3 Removing brackets from expressions of the form $\left(a+b\right)\left(c+d\right)$

Sometimes it is necessary to consider two bracketed terms multiplied together. In the expression $\left(a+b\right)\left(c+d\right)$ , by regarding the first bracket as a single term we can use the result in Key Point 14 to write it as $\left(a+b\right)c+\left(a+b\right)d$ . Removing the brackets from each of these terms produces $ac+bc+ad+bd$ . More concisely:

We see that each term in the first bracketed expression multiplies each term in the second bracketed expression.

Example 38

Remove the brackets from $\left(3+x\right)\left(2+y\right)$

Solution

$\text{We find}\left(3+x\right)\left(2+y\right)=\left(3+x\right)\left(2\right)+\left(3+x\right)y$

$=\left(3\right)\left(2\right)+\left(x\right)\left(2\right)+\left(3\right)\left(y\right)+\left(x\right)\left(y\right)=6+2x+3y+xy$

Example 39

Remove the brackets from $\left(3x+4\right)\left(x+2\right)$ and simplify your result.

Solution

$\left(3x+4\right)\left(x+2\right)=\left(3x+4\right)\left(x\right)+\left(3x+4\right)\left(2\right)$

$=3x^2+4x+6x+8=3x^2+10x+8$

Example 40

Remove the brackets from ${\left(a+b\right)}^2$ and simplify your result.

Solution

When a quantity is squared it is multiplied by itself. Thus

Task!

Remove the brackets from the following expressions and simplify the results.

1. $\left(x+7\right)\left(x+3\right)$ ,
2. $\left(x+3\right)\left(x-2\right)$ ,

Answer

Answer

Example 41

Explain the distinction between $\left(x+3\right)\left(x+2\right)$ and $x+3\left(x+2\right)$ .

Solution

In the first expression removing the brackets we find

In the second expression we have

$x+3\left(x+2\right)=x+3x+6=4x+6$

Note that in the second expression the term $\left(x+2\right)$ is only multiplied by 3 and not by $x$ .

Example 42

Remove the brackets from $\left(s^2+2s+4\right)\left(s+3\right)$ .

Solution

Each term in the first bracket must multiply each term in the second. Working through all combinations systematically we have