need help

[shadamon161]

Multiply: 9-n^2 n^2+2n-15

-------- * ---------

n^2-5n+6 n^2+8n+15
I factored everything but its not coming out to the answer its suppose to be

 

[lookagain]

Multiply: 9-n^2 n^2+2n-15

-------- * ---------

n^2-5n+6 n^2+8n+15
I factored everything but its not coming out to the answer its suppose to be

\(\displaystyle \bigg(\frac{9 - n^2}{n^2 - 5n + 6}\bigg)\bigg(\frac{n^2 + 2n - 15}{n^2 + 8n + 15}\bigg)\)


shadamon161,

you have to show us the factoring you claim so you can be helped.

 

[Deleted member 4993]

Multiply:
9-n^2 n^2+2n-15

-------- * ---------

n^2-5n+6 n^2+8n+15
I factored everything but its not coming out to the answer its suppose to be

I assume you have:

\(\displaystyle \frac{9-n^2}{n^2-5n+6} * \frac{n^2+2n-15}{n^2+8n+15}\)

\(\displaystyle = \ \frac{(3-n)(3+n)}{(n-3)(n-2)}*\frac{(n+5)(n-3)}{(n+5)(n+3)}\)

Now what did you get from here? also explain your steps...

 

[shadamon161]

The answer is 3-n over n-2 cause I crossed out all like terms

 

[JeffM]

The answer is 3-n over n-2 cause I crossed out all like terms

And the answer it is supposed to be is ?

Here is the issue. It is very difficult to be sure exactly what the problem is because you did not express it in standard notation. You did not say what answer you had found or how you got it. Moreover, you did not tell us what the answer is that the book gives. So how can we confirm whether you did the problem correctly or whether the book's answer is wrong?

 

[soroban]

Hello, shadamon161!

\(\displaystyle \text{Multiply: }\:\frac{9-n^2}{n^2-5x+6}\cdot\frac{n^2+2n-15}{n^2+8n+15}\)

I factored everything but its not coming out to the answer its suppose to be.

\(\displaystyle \text{If your answer came out to be: }\:\frac{7\heartsuit +K\spadesuit}{B\flat}\)
. . you probably made an error in converting miles-per-hour to furlongs-per-fortnight.


\(\displaystyle \text{Factor everything: }\:\begin{Bmatrix}9-n^2 &=& -(n^2-9) &=& -(n-3)(n+3) \\ n^2-5n+6 &=& (n-2)(n-3) \\ n^2+2n-15 &=& (n-3)(n+5) \\ n^2+8x+15 &=& (n+3)(n+5) \end{Bmatrix}\)


\(\displaystyle \text{The problem becomes: }\:\frac{-(\rlap{/////}n-3)(\rlap{/////}n+3)}{(n-2)(\rlap{/////}n-3)}\cdot \frac{(n-3)(\rlap{/////}n+5)}{(\rlap{/////}n+3)(\rlap{/////}n+5)} \;=\;\frac{-(n-3)}{n-2} \;=\;\frac{3-n}{n-2}\)


So I agree with your answer.

What did "they" say the answer is?

 

[mmm4444bot]

I crossed out all like terms

A note about terminology.

"Like-terms" is the wrong name to use, in describing what you did.

We do not say, "crossed out all like terms". We say, "cancelled the common factors".

In other words, for example, the expression (n + 5) is called a factor, and when this factor appears in both a numerator and a denominator, it is called a common factor.

You did not cancel like-terms.

 

[shadamon161]

Thanks for the information and the answer in the answer key says 3-n over n-2