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

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

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?

exact same remainder theorem

i get -64n+246=-4

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

thnk u satellite

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

lol ok

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

uh why are u doing synthetic division?

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

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

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 \[x+\frac{5}{2}\]

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

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

ok zarkon what snap method did you use?

satellite ur way i get -4368/125

how did zarkon get 16?

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

CALLING ON ZARKON!!

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

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

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

Lol

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 :)

ack .. remainder = -4 so -125n + 2068 = -4 then?

^according to u it very close n=16.576

@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

Zarkon u there?

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

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

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

i tried that but it didn't work!

it works

wow i marvel at my own stupidity sometimes

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

thnkx zarkon

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

\[\frac{8x^4+nx^3-8x^2-x-19}{2x+5}=g(x)+\frac{-4}{2x+5}\] \[8x^4+nx^3-8x^2-x-19=g(x)(2x+5)-4\] evaluate at x=-5/2 and solve for n I believe this is what satellite73 was doing ?

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

\[\large 8x^4+nx^3-8x^2-x-19=-4\] \[\large 8\left(-\frac{5}{2}\right)^4+n*\left(-\frac{5}{2}\right)^3-8\left(-\frac{5}{2}\right)^2-\left(-\frac{5}{2}\right)-19=-4\] \[\large 8\left(\frac{625}{16}\right)+n*\left(-\frac{125}{8}\right)-8\left(\frac{25}{4}\right)-\left(-\frac{5}{2}\right)-19=-4\] \[\large \frac{5000}{16}-\frac{125n}{8}-\frac{200}{4}+\frac{5}{2}-19=-4\] \[\large \frac{625}{2}-\frac{125n}{8}-50+\frac{5}{2}-19=-4\] \[\large -\frac{125n}{8}+246=-4\] \[\large -\frac{125n}{8}=-4-246\] \[\large -\frac{125n}{8}=-250\] \[\large -125n=-250*8\] \[\large -125n=-2000\] \[\large n=-\frac{2000}{-125}\] \[\large n=16\]

thnk u jim

np

Solved with two Mathematica statements. Refer to the attachment.

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