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sabika13 Group Title

Prove that (x-a) is a factor of x^3 - (a + b + c)x^2 + (ab +bc + ca)x-abc. How do I prove this???

  • 2 years ago
  • 2 years ago

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  1. satellite73 Group Title
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    show that if you replace \(x\) by \(a\) you get 0

    • 2 years ago
  2. satellite73 Group Title
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    then you know if \(a\) is a root of a polynomial \(p(x)\) then you can factor as \(p(x)=(x-a)(q(x))\)

    • 2 years ago
  3. sabika13 Group Title
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    yeah i tried that, but i couldnt end up with a zero..

    • 2 years ago
  4. satellite73 Group Title
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    if you do not get 0, then you cannot factor

    • 2 years ago
  5. sabika13 Group Title
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    i probably made a mistake.. because it has to equal to zero

    • 2 years ago
  6. satellite73 Group Title
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    \[a^3 - (a + b + c)a^2 + (ab +bc + ca)a-abc\] is a startr

    • 2 years ago
  7. satellite73 Group Title
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    yeah it works, try again

    • 2 years ago
  8. sabika13 Group Title
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    a^3 - (a + b +c)a^2 + (ab+ bc +ca)a -abc = a^3 - a^3 + a^3b + a^3 c + a^2b + bc+ca^ - abc

    • 2 years ago
  9. sabika13 Group Title
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    ohhh i see it, i messed up

    • 2 years ago
  10. satellite73 Group Title
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    hmmm forgot the distributive property for the second term

    • 2 years ago
  11. satellite73 Group Title
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    and some other mistakes too, but you can clean it up i am sure

    • 2 years ago
  12. sabika13 Group Title
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    okay i tried and i ended up like this: 2a^2 + 2a^2c

    • 2 years ago
  13. satellite73 Group Title
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    lets go slow

    • 2 years ago
  14. sabika13 Group Title
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    ohh its negative.. i got it

    • 2 years ago
  15. satellite73 Group Title
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    first step \[a^3 - (a + b + c)a^2 + (ab +bc + ca)a-abc\] second step ok

    • 2 years ago
  16. sabika13 Group Title
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    yeah i didnt see the negative sign infront of the first bracket, thank you so much!!

    • 2 years ago
  17. satellite73 Group Title
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    yw (you did all the work)

    • 2 years ago
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