Question

can you check my answer?
what is the remainder when 3x^4+2x^3-x^2+2x-9)/(x+2)?

Answer 1

i got 5

Answer 2

I am getting 15 as the remainder.

Answer 3

I'll recheck it.

Answer 4

I think it's 15.

Answer 5

i did it again and your right

Answer 6

Yeah, good work!

Answer 7

thnks :)

Answer 8

yw

Answer 9

can u help with one more? @SolomonZelman

Answer 10

I'll try my best.

Answer 11

ok i hate this question i always get it wrong.
what are possible number of positive negative and complex zeros of f(x)=x^6-x^5-x^4+4x^3-12x^2+12?

Answer 12

\[f(x)=x^6-x^5-x^4+4x^3-12x^2+12\] ( just to see it clearly)

Answer 13

thats right

Answer 14

positive 3 or 1 negative 2 or 0 complex 4 2 or 0
positive 3 or 1 negative 3 or 1 complex 4 2 or 0
positive 4 2 or 0 negative 2 or 0 complex 6 4 2 or 0
positive 2 or 0 negative 2 or 0 complex 6 4 or 2

Answer 15

What I am think of is this.
\[f(x)=x^5(x-1)-x^4+4x^3-12(x^2-1)\]\[f(x)=x^5(x-1)-x^4+4x^3-12(x-1)(x+1)\]\[(x^5+12x+12)(x-1)-x^4+4x^3\]\[(x^5+12x+12)(x-1)-x^3(x+4)\] but factoring would work.

Answer 16

I would suggest plugging in the choices.

Answer 17

I meant but factoring would NOT work....

Answer 18

im thinking d

Answer 19

@SolomonZelman

Answer 20

\[f(x)=x^6-x^5-x^4+4x^3-12x^2+12\], IDK sorry.

Answer 21

If you had one zero given I might have been able to do it.

Answer 22

what do u think it is??

Answer 23

@SolomonZelman

Answer 24

Positive 3 doesn't work so not A or B.

Answer 25

thnk you :)

Answer 26

And positive 4 doesn't work, cuz x^6 is a HUGE number it won't equal zero is x=4

Answer 27

So I therefore think D.

Answer 28

yay!!!!!!!!! thnks your really the best!!

Answer 29

Yeah, sometimes, I can do the problem, by elimination... my method!!! YW!