Chapter 2: Expansion Formulas

Algebra Introduction - Chapter 2

Introductory (high school level)

What Is Expansion?

$(x+2)(x+3)$ and $x^2+5x+6$ give the same value for every $x$: they are two ways of writing the same polynomial, and what differs is what each form is good for. The product form shows at a glance where the expression becomes $0$, while the sum form is convenient for substituting a value of $x$ and computing. Being able to move between the two at will is the foundation for solving equations.

This chapter covers moving from the product form to the sum form. The reverse transformation is covered in Chapter 3: Factorisation.

Expansion

Rewriting a product of polynomials as a sum by means of the distributive law is called expansion.

Basic example: the distributive law

\begin{align} a(b + c) &= ab + ac \\ (a + b)(c + d) &= ac + ad + bc + bd \end{align}

$($

$a$

$+$

$b$

$)$

$($

$c$

$+$

$d$

$)$

$ac$

$+$

$ad$

$+$

$bc$

$+$

$bd$

Figure 1: Expanding $(a+b)(c+d)$. Each term of the left bracket is multiplied by each term of the right bracket, so $2 \times 2 = 4$ terms are produced.

Formula 1: Product of a Sum and a Difference

Theorem: product of a sum and a difference

$$(a + b)(a - b) = a^2 - b^2$$

Proof

Step 1 (apply the distributive law)

By the distributive law, multiply each term of $(a+b)$ by $(a-b)$:

$$(a + b)(a - b) = a(a - b) + b(a - b)$$

Step 2 (expand)

Expand each term further:

$$= a \cdot a + a \cdot (-b) + b \cdot a + b \cdot (-b)$$ $$= a^2 - ab + ab - b^2$$

Step 3 (collect like terms)

Since $-ab + ab = 0$, the middle terms cancel:

$$= a^2 - b^2$$

■

$a$

$b^2$

$a-b$

$b$

$a$

$b$

$a+b$

$a-b$

Figure 2: Removing a square of side $b$ from a square of side $a$ leaves area $a^2 - b^2$. Cutting the remainder into the two rectangles 1 and 2 and rearranging them gives a rectangle of height $a-b$ and width $a+b$. The area is unchanged, so $(a+b)(a-b) = a^2 - b^2$.

Example 1

Expand $(x + 3)(x - 3)$.

Solution: apply the formula with $a = x$, $b = 3$:

$$(x + 3)(x - 3) = x^2 - 3^2 = x^2 - 9$$

Example 2

Expand $(2x + 5)(2x - 5)$.

Solution: with $a = 2x$, $b = 5$:

$$(2x + 5)(2x - 5) = (2x)^2 - 5^2 = 4x^2 - 25$$

Formula 2: Squares of a Binomial

Theorem: square of a sum

$$(a + b)^2 = a^2 + 2ab + b^2$$

Proof

Step 1 (write the square as a product)

$$(a + b)^2 = (a + b)(a + b)$$

Step 2 (expand with the distributive law)

$$= a \cdot a + a \cdot b + b \cdot a + b \cdot b$$ $$= a^2 + ab + ab + b^2$$

Step 3 (collect like terms)

$$= a^2 + 2ab + b^2$$

■

$a$

$b$

$a$

$b$

$a^2$

$ab$

$b^2$

$(a+b)^2$

$= a^2 + 2ab + b^2$

Figure 3: Dividing a square of side $a+b$ into four parts gives a square of area $a^2$, two rectangles of area $ab$, and a square of area $b^2$. The middle term $2ab$ corresponds to those two rectangles.

Theorem: square of a difference

$$(a - b)^2 = a^2 - 2ab + b^2$$

Proof

Step 1 (write the square as a product)

$$(a - b)^2 = (a - b)(a - b)$$

Step 2 (expand with the distributive law)

$$= a \cdot a + a \cdot (-b) + (-b) \cdot a + (-b) \cdot (-b)$$ $$= a^2 - ab - ab + b^2$$

Step 3 (collect like terms)

$$= a^2 - 2ab + b^2$$

■

$a$

$b$

$(a-b)^2$

$ab$

$b^2$

$(a-b)^2$

$= a^2 - 2ab + b^2$

The hatched $b^2$ is

removed twice; add it back once

Figure 4: Removing a vertical strip of width $b$ (area $ab$) and a horizontal strip of height $b$ (area $ab$) from a square of side $a$ leaves $(a-b)^2$ at the top left. The two strips overlap in the $b^2$ square at the bottom right, so subtracting $ab$ twice removes $b^2$ twice over. Adding that one copy back is exactly the $+b^2$ in $(a-b)^2 = a^2 - 2ab + b^2$.

Example 3

Expand $(x + 4)^2$.

Solution:

\begin{align} (x + 4)^2 &= x^2 + 2 \cdot x \cdot 4 + 4^2 \\ &= x^2 + 8x + 16 \end{align}

Example 4

Expand $(3x - 2)^2$.

Solution:

\begin{align} (3x - 2)^2 &= (3x)^2 - 2 \cdot 3x \cdot 2 + 2^2 \\ &= 9x^2 - 12x + 4 \end{align}

A common mistake

Wrong: $(a + b)^2 = a^2 + b^2$

Right: $(a + b)^2 = a^2 + 2ab + b^2$

Never forget the middle term $2ab$.

Formula 3: $(x + a)(x + b)$

Theorem: expansion of $(x + a)(x + b)$

$$(x + a)(x + b) = x^2 + (a + b)x + ab$$

Proof

Step 1 (expand with the distributive law)

\begin{align} (x + a)(x + b) &= x \cdot x + x \cdot b + a \cdot x + a \cdot b \\ &= x^2 + bx + ax + ab \end{align}

Step 2 (collect the $x$ terms)

$$= x^2 + (a + b)x + ab$$

■

How to remember it

The coefficient of $x^2$ is $1$, the coefficient of $x$ is the sum of $a$ and $b$, and the constant term is the product of $a$ and $b$.

Example 5

Expand $(x + 3)(x + 5)$.

Solution: with $a = 3$, $b = 5$:

\begin{align} (x + 3)(x + 5) &= x^2 + (3 + 5)x + 3 \cdot 5 \\ &= x^2 + 8x + 15 \end{align}

Example 6

Expand $(x - 2)(x + 7)$.

Solution: with $a = -2$, $b = 7$:

\begin{align} (x - 2)(x + 7) &= x^2 + ((-2) + 7)x + (-2) \cdot 7 \\ &= x^2 + 5x - 14 \end{align}

Example 7

Expand $(x - 4)(x - 6)$.

Solution: with $a = -4$, $b = -6$:

\begin{align} (x - 4)(x - 6) &= x^2 + ((-4) + (-6))x + (-4) \cdot (-6) \\ &= x^2 - 10x + 24 \end{align}

Formula 4: Cubes of a Binomial

Theorem: cube of a sum

$$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$$

Proof

Step 1 (split off one factor)

$$(a + b)^3 = (a + b)^2 (a + b)$$

Step 2 (expand $(a+b)^2$)

By the formula already proved:

$$(a + b)^2 = a^2 + 2ab + b^2$$

Hence:

$$(a + b)^3 = (a^2 + 2ab + b^2)(a + b)$$

Step 3 (expand with the distributive law)

\begin{align} &= a^2 \cdot a + a^2 \cdot b + 2ab \cdot a + 2ab \cdot b + b^2 \cdot a + b^2 \cdot b \\ &= a^3 + a^2b + 2a^2b + 2ab^2 + ab^2 + b^3 \end{align}

Step 4 (collect like terms)

$$= a^3 + (1 + 2)a^2b + (2 + 1)ab^2 + b^3$$ $$= a^3 + 3a^2b + 3ab^2 + b^3$$

■

Theorem: cube of a difference

$$(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$$

Proof

Replace $b$ by $-b$ in the formula for $(a+b)^3$:

\begin{align} (a + (-b))^3 &= a^3 + 3a^2(-b) + 3a(-b)^2 + (-b)^3 \\ &= a^3 - 3a^2b + 3ab^2 - b^3 \end{align}

■

Example 8

Expand $(x + 2)^3$.

Solution:

\begin{align} (x + 2)^3 &= x^3 + 3 \cdot x^2 \cdot 2 + 3 \cdot x \cdot 2^2 + 2^3 \\ &= x^3 + 6x^2 + 12x + 8 \end{align}

Summary of the Formulas

The multiplication formulas

$(a + b)(a - b) = a^2 - b^2$
$(a + b)^2 = a^2 + 2ab + b^2$
$(a - b)^2 = a^2 - 2ab + b^2$
$(x + a)(x + b) = x^2 + (a + b)x + ab$
$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$
$(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$

Exercises

Exercise 1

Expand the following.

$(x + 5)(x - 5)$
$(3a + 2b)(3a - 2b)$

Exercise 2

Expand the following.

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

Exercise 3

Expand the following.

$(x + 2)(x + 8)$
$(x - 3)(x + 4)$
$(x - 5)(x - 7)$

Exercise 4

Expand $(x + 1)^3$.

Show solutions

Solution to Exercise 1

$(x + 5)(x - 5) = x^2 - 25$
$(3a + 2b)(3a - 2b) = (3a)^2 - (2b)^2 = 9a^2 - 4b^2$

Solution to Exercise 2

$(x + 6)^2 = x^2 + 12x + 36$
$(2x - 3)^2 = 4x^2 - 12x + 9$

Solution to Exercise 3

$(x + 2)(x + 8) = x^2 + 10x + 16$
$(x - 3)(x + 4) = x^2 + x - 12$
$(x - 5)(x - 7) = x^2 - 12x + 35$

Solution to Exercise 4

\begin{align} (x + 1)^3 &= x^3 + 3x^2 \cdot 1 + 3x \cdot 1^2 + 1^3 \\ &= x^3 + 3x^2 + 3x + 1 \end{align}

Further Topics

Where to go next

The binomial theorem: the expansion of $(a+b)^n$ has coefficients given by Pascal’s triangle, that is the binomial coefficients $\binom{n}{k}$. The cases $(a+b)^2$ and $(a+b)^3$ in this chapter are $n=2$ and $n=3$.
Factorisation: reading an expansion formula from right to left gives a factorisation formula. Continue with Chapter 3: Factorisation.
Quadratic equations: once expansion and factorisation are fluent, the solutions can be read straight off the form $(x-\alpha)(x-\beta)=0$. This is covered in Chapter 7: Quadratic Equations.

Frequently Asked Questions

Q1: What are the main expansion formulas?

A: The main multiplication formulas are $(a+b)^2 = a^2 + 2ab + b^2$, $(a-b)^2 = a^2 - 2ab + b^2$, $(a+b)(a-b) = a^2 - b^2$, and $(x+a)(x+b) = x^2 + (a+b)x + ab$. All of them follow from the distributive law, but memorising them makes calculation much faster.

Q2: What is the expansion of $(a+b)^3$?

A: $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$. The coefficients $1, 3, 3, 1$ are the third row of Pascal’s triangle. Replacing $b$ by $-b$ gives $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$.

Q3: How are expansion and factorisation related?

A: Expansion turns a product into a sum, and factorisation turns a sum back into a product, so they are inverse operations. Reading an expansion formula from right to left gives a factorisation formula.