Obtain all the zeroes of the polynomial X^4 - 3√2 x³ + 3x² +3 √2x - 4 = 0 , if two of it's zeroes are √2 and 2√2 ... Guys , Solve out !!!

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Let me try.....
is x in root here 3 √2x
it is not possible to solve bcoz it can't be solved here
but if u say i can solve it on paper and upload the file
Yeah I have also solved one question the same way and uploaded.You can do so

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2015-10-01T21:29:00+05:30

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P(x) =  x⁴ - 3 √2  x³ + 3 x² + 3√2  x - 4

given that (x - √2) , (x - 2√2) are factors of P(x) as √2 and 2√2 are two zeroes of P(x) = 0.
 
(x -√2) (x - 2√2) =  x²  -3√2 x + 4  is a factor of P(x).  let A be a constant.  We can write the constant term in the second factor by :  -4/4 = -1... dividing the constant terms.

let  (x² - 3√2 x + 4) (x² + A x -1 )  =  P(x)

Now compare the coefficients of x³ :  A - 3√2 = -3√2    =>  A = 0
             coefficient of  x :  4A + 3√2 = 3√2        => A = 0

so  the other factors  are :  x² - 1 = 0
       so  x = 1  and -1 are the other factors.

This method of multiplying the factors and comparing coefficients is simple.
2 5 2
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There is a simple method. I overlooked....
sum of all four roots of P(x) is the - of coefficient of x cube. so it is 3√2. The sum of given two roots is 3 √2. So sum of two more roots is 0.
Now product of four roots is the constant term = -4... product of two given roots is 4. so product of two more roots is -4/4 = -1.....
Hence the polynomial is simple : x² - (0) x + (-1)
roots of this are +1 and -1.