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A1.4 · Solve quadratic equations by factoring

Learn to solve quadratic equations by factoring through clear examples and targeted practice.

Ontario Grade 11 Mathematics

Quadratic Functions

Use factors and the zero product property to find the values that make a quadratic equation true.

A quadratic equation can have two solutions, one solution, or no real solutions. In this lesson, you will solve quadratic equations by factoring. The central idea is to rewrite an equation as a product equal to zero. Then you can find the values that make at least one factor zero. You will use familiar skills: expanding brackets, finding common factors, and solving simple equations.

What you will learn

1. Prerequisite bridge: expressions, factors, and equations

An expression is a mathematical phrase without an equals sign. For example, x2+5x+6x^2+5x+6 is an expression. An equation has an equals sign, such as x2+5x+6=0x^2+5x+6=0.
A factor is a quantity multiplied by another quantity. In 3(x+2)3(x+2), the factors are 33 and (x+2)(x+2). Factoring reverses expanding brackets: it rewrites an expression as a product of factors.
A quadratic expression has a variable raised to the power of two as its highest power. A quadratic equation sets a quadratic expression equal to another expression. In this lesson, the equations will be rearranged so that one side is zero.
x2+5x+6=(x+2)(x+3)x^2+5x+6=(x+2)(x+3)

2. The main idea: make a product equal zero

Consider the equation (x−4)(x+1)=0(x-4)(x+1)=0. A product is zero when at least one of its factors is zero. So either x−4=0x-4=0 or x+1=0x+1=0. These give x=4x=4 or x=−1x=-1.
This rule is called the zero product property. It says that if two or more factors multiply to zero, at least one factor must be zero. It lets us turn a factored equation into simpler equations.
Do not use this property on an expression that is not a product equal to zero. For example, x(x+2)=5x(x+2)=5 is not ready for the property because the product is not equal to zero. First, put all terms on one side so that the other side is zero.
(x-4)(x+1)=0\Rightarrow x-4=0 or x+1=0

3. Factoring and reading the solutions

A common type of quadratic is x2+bx+cx^2+bx+c, where bb and cc are numbers. To factor it, look for two numbers whose product is cc and whose sum is bb. Those numbers become the constants in the two brackets.
For instance, in x2+5x+6x^2+5x+6, the numbers 22 and 33 multiply to 66 and add to 55. So the expression factors as (x+2)(x+3)(x+2)(x+3). The factors reveal the solutions when the expression is set equal to zero.
Some quadratics have a number in front of x2x^2. In those cases, the factors may need more than one number in each bracket. You can test a factorization by expanding it. The expanded expression must match the original expression, term by term.
After finding possible solutions, check each one in the original equation. Substituting a solution means replacing the variable with that value. A correct solution makes both sides equal.
x^2+bx+c=(x+m)(x+n), m+n=b, mn=c

4. A reliable solving routine

First, arrange the equation so that one side is zero. Combine like terms if needed. Like terms have the same variable part, such as 3x3x and −5x-5x.
Next, factor the quadratic expression. Look for a common factor first if every term shares one. Then check the factors by expanding.
Set each factor equal to zero and solve. Finally, substitute the results into the original equation. This last check can catch a sign error or an incorrect factorization.
ab=0\Rightarrow a=0 or b=0

Checking the factorization

PartProduct from the bracketsResult
First terms(3x)(x)(3x)(x)3x23x^2
Outer and inner terms(3x)(−2)+(1)(x)(3x)(-2)+(1)(x)−6x+x=−5x-6x+x=-5x
Constant terms(1)(−2)(1)(-2)−2-2

Worked example

Solve a quadratic with a coefficient in front of the squared term

Solve 3x2−5x−2=03x^2-5x-2=0 by factoring.
  1. Check the equation form
    The equation already has zero on one side, so it is ready to factor.
    3x2−5x−2=03x^2-5x-2=0
  2. Factor the expression
    We need factors that expand to 3x2−5x−23x^2-5x-2. The brackets (3x+1)(3x+1) and (x−2)(x-2) work: their first terms produce 3x23x^2, their constant terms produce −2-2, and the middle terms combine to −5x-5x.
    3x2−5x−2=(3x+1)(x−2)3x^2-5x-2=(3x+1)(x-2)
  3. Use the zero product property
    Since the product equals zero, at least one bracket must equal zero. Solve one equation for each bracket. 3x+1=0 or x-2=0
  4. Solve each equation
    Subtracting or adding within each simple equation gives the two possible values of xx. x=-13\frac{1}{3} or x=2
  5. Check the values
    Substitution into the original equation gives zero for each value. For x=2x=2, the terms total 12−10−2=012-10-2=0. For x=−13x=-\frac{1}{3}, they total 13+53−2=0\frac{1}{3}+\frac{5}{3}-2=0.
    3(2)2−5(2)−2=03(2)^2-5(2)-2=0
Answer: The solutions are x=2x=2 and x=−13x=-\frac{1}{3}.
Check: Both values make the original left side equal zero.

Common mistakes and how to avoid them

Setting each term of a sum equal to zero.
Correction: The zero product property applies to factors multiplied together, not to separate terms being added. Factor the expression first.
Using the zero product property before making one side zero.
Correction: Rearrange the equation so the factored product equals zero before setting its factors equal to zero.
Changing a sign while solving a factor equation.
Correction: Solve each factor equation carefully, then substitute each result into the original equation.
Giving only one solution.
Correction: Check every factor. Each factor can give a solution, though the resulting values may sometimes be the same.

Lesson summary

Check your understanding

Question 1

Which pair of factors correctly rewrites x2+7x+12x^2+7x+12?
  1. (x+3)(x+4)(x+3)(x+4)
  2. (x−3)(x−4)(x-3)(x-4)
  3. (x+2)(x+6)(x+2)(x+6)
  4. correctIndex
Show answer and explanation
(x+3)(x+4)(x+3)(x+4)
The constants 33 and 44 multiply to 1212 and add to 77. Expanding the first choice gives x2+7x+12x^2+7x+12.

Question 2

What values solve (x−5)(x+2)=0(x-5)(x+2)=0?
  1. x=5x=5 or x=−2x=-2
  2. x=−5x=-5 or x=2x=2
  3. x=5x=5 or x=2x=2
  4. correctIndex
Show answer and explanation
x=5x=5 or x=−2x=-2
Set each factor equal to zero: x−5=0x-5=0 gives x=5x=5, and x+2=0x+2=0 gives x=−2x=-2.

Question 3

What should you do first to solve x2+3x=10x^2+3x=10 by factoring?
  1. Rewrite it as x2+3x−10=0x^2+3x-10=0.
  2. Set x2=0x^2=0 and 3x=103x=10.
  3. Divide both sides by xx.
  4. correctIndex
Show answer and explanation
Rewrite it as x2+3x−10=0x^2+3x-10=0.
Move all terms to one side to make an equation equal to zero. Then factor and use the zero product property.

Key terms

Quadratic equation
An equation whose highest power of the variable is two.
Factor
A quantity multiplied by another quantity to form a product.
Factoring
Rewriting an expression as a product of factors.
Zero product property
If a product equals zero, at least one of its factors must equal zero.
Solution
A value of the variable that makes an equation true.

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About this lesson

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Mathematics (MCF3M), expectation A1.4. It is a study resource, not an official curriculum publication.

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