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anonymous

  • 5 years ago

Help please, Divide (x³-x+8) / (x-2)

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  1. anonymous
    • 5 years ago
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    x^3-x+8=x^3-2x^2+2x^2-4x+3x-6+14 =x^2(x-2)+2x(x-2)+3(x-2)+14=(x^2+2x+3)(x-2)+14 (x^3-x+6)/(x-2) =x^2+2x+3+[14/(x-2)]

  2. anonymous
    • 5 years ago
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    ...huh?

  3. anonymous
    • 5 years ago
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    what?

  4. anonymous
    • 5 years ago
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    im trying to writ this down on paper but... you lost me @ the second step

  5. anonymous
    • 5 years ago
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    see i have rewrite the expression so that i can take (x-2) common.

  6. anonymous
    • 5 years ago
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    see you have to divide with x-2

  7. anonymous
    • 5 years ago
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    i have introduced -2x^2 to make with x^3 as x^2(x-2). and i have add 2x^2 to nullify it...

  8. anonymous
    • 5 years ago
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    i have written -x as -4x+3x. and 8 as -6 +14.

  9. anonymous
    • 5 years ago
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    ghhosst.............

  10. anonymous
    • 5 years ago
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    have you now understood.

  11. amistre64
    • 5 years ago
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    x^2 +2x +3 -------------- x-2 | x^3 -x +8 -x^3 +2x^2 ----------- 2x^2 -x +8 -2x^2 +4x ------------ 3x +8 -3x +6 -------- 14 <- remainder

  12. anonymous
    • 5 years ago
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    yea :) i c

  13. anonymous
    • 5 years ago
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    the way it was written just threw me off, but i see how its done now thank you :)

  14. anonymous
    • 5 years ago
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    how did yu get x^2 +2x +3 ?

  15. anonymous
    • 5 years ago
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    simply we divided

  16. anonymous
    • 5 years ago
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    nvm.. yu put the answer on the top. cool

  17. anonymous
    • 5 years ago
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    see i had x^3-x+8=x^2(x-2)+2x(x-2)+3(x-2)+14=(x^2+2x+3)(x-2)+14 so taking (x-2 common from the first three terms i have (x^2+2x +3)(x-2)

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