How to expand brackets 1
What are brackets?
A bracket is a symbol which helps us maintain the difference between two terms. Whenever we have to differ one mathematical term from another mathematical term, we use a bracket. For easy understanding, consider a bracket as a wall, a wall parting two rooms or houses. Brackets work in the same way.
There are three commonly used brackets:
$$\left ( \right )$$
$$\left \{ \right \}$$
$$\left [ \right ]$$
E2.2: Expand products of algebraic expressions
Here, we learn about how to expand the bracket and what are the rules for it. An expression can contain several brackets. If $$+$$ or $$-$$ signs, variables and constants are written between the bracket, it means $$+1$$ or $$-1$$ is multiplied by all the members in the bracket and then simplified. If variables and constants are only written in the bracket, it means that the variable or constant is multiplied by all the bracket members.
For example, while simplifying $$2x\left ( 9x-2y \right )$$, there is no $$+$$ or $$-$$ signs between the bracket and $$2x$$. It means that $$2x$$ should be multiplied with $$9x$$ and $$-2y$$ separately. So, the expanded form of the expression is $$18 x^{2}- 4y$$. Take one more example, simplify $$2x - \left ( 9x-2y \right )$$. Here, $$-$$ sign is written between bracket and $$2x$$. It means that the members of bracket are multiplied by $$-1$$. So, the value of $$2x - \left ( 9x-2y \right )$$ becomes $$2x- 9x+ 2y$$. Simplify it to get the final answer, as $$-7x+2y$$.
Worked examples of questions involving brackets
Example 1: Expand $$5x\left ( 2x-3y \right )$$.
Step 1: Multiply $$5x$$ by $$2x$$ and $$-3y$$ separately.
$$5x\left ( 2x-3y \right )=10 x^{2} - 15xy$$
Step 2: Write the final answer.
The expanded form of $$5x\left ( 2x-3y \right )$$ is $$10 x^{2} - 15xy$$.
Example 2: Expand $$m^{3}\left ( 2m- 3m+ 5mn \right )$$.
Step 1: Multiply $$m^3$$ by $$2m$$, $$-3m$$ and $$5mn$$ separately.
$$m^{3} (2m- 3n+ 5mn)= 2m^{4} - 3m^{3}n + 5m^{4}n$$
Step 2: Write the final answer.
The expanded form of $$m^{3} \left ( 2m- 3n+ 5mn \right )$$ is $$2m^{4} - 3m^{3}n + 5m^{4}n $$.
Example 3: Expand and simplify $$7x\left ( x^{2}+4 \right )+ 3x^{3}+ 5x+ 4$$.
Step 1: Expand the bracket, multiplying $$7x$$ by $$x^{2}$$ and $$4$$ separately.
$$7x\left (x^{2}+4 \right )= 7x^{3}+ 28x$$
Step 2: Substitute the value of $$7x\left ( x^{2}+4 \right )$$ in the equation.
$$7x\left ( x^{2}+4 \right )+ 3x^{3}+ 5x+ 4=7x^{3}+ 28x+ 3x^{3}+ 5x+ 4$$
Step 3: Add like terms.
$$7x^{3}+ 28x+ 3 x^{3}+ 5x+ 4=10 x^{3}+ 33x+ 4$$.
Step 4: Write the final answer.
$$7x\left ( x^{2}+4 \right )+ 3x^{3}+ 5x+ 4=10 x^{3}+ 33x+ 4$$.
Example 4: Expand and simplify $$\frac{2}{3}\left ( 9x- 6y \right )+\frac{1}{4}\left ( 8x+ 16y \right )$$.
Step 1: Expand the first bracket, multiplying $$\frac{2}{3}$$ by $$9x$$ and $$-6y$$ separately.
$$\frac{2}{3}\left ( 9x- 6y \right )=\frac{2}{3}\times 9x -\frac{2}{3}\times 6y$$
$$\frac{2}{3}\left ( 9x- 6y \right )=6x- 4y$$
Step 2: Expand the second bracket, multiplying $$\frac{1}{4}$$ by $$8x$$ and $$16y$$ separately.
$$\frac{1}{4}\left ( 8x+ 16y \right )=\frac{1}{4}\times 8x+ \frac{1}{4}\times 16y$$
$$\frac{1}{4}\left ( 8x+ 16y \right )=2x+ 4y$$
Step 3: Add or subtract like terms.
$$\frac{2}{3}\left ( 9x- 6y \right )+\frac{1}{4}\left ( 8x+ 16y \right )=6x- 4y+2x +4y$$
$$\frac{2}{3}\left ( 9x- 6y \right )+\frac{1}{4}\left ( 8x+ 16y \right )=8x$$
Step 4: Write the final answer.
$$\frac{2}{3}\left ( 9x- 6y \right )+\frac{1}{4}\left ( 8x+ 16y \right )=8x$$