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Question

Obtain all zeroes of x4-3x3-7x2 +9x +12 iftwo of its zeroes are + V3.Sum of zeroes is

JEE/Engineering Exams
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13 are zeroes = (x++3)(x - 3) = x2 – 3 is a divisor of x4–3x3 - 7x2 +9x +12.17 x'-3x*-3x -7x +9x+12x-3x-4 4 x -3x 3x³ 4x² +9x + 12 +9x 4x 4x + 12 +12 0 q(x) = x2 – 3x-4=x2-4x+x-4 =x(x-4)+1(x-4) q(x) = (x+1)(x-4) zeroes of q(x) are-1,4 .. Remaining required zeroes are -1,4. Sum=-1+4=3
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