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sabika13

  • 2 years ago

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???

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  1. satellite73
    • 2 years ago
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    show that if you replace \(x\) by \(a\) you get 0

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

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

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

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

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

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

  8. sabika13
    • 2 years ago
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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

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

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

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

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

  13. satellite73
    • 2 years ago
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    lets go slow

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

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

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

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

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