Question

How do you multiply and simplify `\frac { 4a ^ { 2} - 36} { a ^ { 2} + 6a + 9} \cdot \frac { a ^ { 2} + 3a } { 12a }`?

Answer 1

`= ((a-3))/3`

Explanation:

With algebraic fractions, the starting point is to factorise.

`(4a^2-36)/(a^2+6a+9) xx (a^2+3a)/(12a)`

`=(4(a+3)(a-3))/((a+3)(a+3)) xx (a(a+3))/(12a)`

Now that everything is written as factors we can cancel.

`=(cancel4cancel((a+3))(a-3))/(cancel((a+3))cancel((a+3))) xx (cancela(cancel(a+3)))/(cancel12_3cancela)`

`= ((a-3))/3`

This cannot simplify further. Do not be tempted to cancel the `3's`

Answer 2

`(a-3)/3 " "->" "a/3-1`

Slightly different way of looking at cancelling out.

Explanation:

`color(blue)("Consider "4a^2-36)`

It is usually a good idea to 'experiment' with the numbers to see if we can cancel out anything.

It is always worth remembering several standardised forms and what follows is one of them.

It is usually presented using a and b so I will do the same.
`a^2-b^2=(a-b)(a+b)`

We have this form in the part: `4a^2-36`

This can and may be written as: `(2a)^2-6^2`

Giving: `(2a-6)(2a+6)`

The 2 and the 6 are both even so 2 is a factor so we may write this as:

`2(a-3)xx2(a+3)-> color(green)(4(a-3)(a+3))`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Consider "a^2+6a+9)`

Observe that `3+3=6 and 3xx3=9` so we may write this as:

`color(green)((a+3)(a+3)) ....->(a+3)^2`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Consider "(a^2+3a))`
Notice that `a` is in both terms so we may factor that out giving:

`color(green)(a(a+3))`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Putting it all together")`

`(4(a-3)(a+3))/((a+3)(a+3))xx(a(a+3))/(12a)`

This is the same as:

`4xx(a-3)xx(a+3)/(a+3) xx(a+3)/(a+3)xxa/axx1/12`

`4xx(a-3)color(white)(.)xxcolor(white)(.)1" " xx" "1" "xxcolor(white)(.)1xx1/12`

The above is what they mean when they say cancel out: for multiply and divide you turn the values into 1. Any value multiplied by or divided by 1 does not change

`(4(a-3))/12`

But 12 is the same as `4xx3`

`(4(a-3))/(4xx3)" "->" "4/4xx(a-3)/3`

`(a-3)/3 " "->" "a/3-1`