Subsection 1.3.3 Factorising Quadratics
Factorisation is the process of breaking down a quadratic expression into a product of simpler binomials.
Subsubsection 1.3.3.1 Factorising Monic Quadratics
Teacher Resource 1.3.42.
To assist your teaching, we have prepared lesson resources, aligned with this textbook and the CBC. The Lesson Plan links to syllabus learning outcomes and provides suggest time allocations. The Step-by-Step Guide provides more detailed guidance on how to teach the content, including suggested questions to ask learners, and possible answers.
Learner Experience 1.3.9.
Work in groups to explore the following:
Expanding Brackets: Expand the following expressions:
-
\(\displaystyle (x + 3)(x + 4) = ?\)
-
\(\displaystyle (x - 6)(x - 5) = ?\)
-
\(\displaystyle (x + 2)(x + 3) = ?\)
Finding the Pattern: Look at your expanded answers and complete this table:
| Factored Form | Expanded Form | Sum of Constants | Product of Constants |
|---|---|---|---|
| \((x + 3)(x + 4)\) | \(x^2 + 7x + 12\) | \(3 + 4 = ?\) | \(3 \times 4 = ?\) |
| \((x - 6)(x - 5)\) | \(x^2 - 11x + 30\) | \((-6) + (-5) = ?\) | \((-6) \times (-5) = ?\) |
| \((x + 2)(x + 3)\) | \(?\) | \(2 + 3 = ?\) | \(2 \times 3 = ?\) |
The Reverse Challenge: Now try to REVERSE the process. Given \(x^2 + 5x + 6\text{,}\) can you find two numbers that
Discussion Questions:
-
What is the relationship between the sum of the constants in the factors and the coefficient of \(x\text{?}\)
-
What is the relationship between the product of the constants and the constant term?
-
How can you use these relationships to factorise a quadratic expression?
Teacher’s Role: Circulate among groups, asking probing questions. Guide students to discover that for \(x^2 + bx + c\text{,}\) we need two numbers m and n where \(m + n = b\) and \(m \times n = c\text{.}\) Use student discoveries to bridge to formal instruction.
Key Takeaway 1.3.43.
The expression
\begin{equation*}
ax^2 + bx + c,
\end{equation*}
where \(a, b, c\) are constants and \(a \neq 0\text{,}\) is called a quadratic expression.
In such expressions \(a\) is called the coefficient of \(x^2\text{,}\) \(b\) is called the coefficient of \(x\) and \(c\) is called the constant term.
When the coefficient of \(x^2\) is one, the expression is of the form;
\begin{equation*}
x^2 + bx + c.
\end{equation*}
Cosider the following expressions:
Expanding the expressions we get:
\begin{align*}
(x + 3)(x + 4) \amp = x(x + 3) + 4(x + 3)
\end{align*}
\begin{align*}
(x + 3)(x + 4) \amp = x^2 + 3x + 4x + 12
\end{align*}
Collecting like terms to get:
\begin{equation*}
x^2 + 7x + 12.
\end{equation*}
\(\,\)
Expanding the expressions we get:
\begin{align*}
(x - 6)(x - 5) \amp = x(x - 5) - 6(x - 5)
\end{align*}
\begin{align*}
(x - 6)(x - 5) \amp = x^2 - 5x - 6x + 30
\end{align*}
Simplify the middle terms:
\begin{equation*}
x^2 - 11x + 30.
\end{equation*}
From the above examples, we can see that the expressions \(x^2 + 7x + 12\) and \(x^2 - 11x + 30\) are formed from the factors of expressions \((x + 3)(x + 4)\) and \((x - 6)(x - 5)\) respectively.
The factorised form of the expression \(x^2 + bx + c\) is \((x + m)(x + n)\) where \(m\) and \(n\) are the factors of \(c\) whose sum is \(b\text{.}\)
In each case;
-
The sum of the constant terms in the factors is equal to the coefficient of \(x\) in the expression.
-
The product of the constant terms in the factors is equal to the constant term in the expression.
Example 1.3.44.
Solution.
In this case, the coefficient of \(x^2\) is one, the coefficient of \(x\) is 5 and the constant term is 6.
So the factors will be of the form \((x + m)(x + n)\) where \(m\) and \(n\) are the factors of 6 whose sum is 5.
In the expression \(x^2 + 5x + 6\text{,}\) look for two numbers such the numbers \(a\) and \(b\) such that
\begin{equation*}
a + b = 5
\end{equation*}
is coefficient of \(x\) and \(ab = 6\) is the constant term.
In this case, the numbers are 2 and 3.
\begin{equation*}
x^2 + 5x + 6 = x^2 + 2x + 3x + 6.
\end{equation*}
Grouping terms, we get:
\begin{equation*}
x^2 + 5x + 6 = x(x + 2) + 3(x + 2).
\end{equation*}
\begin{equation*}
x^2 + 5x + 6 = (x + 2)(x + 3).
\end{equation*}
Example 1.3.45.
Solution.
Look for two numbers such that the numbers \(a\) and \(b\) such that \(a + b = -9\) and \(ab = 20\text{.}\)
The numbers are -4 and -5.
Then, the expression \(x^2 - 9x + 20\) can be written as:
\begin{equation*}
x^2 - 4x - 5x + 20.
\end{equation*}
\begin{equation*}
x^2 - 9x + 20 = (x - 4)(x - 5).
\end{equation*}
Example 1.3.46.
Solution.
Look for two numbers such that the numbers \(a\) and \(b\) such that \(a + b = 6\) and \(ab = 9\text{.}\)
The numbers are 3 and 3.
Then, the expression \(x^2 + 6x + 9\) can be written as:
\begin{equation*}
x^2 + 3x + 3x + 9.
\end{equation*}
\begin{equation*}
x^2 + 6x + 9 = (x + 3)(x + 3).
\end{equation*}
Checkpoint 1.3.47.
Checkpoint 1.3.48.
Exercises Exercises
1.
Factorise the following expressions:
-
\(\displaystyle x^2 + 4x + 4.\)
-
\(\displaystyle x^2 + 8x + 15.\)
-
\(\displaystyle x^2 - 7x + 12.\)
-
\(\displaystyle x^2 - 6x + 9.\)
-
\(\displaystyle x^2 + 3x + 2.\)
-
\(\displaystyle x^2 - 5x + 6.\)
-
\(\displaystyle x^2 + 2x - 15.\)
-
\(\displaystyle x^2 - 4x - 5.\)
-
\(\displaystyle x^2 - 3x - 10.\)
-
\(\displaystyle x^2 + 7x + 10.\)
-
\(\displaystyle x^2 - 8x - 20.\)
-
\(\displaystyle x^2 + 9x + 20.\)
Answer.
-
We need two numbers that add to \(4\) and multiply to \(4\text{;}\) \(2\) and \(2\) work, so separate the middle term:\begin{align*} x^2 + 4x + 4 \amp = x^2 + 2x + 2x + 4\\ \amp = x(x + 2) + 2(x + 2)\\ \amp = (x + 2)(x + 2) \end{align*}
-
We look for numbers adding to \(8\) and multiplying to \(15\text{;}\) \(3\) and \(5\) meet those conditions, so rewrite the middle term:\begin{align*} x^2 + 8x + 15 \amp = x^2 + 3x + 5x + 15\\ \amp = x(x + 3) + 5(x + 3)\\ \amp = (x + 3)(x + 5) \end{align*}
-
We need numbers summing to \(-7\) and multiplying to \(12\text{;}\) \(-3\) and \(-4\) satisfy this, so split the middle term accordingly:\begin{align*} x^2 - 7x + 12 \amp = x^2 - 3x - 4x + 12\\ \amp = x(x - 3) - 4(x - 3)\\ \amp = (x - 3)(x - 4) \end{align*}
-
We need numbers that add to \(-6\) and multiply to \(9\text{;}\) \(-3\) and \(-3\) do, showing this is a perfect square. Splitting the middle term:\begin{align*} x^2 - 6x + 9 \amp = x^2 - 3x - 3x + 9\\ \amp = x(x - 3) - 3(x - 3)\\ \amp = (x - 3)(x - 3) \end{align*}
-
The numbers must sum to \(3\) and multiply to \(2\text{;}\) \(1\) and \(2\) satisfy this, so separate the middle term with \(1x\) and \(2x\text{:}\)\begin{align*} x^2 + 3x + 2 \amp = x^2 + x + 2x + 2\\ \amp = x(x + 1) + 2(x + 1)\\ \amp = (x + 1)(x + 2) \end{align*}
-
We need numbers that add to \(-5\) and multiply to \(6\text{;}\) \(-2\) and \(-3\) work, so use them to rewrite the middle term:\begin{align*} x^2 - 5x + 6 \amp = x^2 - 2x - 3x + 6\\ \amp = x(x - 2) - 3(x - 2)\\ \amp = (x - 2)(x - 3) \end{align*}
-
Find numbers summing to \(2\) and multiplying to \(-15\text{;}\) \(5\) and \(-3\) fit, so split the middle term with those values:\begin{align*} x^2 + 2x - 15 \amp = x^2 + 5x - 3x - 15\\ \amp = x(x + 5) - 3(x + 5)\\ \amp = (x + 5)(x - 3) \end{align*}
-
We want numbers adding to \(-4\) and multiplying to \(-5\text{;}\) \(-5\) and \(1\) work, so express the middle term as \(-5x + x\text{:}\)\begin{align*} x^2 - 4x - 5 \amp = x^2 - 5x + x - 5\\ \amp = x(x - 5) + 1(x - 5)\\ \amp = (x - 5)(x + 1) \end{align*}
-
Numbers must total \(-3\) and multiply to \(-10\text{;}\) \(-5\) and \(2\) meet this, letting us split the middle term:\begin{align*} x^2 - 3x - 10 \amp = x^2 - 5x + 2x - 10\\ \amp = x(x - 5) + 2(x - 5)\\ \amp = (x - 5)(x + 2) \end{align*}
-
We need numbers adding to \(7\) and multiplying to \(10\text{;}\) \(5\) and \(2\) satisfy this, so rewrite the middle term as \(5x + 2x\text{:}\)\begin{align*} x^2 + 7x + 10 \amp = x^2 + 5x + 2x + 10\\ \amp = x(x + 5) + 2(x + 5)\\ \amp = (x + 5)(x + 2) \end{align*}
-
We search for numbers summing to \(-8\) and multiplying to \(-20\text{;}\) \(-10\) and \(2\) work, so split using those:\begin{align*} x^2 - 8x - 20 \amp = x^2 - 10x + 2x - 20\\ \amp = x(x - 10) + 2(x - 10)\\ \amp = (x - 10)(x + 2) \end{align*}
-
We need numbers that add to \(9\) and multiply to \(20\text{;}\) \(4\) and \(5\) work perfectly, so use them to break the middle term:\begin{align*} x^2 + 9x + 20 \amp = x^2 + 4x + 5x + 20\\ \amp = x(x + 4) + 5(x + 4)\\ \amp = (x + 4)(x + 5) \end{align*}
Subsubsection 1.3.3.2 Factorising Non-Monic Quadratics
Curriculum Alignment
Teacher Resource 1.3.49.
To assist your teaching, we have prepared lesson resources, aligned with this textbook and the CBC. The Lesson Plan links to syllabus learning outcomes and provides suggest time allocations. The Step-by-Step Guide provides more detailed guidance on how to teach the content, including suggested questions to ask learners, and possible answers.
The general form of a quadratic expression is \(ax^2 + bx + c\text{,}\) where \(a\text{,}\) \(b\) and \(c\) are constants.
In this section we are going to discuss how to factorise a quadratic expression when the \(a\) in \(ax^2 + bx + c\) is not equal to \(1\text{.}\)
Consider the following expansions:
\begin{align*}
(3x + 2)(2x + 1) \amp = (3x \times 2x) + (3x \times 1) + (2 \times 2x) + (2 \times 1)\\
\amp = 6x^2 + 3x + 4x + 2\\
\amp = 6x^2 + 7x + 2\\
(3x - 2)(4x - 3) \amp = (3x \times 4x) - (3x \times 3) - (2 \times 4x) + (2 \times 3)\\
\amp = 12x^2 - 9x - 8x + 6\\
\amp = 12x^2 - 17x + 6\\
(2x + 3)(3x - 4) \amp = (2x \times 3x) - (2x \times 4) + (3 \times 3x) - (3 \times 4)\\
\amp = 6x^2 - 8x + 9x - 12\\
\amp = 6x^2 + x - 12
\end{align*}
From the above examples, we can see that the factorisation of a quadratic expression when the coefficient of \(x^2\) is not one is similar to the factorisation of a quadratic expression when the coefficient of \(x^2\) is one.
Factors are multiplied to get the final expresions on the righthand side (RHS) of the equations.
Investigation 1.3.10.
Given the quadratic expressions on the right-hand side of the equation, how can you be able factor them?
Consider the expression \(6x^2 + 7x + 2\text{.}\)
The problem can factorised as follows:
-
Look for two numbers such that:
-
Their product is equal to the product of the coefficient of \(x^2\) and the constant term, i.e. \(6 \times 2 = 12\text{.}\)
-
-
Rewrite the term \(7x\) as the sum of the two numbers found in step (i)(b).\begin{equation*} = 3x(2x + 1) + 2(2x + 1) \end{equation*}\begin{equation*} = (3x + 2)(2x + 1) \end{equation*}
Learner Experience 1.3.11.
Work in groups to define, discuss, and work on the following:
-
Factorization of quadratic expressions
-
Identifying common factors in expressions
-
Factorizing using the method of splitting the middle term
-
Recognizing the difference between factoring by grouping and simple factoring
Copy the following expressions, observe and discuss:
-
Factorize: \(x^2 + 5x + 6\)
-
Factorize: \(x^2 - 7x + 12\)
-
Factorize: \(3x^2 - 15x\)
-
Factorize by grouping: \(x^2 + 4x + 3x + 12\)>
Discussion Questions:
-
Compare the different approaches groups used to factor similar expressions.
-
Discuss how factoring helps in solving quadratic equations.
-
Explore how recognizing common factors and patterns makes factoring easier.
-
How does factoring help us simplify algebraic expressions faster?
-
What challenges do you face when factoring complex expressions?
-
Can you think of any real-world scenarios where factoring is useful (e.g., optimizing areas, engineering problems)?
Teacher’s Role: The teacher circulates among groups, asking probing questions (e.g., "What two numbers multiply to give \(c\) and add to give \(b\text{?}\)", "Can you identify a common factor first?", "How do you know when to use grouping?"). The teacher uses student discoveries to bridge to formal instruction.
Key Takeaway 1.3.50.
What is Factoring Quadratic Expressions?
Factoring quadratic expressions involves expressing a quadratic equation in the form of two binomials:
\((ax + m)(bx + n)\)
The ac-Method (Splitting the Middle Term):
For a quadratic expression \(ax^2 + bx + c\text{,}\) find two numbers m and n such that:
Once we find \(m\) and \(n\text{,}\) we break the middle term \(bx\) into two terms using \(m\) and \(n\text{,}\) then factor by grouping.
Three Factoring Methods:
| Method | When to Use | Example |
|---|---|---|
| Common Factor | When all terms share a common factor | \(3x^2 - 15x = 3x(x - 5)\) |
| Simple Factoring | When \(a = 1\) (leading coefficient is \(1\)) | \(x^2 + 5x + 6 = (x + 2)(x + 3)\) |
| Grouping | When \(a \ne 1\) or expression has \(4\) terms | \(x^2 + 4x + 3x + 12 = (x + 4)(x + 3)\) |
Remember: Always check for common factors FIRST before using other methods. Also, verify your answer by expanding the factors to get the original expression.
Example 1.3.51.
Factorise the expression \(12x^2 - 17x + 6\text{.}\)
Solution.
Multiply the coefficient of \(x^2\) (which is 12) by the constant term (which is 6):
\begin{equation*}
12 \times 6 = 72
\end{equation*}
Find two numbers that multiply to give \(72\) and add up to \(-17\text{.}\) The two numbers are \(-8\) and \(-9\text{.}\)
\begin{equation*}
-9 \times -8 = 72
\end{equation*}
and
\begin{equation*}
-9 + -8 = -17
\end{equation*}
Splt the middle term using these two numbers:
\begin{equation*}
12x^2 - 9x - 8x + 6
\end{equation*}
Factor by grouping:
\begin{equation*}
3x(4x - 3) - 2(4x - 3)
\end{equation*}
Group the factors:
\begin{equation*}
(3x - 2)(4x - 3)
\end{equation*}
Example 1.3.52.
Factorise the expression \(3x^2 - 5x - 2\text{.}\)
Solution.
The expression can be factorised as follows:
-
Look for two numbers such that:
-
Their product is equal to the product of the coefficient of \(x^2\) and the constant term, i.e. \(3 \times -2 = -6\text{.}\)
-
-
Rewrite the term \(-5x\) as the sum of the two numbers found in step 1.\begin{equation*} = 3x(x - 2) + 1(x - 2) \end{equation*}\begin{equation*} = (3x + 1)(x - 2) \end{equation*}
Checkpoint 1.3.53.
Checkpoint 1.3.54.
Exercises Exercises
1.
Factorise the following expressions:
-
-
\(\displaystyle x^2 + 5x + 6\)
-
\(\displaystyle 5x^2 - 9x + 4\)
-
\(\displaystyle x^2 - 7x + 10\)
-
\(\displaystyle 2x^2 + 7x + 3\)
-
-
-
\(\displaystyle 3x^2 + 11x + 6\)
-
\(\displaystyle x^2 - 9x + 20\)
-
\(\displaystyle 4x^2 + 12x + 9\)
-
\(\displaystyle x^2 + 4x + 4\)
-
-
-
\(\displaystyle 2x^2 + 7x + 6\)
-
\(\displaystyle 6x^2 + x - 2\)
-
\(\displaystyle 7x^2 + 10x + 3\)
-
\(\displaystyle 3x^2 - 14x - 5\)
-
-
-
\(\displaystyle 8x^2 + 2x - 3\)
-
\(\displaystyle 2x^2 + 4x - 30\)
-
\(\displaystyle 10x^2 - 7x - 3\)
-
\(\displaystyle 6x^2 + 11x - 10\)
-
Solution.
-
We look for numbers adding to \(5\) and multiplying to \(6\text{;}\) \(2\) and \(3\) satisfy this, so rewrite the middle term:\begin{align*} x^2 + 5x + 6 \amp = x^2 + 2x + 3x + 6\\ \amp = x(x + 2) + 3(x + 2)\\ \amp = (x + 3)(x + 2) \end{align*}
-
We look for numbers adding to \(-9\) and multiplying to \(20\text{;}\) \(-4\) and \(-5\) meet those conditions, so rewrite the middle term:\begin{align*} 5x^2 - 9x + 4 \amp = 5x^2 - 5x - 4x + 4\\ \amp = 5x(x - 1) - 4(x - 1)\\ \amp = (5x - 4)(x - 1) \end{align*}
-
We need numbers summing to \(-7\) and multiplying to \(10\text{;}\) \(-2\) and \(-5\) satisfy this, so split the middle term accordingly:\begin{align*} x^2 - 7x + 10 \amp = x^2 - 2x - 5x + 10\\ \amp = x(x - 2) - 5(x - 2)\\ \amp = (x - 2)(x - 5) \end{align*}
-
We need numbers that add to \(7\) and multiply to \(6\text{;}\) \(6\) and \(1\) satisfy this, so splitting the middle term accordingly:\begin{align*} 2x^2 + 7x + 3 \amp = 2x^2 + 6x + x + 3\\ \amp = 2x(x + 3) + 1(x + 3)\\ \amp = (2x + 1)(x + 3) \end{align*}
-
The numbers must sum to \(11\) and multiply to \(18\text{;}\) \(9\) and \(2\) satisfy this, so separate the middle term with \(9x\) and \(2x\text{:}\)\begin{align*} 3x^2 + 11x + 6 \amp = 3x^2 + 9x + 2x + 6\\ \amp = 3x(x + 3) + 2(x + 3)\\ \amp = (3x + 2)(x + 3) \end{align*}
-
We need numbers that add to \(-9\) and multiply to \(20\text{;}\) \(-5\) and \(-4\) work, so use them to rewrite the middle term:\begin{align*} x^2 - 9x + 20 \amp = x^2 - 5x - 4x + 20\\ \amp = x(x - 5) - 4(x - 5)\\ \amp = (x - 4)(x - 5) \end{align*}
-
Find numbers summing to \(12\) and multiplying to \(36\text{;}\) \(6\) and \(6\) fit, so split the middle term with those values:\begin{align*} 4x^2 + 12x + 9 \amp = 4x^2 + 6x + 6x + 9\\ \amp = 2x(2x + 3) + 3(2x + 3)\\ \amp = (2x + 3)(2x - 3)\\ \amp = (2x + 3)^2 \end{align*}
-
We want numbers adding to \(4\) and multiplying to \(4\text{;}\) \(2\) and \(2\) work, so express the middle term as \(2x + 2x\text{:}\)\begin{align*} x^2 + 4x + 4 \amp = x^2 + 2x + 2x + 4\\ \amp = x(x + 2) + 2(x + 2)\\ \amp = (x + 2)(x + 2)\\ \amp = (x+2)^2 \end{align*}
-
Numbers must total \(7\) and multiply to \(12\text{;}\) \(3\) and \(4\) meet this, letting us split the middle term:\begin{align*} 2x^2 + 7x + 6 \amp = 2x^2 + 4x + 3x + 6\\ \amp = 2x(x + 2) + 3(x + 2)\\ \amp = (2x + 3)(x + 2) \end{align*}
-
We need numbers adding to \(1\) and multiplying to \(-12\text{;}\) \(4\) and \(-3\) satisfy this, so rewrite the middle term as \(4x - 3x\text{:}\)\begin{align*} 6x^2 + x - 2 \amp = 6x^2 + 4x - 3x - 2\\ \amp = 2x(3x + 2) - 1(3x + 2)\\ \amp = (2x - 1)(3x + 2) \end{align*}
-
We search for numbers summing to \(10\) and multiplying to \(21\text{;}\) \(7\) and \(3\) work, so split using those:\begin{align*} 7x^2 + 10x + 3 \amp = 7x^2 + 7x + 3x + 3\\ \amp = 7x(x + 1) + 3(x + 1)\\ \amp = (7x + 3)(x + 1) \end{align*}
-
We need numbers that add to \(-14\) and multiply to \(-15\text{;}\) \(-15\) and \(1\) work perfectly, so use them to break the middle term:\begin{align*} 3x^2 - 14x - 5 \amp = 3x^2 - 15x + x - 5\\ \amp = 3x(x - 5) + 1(x - 5)\\ \amp = (3x + 1)(x - 5) \end{align*}
-
We look for numbers adding to \(2\) and multiplying to \(-24\text{;}\) \(6\) and \(4\) satisfy this, so rewrite the middle term:\begin{align*} 8x^2 + 2x - 3 \amp = 8x^2 + 6x - 4x - 3\\ \amp = 2x(4x + 3) - 1(4x + 3)\\ \amp = (2x - 1)(4x + 3) \end{align*}
-
We look for numbers adding to \(2\) and multiplying to \(-24\text{;}\) \(6\) and \(4\) satisfy this, so rewrite the middle term:\begin{align*} 2x^2 + 4x - 30 \amp = 8x^2 + 10x - 6x - 30\\ \amp = 2x(x + 5) - 6(x + 5)\\ \amp = (2x - 6)(x + 5) \end{align*}
-
We look for numbers adding to \(-7\) and multiplying to \(-30\text{;}\) \(-10\) and \(3\) satisfy this, so rewrite the middle term:\begin{align*} 10x^2 - 7x - 3 \amp = 10x^2 -10x + 3x - 3\\ \amp = 10x(x - 1) + 3(x - 1)\\ \amp = (10x + 3)(x - 1) \end{align*}
-
We look for numbers adding to \(2\) and multiplying to \(-24\text{;}\) \(6\) and \(4\) satisfy this, so rewrite the middle term:\begin{align*} 6x^2 + 11x - 10 \amp = 6x^2 + 15x - 4x - 10\\ \amp = 3x(2x + 5) - 2(2x + 5)\\ \amp = (3x - 2)(2x + 5) \end{align*}
