If two zeroes of polynomial p(x) = x4 – 6×3 – 26×2 + 138x – 35 are 2 + root 3
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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