Hello everyone. I need help with my chemistry homework. I already have two possible answers for this one, I just have to ensure which one of them is the correct one.
Is it 3-amino-5-bromo-1,4-cyclohexadiene carboxylic acid or 1-aceto-3-amino-5-bromo-1,4-hexadiene??
This one is giving me a headache because I’m not sure whether the functional group COOH is a substituent or not.
@Students
Lol is anyone good about finding zeros of a function? I’m confused about checking the zeros lol idk how.
Original problem was p(x)= x to the third, -5x squared, + 2x + 8
How I’m doing it is I find them on calculator first with graphing (got -1, 1, 4) then did synthetic substitution… no problems, got 1, -1, -2 and 0.
Then after this I got a bit confused.
I assume I’m supposed to substitute to check them but idk how.
I saw in my notes something like that so I did this:
(x-4) (x squared + -1x - 2)
I did the quadratic formula to find them and I got -1.906009601 and 2.156009601
So um… wtf? So apparently it’s not -1, it’s negative 2 and I have no idea how to check the 1 to see if it’s a zero…
@HomeworkHelp
Sorry. Terrible at math. I can help with history, though!
I’m very bad at remembering steps lol.
I have dyscalculia
Imma wait to see if there are some good math people on before trying to do those problems again. I’m just gonna skip to the slightly easier problems then do science lol.
I can probably help with science.
I don’t need help with that lol. It’s literally just reading and answering one question. But thanks tho.
Substitute the values you got for x to check if the p(x) is zero
Ok so literally just put the values back into the equation?
Yep!
So p(x) = x^3 - 5x^2 + 2x + 8
Just set whatever values as x
So like if I’m checking -1 I would do it like this:
P(-1)= -1 to the third - 5 (-1) to the second + 2 (-1) + 8?
Also for x^2 - x -2
You get (x - 2)(x + 1)
So x = 2, -1 are zeroes
Yep!
Cool I’ll try that lol.
Just to clarify- after substituting would I factor each? Ik factoring is a part of this but idk really where it comes in lol.
Factoring?
When you substitute, you get an answer
like for example
p(-1) = 0
Ok I’m still confused lol.
I did -2^3 - 5(-2)^2 + 2 (-2) + 8
and I got 24 for that so I’m pretty sure I’m doing something wrong.
Ooooh ok so -2 isn’t a zero since you don’t get that when you factor out the function
So you had
(x - 4)(x^2 - x - 2)
Turns into
(x - 4)(x - 2) (x + 1)
Now set all of them equal to zero
x - 4 = 0
x - 2 = 0
x + 1 = 0
Which is
x = 4
x = 2
x = -1
So when x = 4, 2, -1 the function is zero. Therefore, all of these values are zeroes
Ohh ok I think I’m getting it now lol. Thanks for the help!
No problem! 
