Explain why (x^2 - 9) divided by (x + 3) is not the same as (x - 3)

Let's divide (`x^2-9` ) by (x+3)

This can be divided by many methods,

1) Lets first divide it by factorizing the polynomial (`x^2-9` ),

We know that `a^2-b^2=(a+b)(a-b)`

So `x^2-9=x^2-3^2=(x+3)(x-3)`

Now,`(x^2-9)/(x+3)=((x+3)(x-3))/(x+3)`

`=x-3`

And if you want to divide (`x^2-9` ) by (x-3),

`(x^2-9)/(x-3)=((x+3)(x-3))/(x-3)`

`=(x+3)`

2) We can divide the polynomial by using the long division method:     

        x - 3

      ________

x+3|`x^2-9`

       `x^2+3x`

     ...

Let's divide (`x^2-9` ) by (x+3)

This can be divided by many methods,

1) Lets first divide it by factorizing the polynomial (`x^2-9` ),

We know that `a^2-b^2=(a+b)(a-b)`

So `x^2-9=x^2-3^2=(x+3)(x-3)`

Now,`(x^2-9)/(x+3)=((x+3)(x-3))/(x+3)`

`=x-3`

And if you want to divide (`x^2-9` ) by (x-3),

`(x^2-9)/(x-3)=((x+3)(x-3))/(x-3)`

`=(x+3)`

2) We can divide the polynomial by using the long division method:     

        x - 3

      ________

x+3|`x^2-9`

       `x^2+3x`

      ________

              `-3x-9`

               `-3x-9`

              __________

                    0

               __________

 So, (`x^2-9` ) divided by (x+3) yields (x-3)

and (`x^2-9` ) divided by (x-3) yields (x+3)