Question

How do you multiply and simplify `\frac { x ^ { 2} - 9} { x ^ { 2} + 10x + 21} \cdot \frac { x ^ { 2} } { x ^ { 2} - 3x }`?

Answer 1

`x/(x-7)`

Explanation:

`color(brown)("It is usually constructive to look for parts of an expression that can be")` `color(brown)("cancelled out.")`

Comparing the two denominators.

Consider `x^2+10x+21`
`" "`Note that `3xx7=21` and that `3+7=10`.

Consider `x^2-3x`
`" "`Note that if we factor out an `x` we have `(x-3)`
`" "`and we will have an `x+3` when we factorise `x^2+10x+21`

We may, or may not be able to use this. Lets have a 'play' and see what happens.
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Write as: `" "(x^2-9)/((x+7)(x+3))xx (cancel(x) xx x)/(cancel(x)(x-3))`

Giving:`" "(x^2-9)/((x+7)(x+3))xx x/(x-3)`

Consider the numerator: `x^2-9`

This is the same as `x^2-3^2" "->" "(x-3)(x+3)`

So now we have:

`((x-3)cancel((x+3)))/((x+7)cancel((x+3)))xx x/(x-3)`

Giving:

`(x-3)/(x+7)xxx/(x-3)`

`(cancel(x-3))/(cancel(x-3))xx x/(x-7)`

`x/(x-7)`