This is a ridiculous math problem below, I can not figure it out. Try it and let me know what you get?

Question

p(x) = 3x^2 - 21x + 30 . q(x) = x^2 + 3x - 10 . f(x) = [p(x)]/[q(x)] .

Each answer is designed to be used only once.

Questions
1. There is a removable discontinuity where x = 3
2. There is a non-removable discontinuity where x = -5
3. A zero for f(x) occurs where x = 2
4. A horizontal asymptote for the graph of y = f(x) is y = none of these

.
Matching Pairs

1. 3, 5, 2 or none of these
2. 3, 5, 2 or none of these
3. 3, 5, 2 or none of these
4. 3, 5, 2 or none of these

example, if you think that the answer to the second question is 2, then select 2 from the list of answers inside the matching answer

Answer 1

are you sure is that all ???????????????
we should match the pair with what ? try to explain more so i can help :) sorry

okay i am gonna tell you what i came up with and you decide :)

F(x) = ( 3x^2 - 21x + 30) / ( x^2 + 3x - 10 )
F(x) = [ 3( x^2 -7x+10 ) ] / ( x^2 + 3x - 10 )
F(x) = [ 3 ( x-2 ) ( x+5 ) ] / [ ( x-2 ) ( x + 5 )
the ( x-2 ) goes with ( x-2 )
F(x) = [ 3 ( x-5 ) ] / ( x+5 )
so F(x) = ( 3x-15) / ( x+5 )
as for Q number one
1) There is a removable discontinuity where x = 3

lim f(x ) =
( x->3)

=lim [ ( 3x-15) / ( x+5 ) ] =
( x->3)

=( 9-15) / ( 3+5 ) = -6/ 8 = -.75
so there isn't a removable discontinuity here so its not true ?

tell me what you think in that ?

Answer 2

Evaluate the three polynomials for x = 2, 3, and -5. If f(x) is 0/0, you can remove the discontinuity using l'Hospital's rule; if it is anything else divided by 0, you can't.

Answer 3

p(x) = 3x^2 - 21x + 30 . q(x) = x^2 + 3x - 10 . f(x) = [p(x)]/[q(x)]

Ans:
f(x)={3x^2 - 21x + 3}/{x^2 + 3x - 10 }
=3 - {30x+27}/{(x-2 )(x + 5 }
(1)None of this
(2) none of this
(3) none of this
(4) Horizontal asymtoes is y=3

Answer 4

1453

Answer 5

the answer is 1

Answer 6

Oh, heck no! this is horrible math.. sorry

Answer 7

god damn! idk