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DeadShot

  • 2 years ago

Determine the zeros of f(x) = x3 - 12x2 + 28x - 9.

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  1. RONNCC
    • 2 years ago
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    try factoring :)

  2. DeadShot
    • 2 years ago
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    so, I'd remove the x^2 from the first two terms, making it x^2 (x - 12 + 28x - 9)?

  3. DeadShot
    • 2 years ago
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    and then combine like terms, making it x^2 (29x - 21)

  4. RONNCC
    • 2 years ago
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    ... you can't just remove x^2 from the front. what about the back? you get x^2(x - 12) + 28x - 9

  5. ZakaullahUET
    • 2 years ago
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    start by hit and tril, put x=1, -1, 2, -2 till u get f(x)=0

  6. DeadShot
    • 2 years ago
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    ok

  7. ZakaullahUET
    • 2 years ago
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    |dw:1359226096145:dw|

  8. ZakaullahUET
    • 2 years ago
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    |dw:1359226195996:dw|

  9. ZakaullahUET
    • 2 years ago
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    try ur self till u get f(x)=0 after that v proceed next

  10. DeadShot
    • 2 years ago
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    ok

  11. ZakaullahUET
    • 2 years ago
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    let "a" is that value i-e f(a)=0 means "a" is a root and x-a is a factor, now divide the original equation by x-a to get another factor(that will b quadratic, factorize that and equate the three factors one by one to get the roots,

  12. ZakaullahUET
    • 2 years ago
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    my laptop battery is low , m about to offline if i didnt reach power on time, good luck for the question

  13. DeadShot
    • 2 years ago
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    I couldn't find any numbers that made f(x)=0

  14. RONNCC
    • 2 years ago
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    ok sigh i don't know what he was doing

  15. RONNCC
    • 2 years ago
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    but via factoring you get (x-9)(x^2-3x+1)

  16. RONNCC
    • 2 years ago
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    therefore x = 9 and you use the quadratic formula to find the roots for x^2 - 3x + 1

  17. DeadShot
    • 2 years ago
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    I'm having trouble using the quadratic formula with (x^2 - 3x + 1)

  18. satellite73
    • 2 years ago
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    \[a=1,b=-3,c=1\] \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

  19. DeadShot
    • 2 years ago
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    so, a=1, b=-3, and c=1?

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