anonymous
  • anonymous
(2x^(4)-x^(3)+1)/(x+3) Use remainder theorem to find the remainder. Please show all work. Thanks.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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amistre64
  • amistre64
the remainder thrm just says that there is an added remaineder to the answer ;)
amistre64
  • amistre64
P(x) = Q(x) + R(x)
amistre64
  • amistre64
we divide these like they were any other numbers.... division is division regardless of what you divide..

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amistre64
  • amistre64
the key is to remember how long division is done; or use synthetic division
anonymous
  • anonymous
i have to use long division but im confused because in order to devide i have to 2x^4 times (x+3) then it would be 2x^4 -x^3 over 2x^4- 6x^3 and then i got confused
amistre64
  • amistre64
we can work thru that are you remembering to subtract your answers?
anonymous
  • anonymous
what do you mean exactly u mean 2x^4- 2x^4 = 0 and then -x^3 - (-6x^3)?
amistre64
  • amistre64
2x^3 <- 2x^3(x+3)= (2x^4+6x^3), but we subtract this result ---------------- to get our next round...... x+3| 2x^4 -x^3 +1 -2x^4 -6x^3 ------------ -7x^3 +1
amistre64
  • amistre64
remember how to do it with numbers? same thing: 3 <- 6(3) = 18....what do we do with the 18? ----- 6| 19 -18 <- we subtract it.... ------- 1
anonymous
  • anonymous
ok i see where i messed up originally now we times (x+3) by 7x^2? and then work the problem?
amistre64
  • amistre64
2x^3 -7x^2 ---------------- x+3| 2x^4 -x^3 +1 -2x^4 -6x^3 ------------ -7x^3 +1 +7x^3 +21x^2 --------------- 21x^2 +1 <- whatever doesnt cancel out gets brought down
anonymous
  • anonymous
not (x+3) times 21x?
amistre64
  • amistre64
the next number need to be the same as this bottom first number...or at least get to equal it.... we need a 21; and x*x=x^2 so....... 21x
amistre64
  • amistre64
thats right ;)
anonymous
  • anonymous
wait never mind it needs to be by -21x^2?
amistre64
  • amistre64
2x^3 -7x^2 +21x ---------------- x+3| 2x^4 -x^3 +1 -2x^4 -6x^3 ------------ -7x^3 +1 +7x^3 +21x^2 --------------- 21x^2 +1 -21x^2 -63x ------------- -63x +1
amistre64
  • amistre64
do you understand that we are subtracting the result?
anonymous
  • anonymous
now just (x+3) times 63? and yes sir/ma'am
amistre64
  • amistre64
x times ??? = -63x ?
anonymous
  • anonymous
63 right so we get 63x+189?
anonymous
  • anonymous
and then remainder is 190?
amistre64
  • amistre64
thats right :) now... -63x - (63x) = 0 right?
anonymous
  • anonymous
Yeah.
amistre64
  • amistre64
2x^3 -7x^2 +21x -63x ---------------- x+3| 2x^4 -x^3 +1 -2x^4 -6x^3 ------------ -7x^3 +1 +7x^3 +21x^2 --------------- 21x^2 +1 -21x^2 -63x ------------- -63x +1 +63x +189 ------------- 190 <- remainder :)
anonymous
  • anonymous
Thanks!
amistre64
  • amistre64
synthetic is similar, but we already work with an opposite so we add stuff.... (x+3)..x=-3 -3|2 -1 0 0 1 0 -6 21 -63 189 ------------------- 2 -7 +21 -63 190 <- remainer 190 ^^^^^^^^^^^ 2x^3 -7x^2 +21x -63

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