roots as Lesson 3.3
Graphing method — zeros
x² − 5x + 6 = 0
Select all roots (x-values), or select none for no real roots.
x² − 5x + 6 = (x − 2)(x − 3)The factors describe the same parabola.(x − 2)(x − 3) = 0 → x = 2 or x = 3Factoring finds the same roots algebraically.
3 × 0 = 0The second factor is zero.0 × 8 = 0The first factor is zero.3 × 8 = 24 ≠ 0Neither factor is zero.
Zero-product property: if AB = 0, then A = 0 OR B = 0. Use it only when the product equals zero.
(x − 4)(x + 2) = 0 → x − 4 = 0 or x + 2 = 0 → x = 4 or −2.
Factor x − 4 → root 4; factor x + 2 → root −2. Each root makes a different bracket zero.
y(x) = (x − 1)² − 9
Five steps to keep every solution
1. Rearrange so one side is zero. 2. Factor completely. 3. Set each variable-containing factor equal to zero. 4. Solve. 5. Verify in the original equation.
Worked example
Example A — Solve x² − 7x + 10 = 0.
Rearrange
The equation already equals zero.
Factor
Numbers with product 10 and sum −7: −5 and −2. (x − 5)(x − 2) = 0.
Set each factor to zero
x − 5 = 0 OR x − 2 = 0.
Solve
x = 5 OR x = 2.
Check
25 − 35 + 10 = 0; 4 − 14 + 10 = 0.
Answer: x = 5 or 2
Worked example
Example B — Solve 2x² + 7x + 3 = 0.
Rearrange
Already zero on one side.
Factor
ac = 6. Split 7x as 6x + x: 2x² + 6x + x + 3 = 2x(x + 3) + (x + 3) = (2x + 1)(x + 3).
Zero-product property
2x + 1 = 0 OR x + 3 = 0.
Solve
2x = −1 → x = −1/2; x = −3.
Check
2(−1/2)² + 7(−1/2) + 3 = 1/2 − 7/2 + 3 = 0. At −3: 18 − 21 + 3 = 0.
Answer: x = −1/2 or −3
Factor → Solve → Graph
x² − 4x − 5 = 0
1. Which pair has product −5 and sum −4?
Worked example
Example C — Solve 3x² − 12x = 0. Why not divide by x?
Factor the GCF
3x(x − 4) = 0.
Set variable factors to zero
x = 0 OR x − 4 = 0. The constant 3 cannot become zero.
Solve
x = 0 OR x = 4.
Check
At 0: 0 − 0 = 0. At 4: 48 − 48 = 0.
Protect every solution
Dividing by x assumes x is not zero and would remove the valid solution x = 0.
Answer: x = 0 or 4
Worked example
Example D — Solve 4x² − 25 = 0.
Recognize the pattern
(2x)² − 5² is a difference of squares.
Factor
(2x − 5)(2x + 5) = 0.
Set factors to zero
2x − 5 = 0 OR 2x + 5 = 0.
Solve
x = 5/2 OR x = −5/2.
Check
For either sign: 4(25/4) − 25 = 0.
Answer: x = ±5/2
Worked example
Example E — Solve x² + 3x = 10.
Rearrange
Subtract 10: x² + 3x − 10 = 0.
Factor
Product −10, sum 3: 5 and −2. (x + 5)(x − 2) = 0.
Set factors to zero
x + 5 = 0 OR x − 2 = 0.
Solve
x = −5 OR x = 2.
Check original equation
(−5)² + 3(−5) = 25 − 15 = 10. 2² + 3(2) = 4 + 6 = 10.
Answer: x = −5 or 2
Worked example
Example F — Solve 2(x + 1)² = 5(x + 1). Do not cancel the bracket.
Rearrange
2(x + 1)² − 5(x + 1) = 0.
Factor the common binomial
(x + 1)[2(x + 1) − 5] = 0.
Simplify
(x + 1)(2x − 3) = 0.
Set factors to zero and solve
x + 1 = 0 → x = −1; 2x − 3 = 0 → x = 3/2.
Check original equation
At −1: 2(0)² = 5(0) = 0. At 3/2: 2(5/2)² = 25/2 and 5(5/2) = 25/2.
Explain
Cancelling x + 1 assumes it is not zero and would lose x = −1.
Answer: x = −1 or 3/2
See it — a repeated root
Graphing method — zeros
x² − 6x + 9 = 0
Select all roots (x-values), or select none for no real roots.
x² − 6x + 9 = (x − 3)² = 0Perfect-square trinomialx − 3 = 0 → x = 3One distinct root; 9 − 18 + 9 = 0. The graph touches once.
When real factoring is unavailable
x² + 4 = 0
Graphing method — zeros
Factoring method
No real linear factors: x² + 4 is always positive.
No crossings and no real linear factors: both methods agree.
Choose a suitable method
Real-world idea
School fundraiser — when does it break even?
A school models ticket profit using P(x) = −2x² + 40x − 128. The input x counts tickets in tens; the output P is profit.
Break-even means profit equals zero. Positive profit occurs where the downward-opening graph lies above the x-axis.
See it — break-even points and positive profit
Graphing method — zeros
P(x) = −2x² + 40x − 128
Select all roots (x-values), or select none for no real roots.
x counts tens of tickets; P(x) is profit. Zeros 4 and 16 mean 40 and 160 tickets. The curve is above zero between them: 41–159 whole tickets, or 50–150 if selling only in tens.
Worked example
School fundraiser — When does P(x) = −2x² + 40x − 128 break even? x counts tickets in tens.
Model and term
Input x = tickets ÷ 10; output P is profit. Break-even means profit zero, so solve P(x) = 0.
Factor completely
−2(x² − 20x + 64) = −2(x − 4)(x − 16) = 0.
Set variable factors to zero
x − 4 = 0 OR x − 16 = 0 → x = 4 or 16.
Check
P(4) = −32 + 160 − 128 = 0. P(16) = −512 + 640 − 128 = 0.
Interpret the units
x = 4 means 40 tickets; x = 16 means 160 tickets.
Profit between roots
The graph opens down and is above zero for 4 < x < 16. That is 40 < tickets < 160: whole-ticket counts 41–159; if selling only in tens, 50–150.
Answer: Break-even at 40 and 160 tickets
Two methods, same solutions
x² − 5x + 6 = 0
Graphing method — zeros
Factoring method
(x − 2)(x − 3) = 0
A factor is zero at x = 2
A factor is zero at x = 3
Both methods give the same solutions. Graphing shows where; factoring explains why.
Step 3 of 4
Practice
Graph check — factors and zeros agree
x² − 5x + 6 = 0
Graphing method — zeros
Factoring method
(x − 2)(x − 3) = 0
A factor is zero at x = 2
A factor is zero at x = 3
Both methods give the same solutions. Graphing shows where; factoring explains why.
Build it
Challenge
Apply it
Explain it
Step 4 of 4
Mastery Check
Lesson 3.4 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.