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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 \(\sf 3x^32x^27x2\). Find the third zero.
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 \(\sf 3x^32x^27x2\). Find the third zero.

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ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Ninth mein hai ya tenth mein?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Plugging @Jhannybean for "teh lulz". The sum of roots is \(\frac{2}{3}\) so the third root is \(\frac{2}{3}\left(2 \frac{1}{3}\right)\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Thus the answer is 1.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3We could also have done it using the product of roots. The product of roots is \(2/3\) so the third root is\[\frac{2/3}{2\cdot (1/3)} = \boxed{1}\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Oh, ninth mein... hmm. I don't think they teach that in ninth  it's taught only in tenth. Here's another method!

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3As I recall, they teach the socalled Factor Theorem in 9th grade. It says that if \(k\) is a root of a polynomial, then \(xk\) must divide it. We know that \(2, 1/3\) are roots. Let the third root be \(k\). Then the polynomial can be written as\[3(x2)(x+1/3)(xk) = 3x^3  2x^2  7x  2\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Now comes the interesting part. Is he trying to understand the solution for himself or for his schoolexams (as he has to write in his 9th examsheet)?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3No. That's not it.\[3(x2)(x+1/3)(xk) = 3x^2  2x^2  7x  2\]\[\Rightarrow xk = \frac{3x^2  2x^2  7x  2}{3(x2)(x+1/3)}\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3\[x k = \frac{3x^3  2x^2  7x  2}{(x2)(3x+1)} = \frac{3x^3  2x^2  7x  2}{3x^2  5x  2}\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3dw:1443444259343:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3dw:1443444319443:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3\[xk = x+1\]\[k = 1\]

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0Ok, I get it but why have you multiplied \[(x2)(x+\left(\begin{matrix}1 \\ 3\end{matrix}\right))(xk)\] by 3

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Good 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 \(x^3  2/3 x^2  7/3 x  2/3\).

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3If 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 \((xa_1)(xa_2)(xa_3)\). What this does is it reduces the degree of a polynomial and smalldegree polynomials are always easier to solve.

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0Okay....I get it...give me a min..

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0All right I got it. Thanks a bunch. c:

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0Ab samjhaunga na usko c:

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0So 3 se multiply kiya taaki fraction me na rahe, right?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3As a personal exercise, instead of long division, try expanding. \[3(x2)(x+1/3)(xk) = 3x^3  2x^2  7x  2 \]\[=(x2)(3x+1)(xk) \]\[= 3x^3 + (5  3k ) x^2 +(k+6k2)x +2k \]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Now compare coefficients:\[5  3k = 2\]\[5k 2 = 7\]\[2k = 2\]All three equations tell you only one thing :)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Basically if you uniquely want to determine a polynomial function, you need to need the roots, but you ALSO need to know the leading coefficient.\[(x1)(x2)\]\[2(x1)(x2)\]\[3(x1)(x2)\]\[4(x1)(x2)\]\[\vdots \]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)\[x^2  3x + 2\]\[2x^2  6x + 4\]\[3x^2  9x + 6\]\[4x^2  12x + 8\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3do you see how the leading coefficient (coefficient of highest power) is the same as the number that is multiplied outside now?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3which is why\[(x2)(x+1/3)(xk) \]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!

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3*Manmohan Singh shtyle* Theek hai?

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0Ok, just to be sure. Suppose 2x+1 is a factor of \(\sf 2x^2x1\) then other zero will be +1 ?
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