If two zeroes of the polynomial x⁴-6x³-26x²+138x-35 are 2+-√3. find other zeroes.

If two zeroes of the polynomial x⁴-6x³-26x²+138x-35 are 2+-√3. f

1.	If two zeroes of the polynomial x⁴-6x³-26x²+138x-35 are 2+-√3. find other zeroes.
Answer» Two zeros are {tex}2\\pm\\sqrt3{/tex}Sum of Zeroes\xa0{tex}2 + \\sqrt { 3 } + 2 - \\sqrt { 3 } = 4{/tex}and product of zeroes =\xa0{tex}( 2 + \\sqrt { 3 } ) ( 2 - \\sqrt { 3 } ) = 4 - 3 = 1{/tex}Hence quadratic polynomial formed out of this will be a factor of given polynomial,So, x2 - (sum of zeroes)x + product of zeroes= x2 - 4x + 1 will be a factor of given polynomial,Divide given polynomial with x2 - 4x + 1 to get other zeroes.Now,x2 -2x - 35= x2 - 7x + 5x - 35= x(x - 7) + 5(x - 7)= (x - 5) (x - 7){tex}\\therefore {/tex}\xa0Zeros arex = 7 and x = -5{tex}\\therefore {/tex} Other two zeros are 7 and -5\xa0 Let the remaining two zeros be X and Y,As,Sum of roots =6 , product of roots = -35We get 2 equations,2+√3 + 2-√3 + x +y = 6x+y= 2-----(1)(2+√3)(2-√3)(xy) = -35xy=-35x=-35/y -----(2)Subs (2) in (1)y-35/y =2y²-2y-35=0y=[2±√(144)]/2y=7 y=-5x=-5 x=7Hence the roots are (2±√3),(-5),(7) .