# 2. Multiplication of Algebraic Expressions

Einstein's famous equation involves multiplying algebraic terms

When we multiply algebraic expressions, we need to remember the Index Laws from the Numbers chapter.

Let's see how algebra multiplication works with a series of examples.

## Example 1

Multiply *x*^{3}(*x*^{4} + 5*a*)

Answer

We expand this out using the distributive law:

x^{3}(x^{4}+ 5a)=

x^{7}+ 5ax^{3}

We cannot simplify this answer any further. We present our answer in alphabetical order since it makes it easier to read when the problems become more involved.

## Example 2

Multiply (*x* +
5)(*a* − 6)

Answer

We multiply this out as follows. We take each term of the first bracket and multiply them by the second bracket. Then we expand out the result.

(

x+ 5)(a− 6)=

x(a− 6) + 5(a− 6)=

ax− 6x+ 5a− 30

We cannot do any more with this answer. There are no like terms, so we cannot simplify it in any way.

## Example 3

Multiply (2*x* +
3)(*x*^{2} − *x* − 5)

Answer

We take the 2 terms of the first bracket and multiply both of them by the second bracket.

(2*x* + 3)(*x*^{2}− *x* − 5)

= (2*x*)(*x*^{2} − *x* − 5) +
(3)(*x*^{2} − *x* − 5)

= (2*x*^{3} − 2*x*^{2} − 10*x*) + (3*x*^{2} − 3*x* − 15)

= 2*x*^{3} + *x*^{2} −
13*x* − 15

This time we could collect together some like terms. There was:

−2

x^{2}+ 3x^{2}=x^{2}

and

−10

x− 3x= −13x

## Example 4

Multiply `(x - 3)^2`

Answer

(*x* − 3)^{2}

= (*x* − 3)(*x* − 3)

= *x*(*x* − 3) − 3(*x*
− 3)

= *x*^{2} − 3*x* −
3*x* + 9

= *x*^{2} − 6*x* + 9

## Important Note:

`(x - 3)^2`

is NOT equal to

`x^2 - 9`

Please take note of this! Many students confuse this idea and it's not surprising because sometimes the way we write math is not consistent. (See also Towards more meaningful math notation for a discussion on this issue.)

## Example 5

Simplify 5*x*[−4 + 10(*x* − *y*)] + 7*x*

Answer

5*x*[−4 + 10(*x* − *y*)] +
7*x*

= 5*x*[−4 + 10*x* − 10*y*] +
7*x*

= −20*x* + 50*x*^{2} − 50*xy* +
7*x*

= −13*x* + 50*x*^{2} − 50*xy*

## Example 6

Simplify (*p* − 1)(*c* − 1) + 2

Answer

*p*(*c* − 1) − 1(*c* − 1) + 2

= *pc* − *p* − *c* + 1 + 2

= *pc* − *p* − *c* + 3