obtain all zero of x^4-6x^3-26x^2+138x-35 if two of its zero are 2+√3/2 and 2-√3

Question

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Answer 1

I assume the two zeroes are 2+√3 and 2-√3. We multiply the two known zeroes together: (x-2-√3)(x-2+√3)=(x-2)^2-3=x^2-4x+4-3=x^2-4x+1. Now divide this quadratic into the given quartic: x^2-2x-35=(x-7)(x+5). So the other two zeroes are 7 and -5.

ALGEBRAIC LONG DIVISION

                  x^2-2x-35                        

x^2-4x+1 ) x^4-6x^3-26x^2+138x-35

             -    x^4-4x^3+   x^2

                       -2x^3-27x^2+138x

                  -     -2x^3+ 8x^2-     2x

                                -35x^2+140x-35

                           -     -35x^2+140x-35      

                                         0