Find the remainder when f(x) is divided by g(x)=x+1 f(x)=8x^63-12x^57+16x^48+5x^23-7x^13-17
use the remainder theorem
if \[\large f(x)=(x-a)\cdot q(x)+r(x) \] then \[\large f(a)=(a-a)\cdot g(a)+r(a)=r(a) \]
okay mind talking me through it as I set up the problem?
plugin x = -1 in f(x)
f(x)=8x^63-12x^57+16x^48+5x^23-7x^13-17 f(-1) = ?
So just sub in (-1) every place there is an x?
yes!
5?
\[8(-1)^{63}-12(-1)^{57}+16(-1)^{48}+5(-1)^{23}-7(-1)^{13}-17\]
f(x)=8x^63-12x^57+16x^48+5x^23-7x^13-17 f(-1) = -8 + 12 + 16 -5 + 7 - 17 = 5
good job !
in simple terms, remainder theorem says this : the remainder when u divide any polynomial \(f(x)\) by a linear polynomial \((x-\color{red}{a})\) is same as \(f(\color{red}{a})\)
Ill have to go through my notes again for that one. Seemed so simple it confused me lol. So my answer would just be 5? Even though g(x)=x+1?
yes, 5 is the remainder
So it could be set up how we did earlier with long division, but this way is just A LOT easier? or no?
if u setup long division, you wont be able to finish it in ur life time -.- try it :)
but when u finish, you wud get the same 5 as remainder
Hahaha okay I was just making sure. God that would be cruel punishment for any man (and woman :p)
What about proving roots?
lol yes, nevertheless its a good idea to try long division and see why its really called "long division"
Lmao
you will get 63 rows in ur long division by the time u finish and see the remainder
Knowing me I would miss a negative and completely ruin the problem .-.
long division is insanely long for this particular problem, try it and see lol
no no I trust you lol. Now for trying to prove roots I cannot find anything in my notes and I like to be "ahead of the game" I guess you can say so mind helping with a practice problem from book?
Or I can open new question
shoot..
Prove g(x) has a root between (1,2) \[g(x)=x^5-4x^4+7x^3-2x^2+x-5\]
intermediate value theorem
let me ask u a simple q :- wat does it mean to have a root between (1, 2) ?
having a root means between (1, 2) means this :- the function must equal "0" for some number between (1, 2)
a graph might help
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