Question

Need help with division problem

Answer 1

Try thinking of what \[x^2-4\] can be factored as and what terms would cancel out.

Answer 2

the factored form would be (x-2)(x+2)

Answer 3

Dont know how this got deleted...

Answer 4

Right. Now how can you simplify the equation?

Answer 5

Well x-2 would cancel right?

Answer 6

Yep! Try factoring the other terms too.

Answer 7

\[(x + 1 - \sqrt{2}) (x + 1 + \sqrt{2})\]

Answer 8

For the top term, try this:

\[x^2+2x+1=(x+1)(x+1)\]

Answer 9

Im really lost

Answer 10

Here's the original equation:
\[\frac{ \frac{ x^2+2x+1 }{ x-2 } }{ \frac{ x^2-1 }{ x^2-4 } }\]
Now we see we have a fraction of fractions... let's put everything in factored form.
Start with:
\[x^2+2x+1 = (x+1)(x+1)\]Let's check this by multiplying it out:
\[(x+1)(x+1) = x*x+(1)x+(1)x+1=x^2+2x+1 ==> sweet\]
The next term is already in the form we want \[x-2\]
The next one can be factored as well \[x^2-1=(x+1)(x-1)\]Let's check this by multiplying it out:
\[(x+1)(x-1)=x*x*+(-1)x+(1)x+(1)(-1)=x^2-1 ==> sweet\]
Onto the last one which you have already done:
\[x^2-4=(x+2)(x-2)\]Now let's check again:
\[(x+2)(x-2)=x*x+2x+(-2)x+(-2)(2)=x^2-4 ==> sweet\]
Now we have a form where we can cancel some terms
\[=\frac{ \frac{ (x+1)(x+1) }{ x-2 } }{ \frac{ (x+1)(x-1) }{ (x-2)(x+2) } }\]
Remember this property of dividing fractions as well:
\[\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a}{b}*\frac{d}{c}\]
Hope this helps!

Answer 11

wow this was a great help