Question

Find all the real zeros of the polynomial. Use the quadratic formula if necessary. If a zero has multiplicity greater than one, only enter the root once.
P(x) = x^4 − 8x^3 + 8x^2 + 23x + 6

Answer 1

ew >.> quartic polynomial. Might as well try to guess a root. Use the Rational Roots Theorem to guess one.

Answer 2

Isn't the highest exponent the number of zeros?

Answer 3

Thats right-ish. The highest power tells you how many zeros there are. It is possible that all the zeros could be the same number though. For example:
\[x^4+4x^3+6x^2+4x+1 = 0\]
Has 4 zeros, they just all happen to be -1. When you factor that polynomial it becomes:
\[(x+1)^4 = 0\]

Answer 4

a huge hint is that since some math teacher made this up, either 1 is a zero or -1 is

Answer 5

LOLOL

Answer 6

1 is easy to check, just add the coeffienients:
\[1-8+8+23+6\] nope

Answer 7

i kid you not. it is always either 1 or -1. in this case it is -1

Answer 8

Best advice ive ever heard on this website lol

Answer 9

Hey Satellite, where have you been? Some people were ganging up on me yesterday, i needed someone to help me defend my answer >.>

Answer 10

vacation

Answer 11

what was it?

Answer 12

ah, i hope you had a good vacation! It was nothing remotely hard. It was just a bunch of misinformed people.
Question was, how many solutions are there to:
\[\sqrt{x} = -4\]

Answer 13

none nil zippo. unless you are working over C

Answer 14

right right. They were saying x = 16 was a solution, and i was trying my hardest to give every mathematical reason why it wasnt lol.

Answer 15

what??????

Answer 16

But no one was buying it. It was like I was some crazy guy trying to spread misinformation and lies >.>

Answer 17

reason? how about the fact that
\[\sqrt{16}=4\] and last time i checked
\[4\neq -4\]

Answer 18

I told them how if you square equations you create what are called extraneous solutions, and that x = 16 was extraneous.
I told them that the graphs of f(x) = sqrtx and g(x) = -4 dont intersect. They didnt listen to any of it lol.

Answer 19

oh i know this one. it goes like this:

\[(-4)^2=16\] so
\[\sqrt{16}=\pm4\] i have battled against this before

Answer 20

Yeah, they kept bring that up, and i was like, "yes, i know that, but square roots only return positive values. If you want the negative, you have to denote that with a negative sign. Since this problem doesnt have a negative sign, its not asking for it."
They proceeded to mock me anyways!

Answer 21

None of my friends were around to help (or anyone willing to listen to reasoning, or anyone semi-intelligent for that matter <.<) :(

Answer 22

well let them suffer under their false delusions. apparently the square root function for them is not well defined. i have had to resort to
"you can bank on it" for this argument so let them mock and be mistaken.

now dinner time

Answer 23

lolol, thanks, i just had to get that off my chest haha