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Polynomial Question! (1 Viewer)

shmickmick

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hey just wondering if anyone could help me with a question as im the only guy doing it in my skool..............

for P(x) 2x^3 - x^2 - 6x +3

how do you find that x= 1/2 for the trial and error method i thought it could only be a factor of 3 and 2 eg 2/3 ,3/2, 1 etc

thanks........mick
 

_ShiFTy_

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You couldve have also factorised that equation

2x^3 - x^2 - 6x +3
x^2(2x - 1) - 3(2x - 1)
(2x - 1)(x^2 - 3)


shmickmick said:
how do you find that x= 1/2 for the trial and error method i thought it could only be a factor of 3 and 2 eg 2/3 ,3/2, 1 etc
k
Factor R: -1, 1, -3, 3
Factor S: 1, 2

1/2 is a possible solution...
 

shmickmick

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ok cheers that explains it ...............pretty obvious now lol !
 

YBK

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Shady01 said:
what if you can't factoirse this the way shifty has done? how can we find 1/2????
2x^3 - x^2 - 6x +3

that means there are several possibilities for factors

take for example (2x + 3) = 0

x = -3/2


Now you can see that the numerator (top) is a factor of the constant, and the denominator is a factor of the x's coeficient(bottom).

Therefore, we can apply this to the main problem. The numerator has to be a factor of 3, and 1 is a factor of 3 and the dinominator has to be a factor of 2, and 2 is a factor of 2.

hope that made sense :)

edit: fixed typo :D
 
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davin

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the basic form is that if you have some function Ax^n + (other lower powers of x) + B where A and B are just integers, all the possible factors you need to try are factors of B over factors of A, both positive and negative values.

So, for example, for:
6x^4 + 3x^3 - 5x^2 +x -8
your number for A is 6 and your number for B is -8
the factors of B are +1, -1, +2, -2, +4, -4, +8, -8
the factors of A are +1, -1, +2, -2, +3, -3, +6, -6

the possible roots, then, are all the possible combinations of B over A, in this example:
+1, -1, +1/2, -1/2, +1/3, -1/3, +1/6, -1/6, +2, -2, +2/3, -2/3, +4, -4, +4/3, -4/3, +8, -8, +8/3, -8/3
 

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