Question

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Answer 1

@LilHefner3 @whpalmer4

Answer 2

I would try factoring the polynomials in the left hand part to see if you can simplify the fraction, or failing that, get a common denominator.

Answer 3

I'm not quite sure how to do that.

Answer 4

Okay, to factor

(x^2+3x+2) we want to find two numbers a and b such that
(x+a)(x+b) = x^2 + 3x + 2
(x+a)(x+b) = x^2 + ax + bx + ab = x^2 + (a+b)x + ab
so if we can find a, b such that a*b = 2 and a+b = 3 we're all set. which two numbers multiply to 2 and add to 3?

Answer 5

I think you may be confused. Is this an example of what to do? Cause the equation above is really different.

Answer 6

This is an example of how to factor. You said you weren't sure how to do that. I don't want to do the problem for you, I want to show you how to do it.

Answer 7

Actually, I am factoring the numerator in my example, I just used x instead of n as the variable.

Answer 8

Oh ok. Hold on, I'll figure it out a sec...

Answer 9

The two numbers are 1 and 2. 1•2 = 2 and 1 + 2 = 3.

Answer 10

Right. So we could factor the numerator to (n+1)(n+2)

How about the denominator?

Answer 11

\[(n^2+6n+8)\]Which two numbers multiply to 8 and add to 6?

Answer 12

4 and 2

Answer 13

Right! A serendipitous result, given that the denominator of the other fraction is \((n+4)\)

We can now write our denominator as \((n+4)(n+2)\) so the whole enchilada becomes
\[\frac{(n+1)(n+2)}{(n+2)(n+4)} - \frac{2n}{n+4}\]and do you see where we can go from here?

Answer 14

From what I see, in the first fraction, we can cancel out the two (n + 2)'s right?

Answer 15

Yep. And that leaves you with a common denominator in the two fractions.

Answer 16

What's your final result?

Answer 17

So I would have (n + 1)2n/n + 4. Then if I combine the numerator I'd have 3n + 1/n + 4!

Answer 18

Thank you so much!!!

Answer 19

Check that work again...

Answer 20

It'd be a shame to get a wrong answer after all of this ;-)

Answer 21

Ok. :)

Answer 22

\[\frac{(n+1)}{(n+4)} - \frac{2n}{(n+4)} = \frac{(n+1)-2n}{(n+4)} = \frac{(1-n)}{(n+4)}\]

Answer 23

You must have overlooked that pesky \(-\) between the fractions when combining.

Answer 24

Sorry I can't see the last bit of your fraction there. It stopped at (n + 1) - 2n/(n + 4).

Answer 25

\[\frac{(n+1)-2n}{(n+4)} = \frac{(1-n)}{(n+4)}\]

Better?

Answer 26

Much. Thank you! :)