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anonymous

  • 5 years ago

(2x^(4)-x^(3)+1)/(x+3) Use remainder theorem to find the remainder. Please show all work. Thanks.

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  1. amistre64
    • 5 years ago
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    the remainder thrm just says that there is an added remaineder to the answer ;)

  2. amistre64
    • 5 years ago
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    P(x) = Q(x) + R(x)

  3. amistre64
    • 5 years ago
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    we divide these like they were any other numbers.... division is division regardless of what you divide..

  4. amistre64
    • 5 years ago
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    the key is to remember how long division is done; or use synthetic division

  5. anonymous
    • 5 years ago
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    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

  6. amistre64
    • 5 years ago
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    we can work thru that are you remembering to subtract your answers?

  7. anonymous
    • 5 years ago
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    what do you mean exactly u mean 2x^4- 2x^4 = 0 and then -x^3 - (-6x^3)?

  8. amistre64
    • 5 years ago
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    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

  9. amistre64
    • 5 years ago
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    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

  10. anonymous
    • 5 years ago
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    ok i see where i messed up originally now we times (x+3) by 7x^2? and then work the problem?

  11. amistre64
    • 5 years ago
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    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

  12. anonymous
    • 5 years ago
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    not (x+3) times 21x?

  13. amistre64
    • 5 years ago
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    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

  14. amistre64
    • 5 years ago
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    thats right ;)

  15. anonymous
    • 5 years ago
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    wait never mind it needs to be by -21x^2?

  16. amistre64
    • 5 years ago
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    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

  17. amistre64
    • 5 years ago
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    do you understand that we are subtracting the result?

  18. anonymous
    • 5 years ago
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    now just (x+3) times 63? and yes sir/ma'am

  19. amistre64
    • 5 years ago
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    x times ??? = -63x ?

  20. anonymous
    • 5 years ago
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    63 right so we get 63x+189?

  21. anonymous
    • 5 years ago
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    and then remainder is 190?

  22. amistre64
    • 5 years ago
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    thats right :) now... -63x - (63x) = 0 right?

  23. anonymous
    • 5 years ago
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    Yeah.

  24. amistre64
    • 5 years ago
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    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 :)

  25. anonymous
    • 5 years ago
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    Thanks!

  26. amistre64
    • 5 years ago
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    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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