Abhisar
  • Abhisar
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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ParthKohli
  • ParthKohli
Ninth mein hai ya tenth mein?
Abhisar
  • Abhisar
9th me
ParthKohli
  • ParthKohli
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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More answers

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

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