Distributive law I

When we have two numbers inside a set of brackets, we can evaluate them before we work out the rest of the equation. However, when we have algebraic terms and can't simplify the terms any further, we can still expand expressions so that they are written without brackets. We just use a slightly different process and multiply each term inside the brackets (both numbers and algebraic terms) by the term outside the brackets.

If we were to think about ,

using the idea that multiplication relates to groups, we can say that $3\left(m+2\right)$3(m+2) is $3$3 groups of $m+2$m+2 -> like this.                         

Now regrouping all the $m$m's together and all the $2$2's together we end up with -> 

Of course this could be a very complicated method if we had to do $11\left(p+4\right)$11(p+4), writing out $11$11groups could be very time consuming.  Fortunately we can take a shortcut once we understand how it relates to groups.  

So $3\left(m+2\right)$3(m+2) is $3$3 groups of the $m$m and $3$3 groups of the $2$2.  

Let's look at how this process works with some examples.

 

Examples

Question 1

Expand and simplify $11\left(p+4\right)$11(p+4).

Think: We need to work out $11\times p$11×p and $11\times4$11×4. 

Do: $11p+44$11p+44

 

Remember your rules for multiplying negative numbers as they will affect your answers.

 

Question 2

Expand and simplify $-9\left(n-2\right)$−9(n−2) .

Think: We need to solve $-9\times n$−9×n and $-9\times\left(-2\right)$−9×(−2).

Do: $-9n+18$−9n+18

 

If there is only a negative sign written outside the brackets, this means you need to multiply everything inside by $-1$−1.

 

Question 3

Evaluate  $-\left(15-y\right)$−(15−y)

Think: We need to solve$-1\times15$−1×15 and  $-1\times\left(-y\right)$−1×(−y)

Do:  $-15+y$−15+y

The same process applies no matter how many terms are inside the brackets. Look for like terms that can be simplified.

 

Question 4

Question 5