If two zeroes of the polynomial x4 – 26x3 -26x2 + 138x –3

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354. If two zeroes of the polynomial x4 – 26x-26x2 + 138x –35 are 2 ± square root of 3, find other two zeroes.


It is given that 2 plus square root of 3 space and space 2 minus square root of 3 are two zeroes
∴      open curly brackets straight x minus left parenthesis 2 plus square root of 3 right parenthesis close curly brackets space space open curly brackets straight x minus left parenthesis 2 minus square root of 3 right parenthesis close curly brackets 
                               equals left parenthesis straight x minus 2 minus square root of 3 right parenthesis space left parenthesis straight x minus 2 plus square root of 3 right parenthesis
                                equals left parenthesis straight x minus 2 right parenthesis squared minus left parenthesis square root of 3 right parenthesis squared
                                equals straight x squared minus 4 straight x plus 1 is a factor of f(x).
Let us now divide f(x) by x2 – 4x + 1.
We have,

It is given that  are two zeroes∴                    
Hence, other two zeroes of f(x) are the zeroes of the polynomial x– 2x – 35, we have
x2 – 2x – 35 = x2–1x + 5x – 35
= (x – 7) (x + 5)
Hence, other two zeroes of f(x) are 7 and -5.


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