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AnilSinghBora
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If two zeroes of the polynomial x4-6×3-26×2+138x-35 are 2 ±√3, find other zeroes. Q.4

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How can i solve this question of polynomials Is it the hard question of class 10th, please suggest me the simplest way of solving this question. If two zeroes of the polynomial x4-6×3-26×2+138x-35 are 2 ±√3, find other zeroes.

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  1. Since this is a polynomial equation of degree 4, hence there will be total 4 roots.

    Let f(x) = x4-6x3-26x2+138x-35

    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,

    x2-4x+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.

    Ncert solutions class 10 chapter 2-10

    So, x4-6x3-26x2+138x-35 = (x2-4x+1)(x2 –2x−35)

    Now, on further factorizing (x2–2x−35) we get,

    x2–(7−5)x −35 = x2– 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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