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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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.

Mathematics
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Ninth mein hai ya tenth mein?
9th me
Plugging @Jhannybean for "teh lulz". The sum of roots is \(\frac{2}{3}\) so the third root is \(\frac{2}{3}-\left(2 -\frac{1}{3}\right)\)

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Thus the answer is -1.
We could also have done it using the product of roots. The product of roots is \(2/3\) so the third root is\[\frac{2/3}{2\cdot (-1/3)} = \boxed{-1}\]
Oh, ninth mein... hmm. I don't think they teach that in ninth - it's taught only in tenth. Here's another method!
Yes..
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\[3(x-2)(x+1/3)(x-k) = 3x^3 - 2x^2 - 7x - 2\]
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)?
Exams..
Is that it? o_O
No. That's not it.\[3(x-2)(x+1/3)(x-k) = 3x^2 - 2x^2 - 7x - 2\]\[\Rightarrow x-k = \frac{3x^2 - 2x^2 - 7x - 2}{3(x-2)(x+1/3)}\]
\[x -k = \frac{3x^3 - 2x^2 - 7x - 2}{(x-2)(3x+1)} = \frac{3x^3 - 2x^2 - 7x - 2}{3x^2 - 5x - 2}\]
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\[x-k = x+1\]\[k = -1\]
Ok, I get it but why have you multiplied \[(x-2)(x+\left(\begin{matrix}1 \\ 3\end{matrix}\right))(x-k)\] by 3
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 \(x^3 - 2/3 x^2 - 7/3 x - 2/3\).
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.
Okay....I get it...give me a min..
All right I got it. Thanks a bunch. c:
Ussey samajh aaya?
Ab samjhaunga na usko c:
So 3 se multiply kiya taaki fraction me na rahe, right?
As a personal exercise, instead of long division, try expanding. \[3(x-2)(x+1/3)(x-k) = 3x^3 - 2x^2 - 7x - 2 \]\[=(x-2)(3x+1)(x-k) \]\[= 3x^3 + (-5 - 3k ) x^2 +(-k+6k-2)x +2k \]
Now compare coefficients:\[-5 - 3k = -2\]\[5k -2 = -7\]\[2k = -2\]All three equations tell you only one thing :)
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)\[x^2 - 3x + 2\]\[2x^2 - 6x + 4\]\[3x^2 - 9x + 6\]\[4x^2 - 12x + 8\]
do you see how the leading coefficient (coefficient of highest power) is the same as the number that is multiplied outside now?
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!
*Manmohan Singh shtyle* Theek hai?
Haha
Ok, just to be sure. Suppose 2x+1 is a factor of \(\sf 2x^2-x-1\) then other zero will be +1 ?
Yes.
Ok, thanks again..
Badiya.
*

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