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Abhisar

  • one year ago

Can some one help me understand this. I need to help my little brother 2 and -1/3 are zeros of the polynomial \(\sf 3x^3-2x^2-7x-2\). Find the third zero.

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  1. ParthKohli
    • one year ago
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    Ninth mein hai ya tenth mein?

  2. Abhisar
    • one year ago
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    9th me

  3. ParthKohli
    • one year ago
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    Plugging @Jhannybean for "teh lulz". The sum of roots is \(\frac{2}{3}\) so the third root is \(\frac{2}{3}-\left(2 -\frac{1}{3}\right)\)

  4. ParthKohli
    • one year ago
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    Thus the answer is -1.

  5. ParthKohli
    • one year ago
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    We could also have done it using the product of roots. The product of roots is \(2/3\) so the third root is\[\frac{2/3}{2\cdot (-1/3)} = \boxed{-1}\]

  6. ParthKohli
    • one year ago
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    Oh, ninth mein... hmm. I don't think they teach that in ninth - it's taught only in tenth. Here's another method!

  7. Abhisar
    • one year ago
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    Yes..

  8. ParthKohli
    • one year ago
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    As I recall, they teach the so-called Factor Theorem in 9th grade. It says that if \(k\) is a root of a polynomial, then \(x-k\) must divide it. We know that \(2, -1/3\) are roots. Let the third root be \(k\). Then the polynomial can be written as\[3(x-2)(x+1/3)(x-k) = 3x^3 - 2x^2 - 7x - 2\]

  9. ParthKohli
    • one year ago
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    Now comes the interesting part. Is he trying to understand the solution for himself or for his school-exams (as he has to write in his 9th exam-sheet)?

  10. Abhisar
    • one year ago
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    Exams..

  11. Abhisar
    • one year ago
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    Is that it? o_O

  12. ParthKohli
    • one year ago
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    No. That's not it.\[3(x-2)(x+1/3)(x-k) = 3x^2 - 2x^2 - 7x - 2\]\[\Rightarrow x-k = \frac{3x^2 - 2x^2 - 7x - 2}{3(x-2)(x+1/3)}\]

  13. ParthKohli
    • one year ago
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    \[x -k = \frac{3x^3 - 2x^2 - 7x - 2}{(x-2)(3x+1)} = \frac{3x^3 - 2x^2 - 7x - 2}{3x^2 - 5x - 2}\]

  14. ParthKohli
    • one year ago
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    |dw:1443444259343:dw|

  15. ParthKohli
    • one year ago
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    |dw:1443444319443:dw|

  16. ParthKohli
    • one year ago
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    \[x-k = x+1\]\[k = -1\]

  17. Abhisar
    • one year ago
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    Ok, I get it but why have you multiplied \[(x-2)(x+\left(\begin{matrix}1 \\ 3\end{matrix}\right))(x-k)\] by 3

  18. ParthKohli
    • one year ago
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    Good question - we have to do that to get a 3 on the \(x^3\) term. Otherwise the cubic expression still has the same roots, but will be \(x^3 - 2/3 x^2 - 7/3 x - 2/3\).

  19. ParthKohli
    • one year ago
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    If you have known roots say \(a_1, a_2, a_3\) and you want to know the rest of the known roots of a given polynomial, it would make more sense to divide the polynomial by \((x-a_1)(x-a_2)(x-a_3)\). What this does is it reduces the degree of a polynomial and small-degree polynomials are always easier to solve.

  20. Abhisar
    • one year ago
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    Okay....I get it...give me a min..

  21. Abhisar
    • one year ago
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    All right I got it. Thanks a bunch. c:

  22. ParthKohli
    • one year ago
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    Ussey samajh aaya?

  23. Abhisar
    • one year ago
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    Ab samjhaunga na usko c:

  24. Abhisar
    • one year ago
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    So 3 se multiply kiya taaki fraction me na rahe, right?

  25. ParthKohli
    • one year ago
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    As a personal exercise, instead of long division, try expanding. \[3(x-2)(x+1/3)(x-k) = 3x^3 - 2x^2 - 7x - 2 \]\[=(x-2)(3x+1)(x-k) \]\[= 3x^3 + (-5 - 3k ) x^2 +(-k+6k-2)x +2k \]

  26. ParthKohli
    • one year ago
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    Now compare coefficients:\[-5 - 3k = -2\]\[5k -2 = -7\]\[2k = -2\]All three equations tell you only one thing :)

  27. ParthKohli
    • one year ago
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    Basically if you uniquely want to determine a polynomial function, you need to need the roots, but you ALSO need to know the leading coefficient.\[(x-1)(x-2)\]\[2(x-1)(x-2)\]\[3(x-1)(x-2)\]\[4(x-1)(x-2)\]\[\vdots \]all of these have roots \(1,2\). If you expand each, you'll see that the difference is the leading coefficient (actually all coefficients, but the importance of the leading coefficient is that it's the same as what has been multiplied outside)\[x^2 - 3x + 2\]\[2x^2 - 6x + 4\]\[3x^2 - 9x + 6\]\[4x^2 - 12x + 8\]

  28. ParthKohli
    • one year ago
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    do you see how the leading coefficient (coefficient of highest power) is the same as the number that is multiplied outside now?

  29. ParthKohli
    • one year ago
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    which is why\[(x-2)(x+1/3)(x-k) \]cannot equal \(3x^2 - 2x^2 +5x -2\) unless we multiply it by 3. it's because the leading coefficient and the number multiplied outside must be the same!

  30. ParthKohli
    • one year ago
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    *Manmohan Singh shtyle* Theek hai?

  31. Abhisar
    • one year ago
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    Haha

  32. Abhisar
    • one year ago
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    Ok, just to be sure. Suppose 2x+1 is a factor of \(\sf 2x^2-x-1\) then other zero will be +1 ?

  33. ParthKohli
    • one year ago
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    Yes.

  34. Abhisar
    • one year ago
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    Ok, thanks again..

  35. ParthKohli
    • one year ago
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    Badiya.

  36. Jhannybean
    • one year ago
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    *

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