Question

n^4-11n^2+30/n^4-7n^2+10

Answer 1

have you done factoring of quadratics yet?

Answer 2

I don't know, sorry, I'm just trying to simplify the rational expression....

Answer 3

well, you'd need to cover that
since is how you'd simplify it, by factoring top and bottom
-> http://www.youtube.com/watch?v=2Q8lG1WaqJY

Answer 4

Okay, thank you, am I correct in thinking that factoring involves the reverse of foil (last term, third term, second term, first) ?

Answer 5

yeap

Answer 6

Okay, maybe I'm not as hopelessly lost as I though, you've been a big help, thank you!

Answer 7

\(\large { \begin{array}{cccl}
({\color{red}{ n^2}})^4&-11n^2&+30\\
&\uparrow &\uparrow \\
&-6-5&-6\cdot -5
\end{array}\implies ({\color{red}{ n^2}}-6)({\color{red}{ n^2}}-5)
\\ \quad \\
---------------------\\
\begin{array}{cccl}
({\color{red}{ n^2}})^4&-7n^2&+10\\
&\uparrow &\uparrow \\
&-2-5&-2\cdot -5
\end{array}\implies ({\color{red}{ n^2}}-2)({\color{red}{ n^2}}-5)}\)

Answer 8

wops.. darn I got the wrong ... lemme fix that quick \(\large { \begin{array}{cccl}
({\color{red}{ n^2}})^2&-11n^2&+30\\
&\uparrow &\uparrow \\
&-6-5&-6\cdot -5
\end{array}\implies ({\color{red}{ n^2}}-6)({\color{red}{ n^2}}-5)
\\ \quad \\
---------------------\\
\begin{array}{cccl}
({\color{red}{ n^2}})^2&-7n^2&+10\\
&\uparrow &\uparrow \\
&-2-5&-2\cdot -5
\end{array}\implies ({\color{red}{ n^2}}-2)({\color{red}{ n^2}}-5)}\)

Answer 9

then you'd use the factors atop and bottom to see which ones cancel each other

Answer 10

So, if I were to cancel the top and bottoms common factor I would cross out the n squared minus five, leaving me with n squared minus six over n squared minus two as the simplified expression?

Answer 11

yeap

Answer 12

Would it be too much to ask if I worked one out of my own and attached it could you see if I was doing it correctly? If it's any inconvenience I don't want to trouble you.

Answer 13

sure