🇨🇩 Algebra: expressions and equations

Algebraic expressions and factorising

⏱ 25 minOfficial programme: CD-E7-ALG-04

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🎯 In this lesson, you will learn to

Write, transform and evaluate expressions.
Factorise by common factor, grouping and special products.

The lesson

Introduction

Lusamba wants to fence a rectangular garden whose width is $x$ metres and whose length is $2x+3$ metres. Before knowing $x$ , he can already write the perimeter with letters: this is an algebraic expression. In this lesson you learn to write, transform and evaluate such expressions, and then to factorise them.

Explanation

Algebraic expressions

An algebraic expression contains letters that stand for numbers. Conventions: $3a$ means $3\times a$ ; $ab$ means $a\times b$ ; $a^2=a\times a$ .

Evaluating an expression means replacing each letter by its value, then computing in the right order of operations. Put brackets around a negative value: for $x=-2$ , $3x^2=3\times(-2)^2=12$ .
Simplifying means grouping like terms (same letter part): $7x+3-2x+5=5x+8$ and $2a+3b-a=a+3b$ .

Explanation

Expanding

Expanding means turning a product into a sum:
$k(a+b)=ka+kb\qquad (a+b)(c+d)=ac+ad+bc+bd$

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

Example: $(x+5)^2=x^2+10x+25$ .

Explanation

Factorising

Factorising means turning a sum into a product. Three methods:

Taking out a common factor: $6x+9=3\times2x+3\times3=3(2x+3)$ .
Grouping: group the terms two by two to bring out a common factor: $ax+ay+bx+by=a(x+y)+b(x+y)=(x+y)(a+b)$ .
Special products read from right to left: $x^2-9=(x-3)(x+3)$ and $x^2+6x+9=(x+3)^2$ .

Always check a factorisation by expanding it again.

Worked example

Problem: Compute the value of $E=3x^2-2x+5$ for $x=-2$ .

Step 1: Replace $x$ by $(-2)$ : $E=3\times(-2)^2-2\times(-2)+5$ .

Step 2: Power: $(-2)^2=4$ , so $3\times4=12$ .

Step 3: Product: $-2\times(-2)=4$ .

Step 4: $E=12+4+5=21$ .

Answer: for $x=-2$ , $E=21$ .

Worked example

Problem: Factorise $F=2ab+6a+b+3$ .

Step 1: Group: $F=(2ab+6a)+(b+3)$ .

Step 2: In the first group, take out $2a$ : $2ab+6a=2a(b+3)$ .

Step 3: $F=2a(b+3)+1\times(b+3)=(b+3)(2a+1)$ .

Step 4: Check: $(b+3)(2a+1)=2ab+b+6a+3$ .

Answer: $F=(b+3)(2a+1)$ .

Worked example

Problem: Factorise $G=4x^2-25$ and $H=9y^2-12y+4$ .

Step 1: $4x^2=(2x)^2$ and $25=5^2$ : $G$ has the form $a^2-b^2$ with $a=2x$ and $b=5$ .

Step 2: $G=(2x-5)(2x+5)$ .

Step 3: For $H$ : $9y^2=(3y)^2$ , $4=2^2$ and $2\times3y\times2=12y$ . This is the form $a^2-2ab+b^2$ .

Step 4: $H=(3y-2)^2$ .

Answer: $G=(2x-5)(2x+5)$ and $H=(3y-2)^2$ .

Watch out

Watch out: $(a+b)^2$ is not equal to $a^2+b^2$ : the double product $2ab$ is missing. Check with $a=1$ and $b=2$ : $(1+2)^2=9$ whereas $1^2+2^2=5$ . When taking out a common factor, do not forget the term $1$ : $6x+3=3(2x+1)$ , not $3\times2x$ .

Key points

Evaluate: replace each letter (in brackets if negative), then compute.
Simplify: group like terms.
$(a+b)^2=a^2+2ab+b^2$ ; $(a-b)^2=a^2-2ab+b^2$ ; $(a+b)(a-b)=a^2-b^2$ .
Factorise: common factor, grouping, special products.
Always check by expanding again.

✍️ Exercises

Answer one exercise at a time. Your answer is checked right away, with an explanation.

Exercise 1 of 8

★ Easy

Compute the value of $5a-3b$ for $a=4$ and $b=-2$ .

🎯 In this lesson, you will learn to

The lesson

👋 Introduction

📘 Explanation

📘 Explanation

📘 Explanation

✏️ Worked example

✏️ Worked example

✏️ Worked example

⚠️ Watch out

⭐ Key points

✍️ Exercises

Introduction

Explanation

Explanation

Explanation

Worked example

Worked example

Worked example

Watch out

Key points