I am going to assume that you mean
[ (n^2-4)/(n^2-2n) ] * n/(n^2+10n+16 )
Immediately you can see that n = 0 and n = 2 are excluded from the domain, also if
n^2+10 n + 16 = 0 which is if n = -2 or -8
now
[(n-2)(n+2)/n(n-2)] *n /[(n+2)(n+8)]
= 1/(n+8)
n^2-4/n^2-2n * n/n^2+10n+16
simplify and state the domain
3 answers
your expression definitely needs brackets to make sense
(n^2-4)/(n^2-2n) * n/(n^2+10n+16)
= (n+2)(n-2)/(n(n-2) ) * n( (n+2)(n+5)
= 1/(n+5) , n ≠ -2 , 0, 2 , -5
(n^2-4)/(n^2-2n) * n/(n^2+10n+16)
= (n+2)(n-2)/(n(n-2) ) * n( (n+2)(n+5)
= 1/(n+5) , n ≠ -2 , 0, 2 , -5
arggh, go with Damon's factoring of the last part
Whatever made me think that 2x5 = 16 ??????
It's clearly 2 x 8
Whatever made me think that 2x5 = 16 ??????
It's clearly 2 x 8
1 answer