# How to expand brackets 2

Last updated: September 28, 2021

## What are brackets and how should they be used?

Brackets are the most commonly used symbols, such as the parentheses in an algebraic expression, to establish groups or explain the order of the operations to be performed.

A factor of a number is an accurate divisor of the given number. Every factor of a number is smaller than or equal to the given number, i.e., it cannot be greater than the given number.

## E2.2: Use brackets and extract the common factors

Expanding brackets means that each item in the brackets is multiplied by the expression outside the brackets.

For example, in the expression $$3(m+7)$$, multiply both $$m$$ and $$7$$ by $$3$$. So: $$3(m+7)=3\times m+3\times 7=3m+21$$

A common factor is a number that can be divisible by two different numbers without a remainder. Numbers can have multiple common factors. You can find the common factors of more than two numbers.

Factorising is also known as the reverse process of expanding brackets. Factoring an algebraic expression means putting it in parentheses and taking the common factors.

The way of factorising is:

• Find the highest common factor of each of the terms in the expression.

• Write the highest common factor in front of the brackets.

• Fill in each term in the brackets by multiplying out.

### Worked examples of using brackets

Example 1: Expand $$(3x+2)(2x+3)$$.

Step 1: Write the given information.

The given expression is $$(3x+2)(2x+3)$$.

Step 2: Multiply both $$2x$$ and $$3$$ by $$(3x+2)$$.

$$(3x+2)(2x+3)=(3x+2)(2x)+(3x+2)(3)$$

Step 3: Expand the equation.

$$(3x+2)(2x+3)=(3x+2)(2x)+(3x+2)(3)$$

It can be written as

$$(3x+2)(2x+3)=3x(2x)+2(2x)+(3x)(3)+2(3)$$

$$6x^{2}+4x+9x+6$$

$$6x^{2}+13x+6$$

Therefore, $$(3x+2)(2x+3)=6x^{2}+13x+6$$.

Example 2: Factorise $$15a^{3}-30a^{2}-360a$$.

Step 1: Write the given information.

The given expression is $$15a^{3}-30a^{2}-360a$$.

Step 2:  Write the highest common factor.

The highest common factor is $$15a$$.

Step 3: Write the greatest common factor.

$$15a^{3}-30a^{2}-360a=15a(a^{2}-2a-24)$$

Step 4: Substitute $$-2a=4a-6a$$ in the above equation.

$$15a^{3}-30a^{2}-360a=15a(a^{2}+4a-6a-24)$$

$$15a(a(a+4)-6(a+4))$$

Step 5: Write the common factor.

$$15a^{3}-30a^{2}-360a=15a(a+4)(a-6)$$.

Therefore, the factorisation of $$15a^{3}-30a^{2}-360a$$ is $$15a(a+4)(a-6)$$.

Example 3: A rectangular building is located on a plot that measures $$30\text{m}$$ by $$40\text{m}$$. The building must be placed in such a way that the width of the lawn on all the four sides of the building is the same. Local restrictions state that the building cannot occupy more than $$50%$$. What are the dimensions of the largest building that can be built on the site?

Step 1: Write the given values.

Let $$x$$ represent the width of the lawn.

Let $$40-2x$$ represent the length of the building.

Let $$30-2x$$ represent the width of the building.

The area of the lot is $$1200\text{metre square}$$.

Step 2: Write the maximum area of the building.

The maximum area of the building is $$600\text{metre square}$$.

Step 3: Recall the formula for the area of the rectangle.

$$\text{the area of the rectangle}=\text{length}\times \text{width}$$.

Step 4: Substitute the values in the formula.

$$(40-2x)(30-2x)=600$$.

Step 5: Multiply both $$30$$ and $$-2x$$ by $$(40-2x)$$.

$$(40-2x)(30)+(40-2x)(-2x)=600$$.

Step 6: Expand the brackets.

The equation can be written as follows:

$$40(30)-2x(30)+40(-2x)-2x(-2x)=600$$

$$1200-60x-80x+4x^{2}-600=0$$

$$4x^{2}-140x+600=0$$

Step 7: Factorise the equation $$4x^{2}-140x+600=0$$.

The common factor is $$4$$.

Step 8: Write the common factor in front of the brackets.

$$4(x^{2}-35x+150)=0$$

$$4(x—5)(x-30)=0$$

Now, for $$x-5=0$$:

$$x-5=0$$

$$\therefore x=5$$

Now, for $$x-30=0$$:

$$x-30=0$$

$$\therefore x=30$$

It is not possible because $$x=30$$ and the value of the width is $$30-2(30)=-30$$, i.e., negative.

So, the value of $$x$$ is $$5$$.

Step 9: Write the value of the length.

Substitute $$x=5$$ in $$40-2x$$.

$$\text{Length}=40-2(5)=30\text{m}$$

Step 10: Write the value of the width.

Substitute $$x=5$$ in $$30-2x$$.

$$\text{Width}=30-2(5)=20\text{m}$$

Therefore, the length of the building is $$30\text{m}$$ and the width is $$20\text{m}$$.