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Question

If two zeroes of the polynomialx4-6x3-26x2+138x-35 are 2±3,find other zeroes.


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Solution

Solve for the zeroes of the given polynomial

Given x4-6x3-26x2+138x-35.

The given equation is a polynomial of degree 4,hence there will be total 4roots.

Since 2+3and 2-3are zeroes of the given polynomial f(x).

(x(2+3))(x(2-3))=0(x23)(x2+3)=0

On multiplying the above equation we get,x2-4x+1, that this is a factor of a given polynomialf(x).

Now, dividing f(x)by g(x),the quotient will also be a factor off(x) and the remainder will be0.

x2-4x+1x2-2x-35x46x3-26x2+138x-35x44x3+x2-2x3-27x2+138x-35-2x3+8x2-2x-35x2+140x-35-35x2+140x-350

So, x4-6x3-26x2+138x-35

=(x2-4x+1)(x22x35)

On factorization (x22x35)we get,

x2(75)x35

x27x+5x+35=0

x(x7)+5(x7)=0(x+5)(x7)=0

So, the zeroes are given by: x=5andx=7.

Hence, all four zeroes of the given polynomial equation are2+3,2-3,5 and7.


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