Functions and Algebra: Manipulate and simplify algebraic expressions

# Unit 1: Simplifying algebraic expressions

Natashia Bearam-Edmunds

### Unit outcomes

By the end of this unit you will be able to:

• Find products of two binomials.
• Find products of binomials with trinomials.

## What you should know

Before you start this unit, make sure you can:

Work with real numbers and understand the basic exponent rules. Revise exponents in Subject outcome 1.2, Unit 2: Introduction to exponents.

## Introduction

An algebraic expression is made up of constants and variables joined together by addition, subtraction, multiplication and division.

In the expression $\scriptsize 3x+2$, $\scriptsize 2$ is a constant because it does not vary and $\scriptsize x$ is called a variable because its value can change. The $\scriptsize 3$ in front of the variable $\scriptsize x$ is called the coefficient of $\scriptsize x$. The $\scriptsize 3x$ and $\scriptsize +2$ are called the terms of the algebraic expression. Terms are separated by addition or subtraction. In this algebraic expression we have two terms.

We often simplify an algebraic expression to make it easier to work with or calculate, or to use it in some other way. To do so, we use the properties of real numbers.

### Note

The following properties are true for real numbers $\scriptsize a,b$ and $\scriptsize c$.

 Addition Multiplication Commutative property $\scriptsize a+b=b+a$ $\scriptsize a\cdot b=b\cdot a$ Associative property $\scriptsize a+\left( {b+c} \right)=\left( {a+b} \right)+c$ $\scriptsize a\left( {bc} \right)=\left( {ab} \right)c$ Distributive property $\scriptsize \begin{array}{*{20}{l}} {a\cdot \left( {b+c} \right)=a\cdot b+a\cdot c} \end{array}$ Identity property There is a unique real number called the additive identity, $\scriptsize 0$ such that, for any real number$\scriptsize a$ $\scriptsize a+0=a$. There is a unique real number called the multiplicative identity, $\scriptsize 1$, such that, for any real number$\scriptsize a$ $\scriptsize a\cdot 1=a$. Inverse property Every real number $\scriptsize a$ has an additive inverse, or opposite, written as $\scriptsize -a$, such that$\scriptsize a+(-a)=0$. Every non-zero real number $\scriptsize a$has a multiplicative inverse, or reciprocal, written as $\scriptsize \displaystyle \frac{1}{a}$, such that $\scriptsize a\cdot \displaystyle \frac{1}{a}=1$.

## Products

Remember that a product is the result of multiplication. When you multiply brackets that contain terms you end up with algebraic expressions. Mathematical expressions are just like sentences and each part has a special name.

$\scriptsize 2{{x}^{2}}-3x+4$ is called an expression and is made up of the following parts.
Terms: $\scriptsize 2{{x}^{2}},\text{ }-3x,\text{ }+4$
Variable: $\scriptsize x$
Constant: $\scriptsize +4$
Coefficients (numbers in front) of the variable: $\scriptsize 2$ and $\scriptsize -3$
Exponents: $\scriptsize 2$ and $\scriptsize 1$

The following are words used to describe specific expressions that you will come across often. You need to learn the definitions of these terms.

A monomial is an expression with one term, for example $\scriptsize 2x$ or $\scriptsize 3(xy)$. Remember that brackets do not separate terms.
A binomial is an expression with two terms, for example $\scriptsize x+y$ or $\scriptsize 2a-b$.
A trinomial is an expression with three terms, for example $\scriptsize x+y+z$ or $\scriptsize 2{{x}^{2}}-3x+4$.
You can remember it as ‘mono’ means one, ‘bi’ means two and ‘tri’ means three.

### Multiplying a monomial and a binomial

In the next example we learn how to multiply a monomial by a binomial.

### Example 1.1

Simplify:

1. $\scriptsize 3x(x-10)$
2. $\scriptsize 2a(a-2)-3({{a}^{2}}-a)$

Solutions

1. Multiply the $\scriptsize 3x$ by each term inside the bracket.
$\scriptsize 3x(x-10)=3x\cdot x-3x\cdot 10$
Be careful with the signs. First multiply the signs, positive by positive and positive by negative, before multiplying the numbers.
$\scriptsize \begin{array}{l}3x(x-10)&=3x\cdot x-3x\cdot 10\\&=3{{x}^{2}}-30x\end{array}$
2. In this example we multiply two separate monomials by two binomials. After getting rid of the brackets, we need to add like terms together. Notice that a negative multiplied by a negative gives a positive answer.
$\scriptsize \begin{array}{l}2a(a-2)-3({{a}^{2}}-a)&=2a\cdot a-2a\cdot 2-3\cdot {{a}^{2}}-3(-a)\\&=2{{a}^{2}}-4a-3{{a}^{2}}+3a\\&=-{{a}^{2}}-a\end{array}$

### Exercise 1.1

Simplify:

1. $\scriptsize 2a(a\text{ }+\text{ }3)$
2. $\scriptsize -6(t-1)+2(t+3)$
3. $\scriptsize (a-b)2$

The full solutions are at the end of the unit.

### Multiply a binomial by a binomial

Here we multiply (or expand) two linear (highest power of the variable is one) binomials.
We can use the word FOIL (First Outers Inners Last) to help us remember that each term in the first bracket must be multiplied by each term in the second bracket.

In general:

$\scriptsize \begin{array}{l}(ax+b)(cx+d)&=ax\cdot cx+ax\cdot d+b\cdot cx+b\cdot d\\&=ac{{x}^{2}}+axd+bcx+bd\\&=ac{{x}^{2}}+x(ad+bc)+bd\end{array}$

When we multiply brackets we generally write the products with variables in alphabetical order, for example $\scriptsize acd$ not $\scriptsize dca$. And we write terms in descending order of the first variable, for example $\scriptsize {{x}^{2}}+2x+1$ is written in descending powers of $\scriptsize x$.

### Note

When you have an internet connection, watch the video called “Multiplying binomials” to learn more about FOIL.

Multiplying binomials (Duration: 05.47)

### Activity 1.1: Practise using FOIL

Time required: 15 minutes

What you need:

• pen and paper

What to do:

1. Complete the table by finding the following products:
 Multiply brackets Add like terms $\scriptsize (ax+b)(ax+b)$ $\scriptsize (a-b)(a-b)$ $\scriptsize (x+2)(x+2)$ $\scriptsize (ax-b)(ax+b)$ $\scriptsize (a-b)(a+b)$ $\scriptsize (x+2)(x-2)$
2. In parts a) to c) what do you notice about the brackets before you multiply?
3. In parts a) to c) what happened to the first terms and last terms in the simplified expressions?
4. In parts a) to c) what do you notice about the middle term in the simplified expressions?
5. In parts d) to f) what do you notice about the brackets before you multiply?
6. In parts d) to f) what happened to the first terms and last terms in the simplified expressions?
7. In parts d) to f) what happened to the middle in the simplified expressions?

What did you find?

1. .
 Multiply brackets Add like terms $\scriptsize (ax+b)(ax+b)$ $\scriptsize {{a}^{2}}{{x}^{2}}+axb+axb+{{b}^{2}}$ $\scriptsize {{a}^{2}}{{x}^{2}}+2axb+{{b}^{2}}$ $\scriptsize (a-b)(a-b)$ $\scriptsize {{a}^{2}}-ab-ab+{{b}^{2}}$ $\scriptsize {{a}^{2}}-2ab+{{b}^{2}}$ $\scriptsize (x+2)(x+2)$ $\scriptsize {{x}^{2}}+2x+2x+4$ $\scriptsize {{x}^{2}}+4x+4$ $\scriptsize (ax-b)(ax+b)$ $\scriptsize {{a}^{2}}{{x}^{2}}-abx+abx-{{b}^{2}}$ $\scriptsize {{a}^{2}}{{x}^{2}}-{{b}^{2}}$ $\scriptsize (a-b)(a+b)$ $\scriptsize {{a}^{2}}-ab+ab-{{b}^{2}}$ $\scriptsize {{a}^{2}}-{{b}^{2}}$ $\scriptsize (x+2)(x-2)$ $\scriptsize {{x}^{2}}-2x+2x-4$ $\scriptsize {{x}^{2}}-4$
2. The brackets are identical in a) to c).
3. In parts a) to c) the first terms have been squared and the last terms have been squared once the expression has been simplified.
4. In parts a) to c) the middle terms are repeated when you multiply the brackets out. So the final answer to the middle term is twice the product of the first and second terms.
5. The brackets contain the same terms but with opposite signs.
6. In parts d) to f) the first terms and last terms of the brackets have been squared and are separated by a minus sign.
7. In parts d) to f) because the middle terms, when the brackets are multiplied out, are the same but with opposite signs they add up to zero when you simplify.

The product of two identical binomials is known as the square of the binomial and is written as:

$\scriptsize \begin{array}{l}{{(a+b)}^{2}}&=(a+b)(a+b)\\&={{a}^{2}}+2ab+{{b}^{2}}\end{array}$

after simplifying and adding the two middle terms.

If the two terms are of the form $\scriptsize a+b$ and $\scriptsize a-b$, then their product is $\scriptsize {{a}^{2}}-{{b}^{2}}$. We simply square the first term, square the second term and separate with a minus sign. The middle term will always cancel out as you saw in Activity 1.1. This product gives us a difference of two squares.

### Exercise 1.2

Simplify:

1. $\scriptsize (a-4)(a-4)$
2. $\scriptsize (x-y)(x+2y)$
3. $\scriptsize (c-2d)(c+2d)$
4. $\scriptsize 2(a+2b)(a+2b)$
5. $\scriptsize 3{{(x-3)}^{2}}+2(x+3)(x-3)$

The full solutions are at the end of the unit.

### Multiply a binomial and a trinomial

You have already seen that a trinomial is an expression with three terms, for example $\scriptsize a{{x}^{2}}+bx+c$. Now we will learn how to find the product of a binomial and a trinomial.

To find the product of a binomial and a trinomial, multiply each term of the binomial by each term of the trinomial.

$\scriptsize \begin{array}{l}(a+b)(x+y+z) & =a(x+y+z)+b(x+y+z)\\ & =ax+ay+az+bx+by+bz\end{array}$

Let’s see how this works in an example.

### Example 1.2

Find the product:

1. $\scriptsize (x-1)({{x}^{2}}-2x+1)$
2. $\scriptsize (a-2b)(2a+b-3)$
3. $\scriptsize (-3-x)(2{{x}^{2}}+x-3)$
4. $\scriptsize (2{{y}^{2}}-y+3)(y-1)$

Solutions

1. Multiply each term in the first bracket by each term in the second bracket. Remember to multiply the signs carefully. Then collect all like terms.
.
$\scriptsize \begin{array}{l}(x-1)({{x}^{2}}-2x+1)&=x({{x}^{2}}-2x+1)-1({{x}^{2}}-2x+1)\\&=x\cdot {{x}^{2}}-x\cdot 2x+x\cdot 1-1\cdot {{x}^{2}}-1(-2x)-1\cdot 1\\&={{x}^{3}}-2{{x}^{2}}+x-{{x}^{2}}+2x-1\\&={{x}^{3}}-3{{x}^{2}}+3x-1\end{array}$
2. .
$\scriptsize \begin{array}{l}(a-2b)(2a+b-3)&=a(2a+b-3)-2b(2a+b-3)\\&=2{{a}^{2}}+ab-3a-4ab-2{{b}^{2}}+6b\\&=2{{a}^{2}}-3ab-3a-2{{b}^{2}}+6b\end{array}$
3. Be careful with the negative signs in this example.
$\scriptsize \begin{array}{l}(-3-x)(2{{x}^{2}}+x-3)&=-3(2{{x}^{2}}+x-3)-x(2{{x}^{2}}+x-3)\\&=-6{{x}^{2}}-3x+9-2{{x}^{3}}-{{x}^{2}}+3x\\&=-2{{x}^{3}}-7{{x}^{2}}+9\end{array}$
4. Even though the trinomial is in the first bracket and the binomial in the second, the rules to multiply out the brackets stay the same. If you prefer you can use the commutative property of real numbers $\scriptsize a\cdot b=b\cdot a$ to rearrange the brackets before multiplying.
$\scriptsize \begin{array}{l}(2{{y}^{2}}-y+3)(y-1)&=(y-1)(2{{y}^{2}}-y+3)\\&=y(2{{y}^{2}}-y+3)-1(2{{y}^{2}}-y+3)\\&=2{{y}^{3}}-{{y}^{2}}+3y-2{{y}^{2}}+y-3\\&=2{{y}^{3}}-3{{y}^{2}}+4y-3\end{array}$

## Summary

In this unit you have learnt the following:

• How to use the properties of real numbers to work with algebraic expressions.
• How to identify the parts that make up an algebraic expression.
• How to multiply a monomial and a binomial.
• How to multiply a binomial and a binomial.
• How to multiply a binomial and a trinomial.

# Assessment

#### Suggested time to complete: 35 minutes

1. Expand the following products:
1. $\scriptsize 2x(x-3)$
2. $\scriptsize (a-2b)2b$
3. $\scriptsize (-7+x)(7+x)$
4. $\scriptsize -(2xy-2)(2xy-2)$
5. $\scriptsize {{(2t-3)}^{2}}$
6. $\scriptsize (-2{{y}^{2}}-4y+11)(3-y)$
2. Simplify:
1. $\scriptsize {{(2x-3)}^{2}}-{{(x-2)}^{2}}$
2. $\scriptsize 2(3a+b)(3a-b)-{{(3a-b)}^{2}}$
3. $\scriptsize (2{{a}^{2}}-a-1)({{a}^{2}}+3a-2)$
4. $\scriptsize (\displaystyle \frac{x}{3}-\displaystyle \frac{3}{x})(\displaystyle \frac{x}{4}+\displaystyle \frac{4}{x})$
5. $\scriptsize \displaystyle \frac{1}{2}\left( {10x-12y} \right)+\displaystyle \frac{1}{3}\left( {15x-18y} \right)$

The full solutions are at the end of the unit.

# Unit 1: Solutions

### Exercise 1.1

1. $\scriptsize 2a(a\text{ }+\text{ }3)=2{{a}^{2}}+6a$
2. After multiplying brackets, simplify by adding like terms.
$\scriptsize \begin{array}{l}-6(t-1)+2(t+3)&=-6\cdot t-6(-1)+2\cdot t+2(3)\\&=-6t+6+2t+6\\&=-4t+12\end{array}$
3. Even though the $\scriptsize 2$comes after the bracket, remember that multiplication is commutative.
$\scriptsize (a-b)2=2a-2b$

Back to Exercise 1.1

### Exercise 1.2

1. .
$\scriptsize \begin{array}{l}(a-4)(a-4)&={{a}^{2}}-2(4a)+16\\&={{a}^{2}}-8a+16\end{array}$
2. .
$\scriptsize \begin{array}{l}(x-y)(x+2y)&={{x}^{2}}+2xy-xy-2{{y}^{2}}\\&={{x}^{2}}+xy-2{{y}^{2}}\end{array}$
3. .
$\scriptsize \begin{array}{l}(c-2d)(c+2d)&={{(c)}^{2}}-{{(2d)}^{2}}\\&={{c}^{2}}-4{{d}^{2}}\end{array}$
4. .
$\scriptsize \begin{array}{l}2(a+2b)(a+2b)&=2({{a}^{2}}+4ab+4{{b}^{2}})\\&=2{{a}^{2}}+8ab+8{{b}^{2}}\end{array}$
5. .
$\scriptsize \begin{array}{l}3{{(x-3)}^{2}}+2(x+3)(x-3)&=3({{x}^{2}}-6x+9)+2({{x}^{2}}-9)\\&=3{{x}^{2}}-18x+27+2{{x}^{2}}-18\\&=5{{x}^{2}}-18x+9\end{array}$

Back to Exercise 1.2

### Unit 1: Assessment

1. .
1. $\scriptsize 2x(x-3)=2{{x}^{2}}-6x$
2. $\scriptsize (a-2b)2b=2ab-4{{b}^{2}}$
3. $\scriptsize (-7+x)(7+x)={{x}^{2}}-49$
4. .
$\scriptsize \begin{array}{l}-(2xy-2)(2xy-2)&=-\left( {4{{x}^{2}}{{y}^{2}}-8xy+4} \right)\\&=-4{{x}^{2}}{{y}^{2}}+8xy-4\end{array}$
5. $\scriptsize {{(2t-3)}^{2}}=4{{t}^{2}}-12t+9$
6. .
$\scriptsize \begin{array}{l}(-2{{y}^{2}}-4y+11)(3-y)&=-6{{y}^{2}}-12y+33+2{{y}^{3}}+4{{y}^{2}}-11y\\&=2{{y}^{3}}-2{{y}^{2}}-23y+33\end{array}$
2. .
1. .
$\scriptsize \begin{array}{l}{{(2x-3)}^{2}}-{{(x-2)}^{2}}&=4{{x}^{2}}-12x+9-({{x}^{2}}-4x+4)\\&=4{{x}^{2}}-12x+9-{{x}^{2}}+4x-4\\&=3{{x}^{2}}-8x+5\end{array}$
2. .
$\scriptsize \begin{array}{l}2(3a+b)(3a-b)-{{(3a-b)}^{2}}&=2(9{{a}^{2}}-{{b}^{2}})-(9{{a}^{2}}-6ab+{{b}^{2}})\\&=18{{a}^{2}}-2{{b}^{2}}-9{{a}^{2}}+6ab-{{b}^{2}}\\&=9{{a}^{2}}+6ab-3{{b}^{2}}\end{array}$
3. .
$\scriptsize \begin{array}{l}(2{{a}^{2}}-a-1)({{a}^{2}}+3a-2)&=2{{a}^{2}}({{a}^{2}}+3a-2)-a({{a}^{2}}+3a-2)-1({{a}^{2}}+3a-2)\\&=2{{a}^{4}}+6{{a}^{3}}-4{{a}^{2}}-{{a}^{3}}-3{{a}^{2}}+2a-{{a}^{2}}-3a+2\\&=2{{a}^{4}}+5{{a}^{3}}-8{{a}^{2}}-a+2\end{array}$
4. .
$\scriptsize \begin{array}{l}\left( {\displaystyle \frac{x}{3}-\displaystyle \frac{3}{x}} \right)\left( {\displaystyle \frac{x}{4}+\displaystyle \frac{4}{x}} \right)&=\displaystyle \frac{x}{3}\cdot \displaystyle \frac{x}{4}+\displaystyle \frac{x}{3}\cdot \displaystyle \frac{4}{x}-\displaystyle \frac{3}{x}\cdot \displaystyle \frac{x}{4}-\displaystyle \frac{3}{x}\cdot \displaystyle \frac{4}{x}\\&=\displaystyle \frac{{{{x}^{2}}}}{{12}}+\displaystyle \frac{{4x}}{{3x}}-\displaystyle \frac{{3x}}{{4x}}-\displaystyle \frac{{12}}{x}\\&=\displaystyle \frac{{{{x}^{2}}}}{{12}}+\displaystyle \frac{4}{3}-\displaystyle \frac{3}{4}-\displaystyle \frac{{12}}{x}\\&=\displaystyle \frac{{{{x}^{2}}}}{{12}}+\displaystyle \frac{7}{{12}}-\displaystyle \frac{{12}}{x}\end{array}$
5. .
$\scriptsize \begin{array}{l}\displaystyle \frac{1}{2}\left( {10x-12y} \right)+\displaystyle \frac{1}{3}\left( {15x-18y} \right)&=5x-6y+5x-6y\\&=10x-12y\end{array}$

Back to Unit 1: Assessment