If two zeroes of the polynomial x^46x^326x^2+138x35 are 2 ±√3, find other zeroes.
If two zeroes of the polynomial x^{4}6x^{3}26x^{2}+138x35 are 2 ±√3, find other zeroes.
Since this is a polynomial equation of degree 4, hence there will be total 4 roots.
Let f(x) = x^{4}6x^{3}26x^{2}+138x35
Since 2 +√3 and 2√3 are zeroes of given polynomial f(x).
∴ [x−(2+√3)] [x−(2√3)] = 0
(x−2−√3)(x−2+√3) = 0
On multiplying the above equation we get,
x^{2}4x+1, this is a factor of a given polynomial f(x).
Now, if we will divide f(x) by g(x), the quotient will also be a factor of f(x) and the remainder will be 0.
So, x^{4}6x^{3}26x^{2}+138x35 = (x^{2}4x+1)(x^{2} –2x−35)
Now, on further factorizing (x^{2}–2x−35) we get,
x^{2}–(7−5)x −35 = x^{2}– 7x+5x+35 = 0
x(x −7)+5(x−7) = 0
(x+5)(x−7) = 0
So, its zeroes are given by:
x= −5 and x = 7.
Therefore, all four zeroes of given polynomial equation are: 2+√3 , 2√3, −5 and 7.

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