Removing brackets from expressions a(b + c) and a(b − c)

2 Removing brackets from expressions $a (b + c)$ and $a (b - c)$

Removing brackets means multiplying out . For example $5 (2 + 4) = 5 \times 2 + 5 \times 4 = 10 + 20 = 30$ . In this simple example we could alternatively get the same result as follows: $5 (2 + 4) = 5 \times 6 = 30$ . That is:

$5 (2 + 4) = 5 \times 2 + 5 \times 4$

In an expression such as $5 (x + y)$ it is intended that the 5 multiplies both $x$ and $y$ to produce $5 x + 5 y$ . Thus the expressions $5 (x + y)$ and $5 x + 5 y$ are equivalent. In general we have the following rules known as distributive laws :

Key Point 14

a (b + c) = a b + a c

a (b - c) = a b - a c

Note that when the brackets are removed both terms in the brackets are multiplied by $a$ .

As we have noted above, if you insert numbers instead of letters into these expressions you will see that both left and right hand sides are equivalent. For example

$4 (3 + 5) has the same value as 4 (3) + 4 (5), that is 32$

and

$7 (8 - 3) has the same value as 7 (8) - 7 (3), that is 35$

Example 31

Remove the brackets from

$9 (2 + y)$ ,
$9 (2 y)$ .

Solution

In the expression $9 (2 + y)$ the 9 must multiply both terms in the brackets: $\begin{array}{rcl} 9 (2 + y) & = & 9 (2) + 9 (y) \\ = & 18 + 9 y \end{array}$
Recall that $9 (2 y)$ means $9 \times (2 \times y)$ and that when multiplying numbers together the presence
of brackets is irrelevant. Thus $9 (2 y) = 9 \times 2 \times y = 18 y$

The crucial distinction between the role of the factor 9 in the two expressions $9 (2 + y)$ and $9 (2 y)$ in Example 31 should be noted.

Example 32

Remove the brackets from $9 (x + 2 y)$ .

Solution

In the expression $9 (x + 2 y)$ the 9 must multiply both the $x$ and the $2 y$ in the brackets. Thus

\begin{array}{rcl} 9 (x + 2 y) & = & 9 x + 9 (2 y) \\ = & 9 x + 18 y \end{array}

Task!

Remove the brackets from $9 (2 x + 3 y)$ .

Remember that the 9 must multiply both the term $2 x$ and the term $3 y$ :

$18 x + 27 y$

Example 33

Remove the brackets from $- 3 (5 x - z)$ .

Solution

The number $- 3$ must multiply both the $5 x$ and the $z$ .

\begin{array}{rcl} - 3 (5 x - z) & = & (- 3) (5 x) - (- 3) (z) \\ = & - 15 x + 3 z \end{array}

Task!

Remove the brackets from $6 x (3 x - 2 y)$ .

$6 x (3 x - 2 y) = 6 x (3 x) - 6 x (2 y) = 18 x^{2} - 12 x y$

Example 34

Remove the brackets from $- (3 x + 1)$ .

Solution

Although the $1$ is unwritten, the minus sign outside the brackets stands for $- 1$ . We must therefore consider the expression $- 1 (3 x + 1)$ .

\begin{array}{rcl} - 1 (3 x + 1) & = & (- 1) (3 x) + (- 1) (1) \\ = & - 3 x + (- 1) \\ = & - 3 x - 1 \end{array}

Task!

Remove the brackets from $- (5 x - 3 y)$ .

$- (5 x - 3 y)$ means $- 1 (5 x - 3 y)$ .

$- 1 (5 x - 3 y) = (- 1) (5 x) - (- 1) (3 y) = - 5 x + 3 y$

Task!

Remove the brackets from $m (m - n)$ .

In the expression $m (m - n)$ the first $m$ must multiply both terms in the brackets:

$m^{2} - m n$

Example 35

Remove the brackets from the expression $5 x - (3 x + 1)$ and simplify the result by collecting like terms.

Solution

The brackets in $- (3 x + 1)$ were removed in Example 34 on page 46.

\begin{array}{rcl} 5 x - (3 x + 1) & = & 5 x - 1 (3 x + 1) \\ = & 5 x - 3 x - 1 \\ = & 2 x - 1 \end{array}

Example 36

Show that $\frac{- x - 1}{4}$ , $\frac{- (x + 1)}{4}$ and $- \frac{x + 1}{4}$ are all equivalent expressions.

Solution

Consider $- (x + 1)$ . Removing the brackets we obtain $- x - 1$ and so

$\frac{- x - 1}{4} is equivalent to \frac{- (x + 1)}{4}$

A negative quantity divided by a positive quantity will be negative. Hence

$\frac{- (x + 1)}{4} is equivalent to - \frac{x + 1}{4}$

You should study all three expressions carefully to recognise the variety of equivalent ways in which we can write an algebraic expression.

Sometimes the bracketed expression can appear on the left, as in $(a + b) c$ . To remove the brackets here we use the following rules:

Key Point 15

(a + b) c = a c + b c

(a - b) c = a c - b c

Note that when the brackets are removed both the terms in the brackets multiply $c$ .

Example 37

Remove the brackets from $(2 x + 3 y) x$ .

Solution

Both terms in the brackets multiply the $x$ outside. Thus

\begin{array}{rcl} (2 x + 3 y) x & = & 2 x (x) + 3 y (x) \\ = & 2 x^{2} + 3 y x \end{array}

Task!

Remove the brackets from

$(x + 3) (- 2)$ ,
$(x - 3) (- 2)$ .

Both terms in the bracket must multiply the $- 2$ , giving $- 2 x - 6$

$- 2 x + 6$

2 Removing brackets from expressions a ( b + c ) and a ( b − c )

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2 Removing brackets from expressions $a (b + c)$ and $a (b - c)$