Question

(8x^4+nx^3-8x^2-x-19)/(2x+5)
solve n
when remainder is -4

Answer 1

answer i have 125/32 but it's saying it's wrong i dont know why

Answer 2

this looks like a pain. I would replace x by
\[-\frac{5}{2}\] and set the result = -4 and solve for n
was that the method you used?

Answer 3

exact same remainder theorem

Answer 4

i get -64n+246=-4

Answer 5

let me try something with pencil and paper and see what happens

Answer 6

thnk u satellite

Answer 7

oh maybe you have to divide by 2. hold on a minute i can't work on the screen

Answer 8

lol ok

Answer 9

because synthetic division only works if the leading coefficient is 1. that is why it probably did not work. i am trying it long hand

Answer 10

uh why are u doing synthetic division?

Answer 11

or long division i thought question was supposed to use remainder theorem

Answer 12

id divide it out to see what the "equation" for the remainder would be; set that equal to -4

Answer 13

yes but if you replace x by 5/2 you are ignoring the fact that you are dividing by 2x+5
it is as if you are dividing by
\[x+\frac{5}{2}\]

Answer 14

ok.............

Answer 15

16

Answer 16

i divided the numerator and denominator by 2, the evaluated as 5/2
so far i have
\[269+\frac{n}{2}\times (\frac{5}{2})^3=-4\] and will try from here

Answer 17

ok zarkon what snap method did you use?

Answer 18

satellite ur way i get -4368/125

Answer 19

how did zarkon get 16?

Answer 20

i calculated wrong let me try again slowly. in the meantime maybe zarkon will show the snap way

Answer 21

CALLING ON ZARKON!!

Answer 22

or actually ny1 who can explain how to get the correct answer

Answer 23

16 is correct. now let me see if my feeble mind can find it quickly

Answer 24

i'm trying to find a nice way to get the answer...I don't want to type up what i did

Answer 25

Lol

Answer 26

4x^3+(n-20)x^2/2+(-5n+84)x/4+(25n -424)/8
-------------------------
2x+5 | 8x^4 +nx^3 -8x^2 -x -19
-8x^4 -20 x^3
---------------
(n-20)x^3 ( -16)x^2/2
-(n-20)x^3 (-5n +100)x^2/2
---------------------------
(-5n +84)x^2/2 (-4)x/4
(25n-420)x/4
--------------
(25n -424)x/4 (-152)/8
(-125n+2120)/8
----------------
(-125n+2068)/8

-125n+2068
----------- = -4
8(2x+5)

i cant tell if i did it right or not :)

Answer 27

ack .. remainder = -4 so

-125n + 2068 = -4 then?

Answer 28

^according to u it very close n=16.576

Answer 29

@zarkon can you just say what you did? i divide numerator and denominator by 2, replace x by -5/2 and then set result = -4

Answer 30

Zarkon u there?

Answer 31

solve \[246-\frac{125n}{8}=-4\]

Answer 32

\[8x^4+nx^3-8x^2-x-19|_{x=-5/2}=246-\frac{125n}{8}\]

Answer 33

nvm my stupidity to the max i was plugging in -4 in the nx^3 FML

Answer 34

i tried that but it didn't work!

Answer 35

it works

Answer 36

wow i marvel at my own stupidity sometimes

Answer 37

oh i guess it helps if you type -5/2 instead of 5/2
<dopeslap>

Answer 38

thnkx zarkon

Answer 39

.... that was my first thought; but figured why not kill a few brain cells along the way :)

Answer 40

\[\frac{8x^4+nx^3-8x^2-x-19}{2x+5}=g(x)+\frac{-4}{2x+5}\]

\[8x^4+nx^3-8x^2-x-19=g(x)(2x+5)-4\]

evaluate at x=-5/2 and solve for n

I believe this is what satellite73 was doing ?

Answer 41

yah but i typed in 5/2 so answer was all screwed up.

Answer 42

\[\large 8x^4+nx^3-8x^2-x-19=-4\]



\[\large 8\left(-\frac{5}{2}\right)^4+n*\left(-\frac{5}{2}\right)^3-8\left(-\frac{5}{2}\right)^2-\left(-\frac{5}{2}\right)-19=-4\]



\[\large 8\left(\frac{625}{16}\right)+n*\left(-\frac{125}{8}\right)-8\left(\frac{25}{4}\right)-\left(-\frac{5}{2}\right)-19=-4\]



\[\large \frac{5000}{16}-\frac{125n}{8}-\frac{200}{4}+\frac{5}{2}-19=-4\]



\[\large \frac{625}{2}-\frac{125n}{8}-50+\frac{5}{2}-19=-4\]



\[\large -\frac{125n}{8}+246=-4\]



\[\large -\frac{125n}{8}=-4-246\]



\[\large -\frac{125n}{8}=-250\]



\[\large -125n=-250*8\]



\[\large -125n=-2000\]



\[\large n=-\frac{2000}{-125}\]



\[\large n=16\]

Answer 43

thnk u jim

Answer 44

np

Answer 45

Solved with two Mathematica statements. Refer to the attachment.