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Difficult Polynomial Questions
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Difficult Polynomial Questions
Last updated at May 29, 2023 by Teachoo
Question 2 Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, β7, β14 respectively. For a cubic Polynomial p(x) = ax3 + bx2 + cx + d With zeroes Ξ±, π½ and Ξ³ We have π + π½ + π = (βπ)/π π"π½" + π½π + ππ = π/π π"π½" π= (βπ )/π Let cubic polynomial be p(x) = ax3 + bx2 + cx + d Sum of zeroes Sum of zeroes = 2 (βπ)/π = 2 Assuming a = 1 (βπ)/1 = 2 b = β2 Sum of product of zeroes Sum of product of zeroes = β7 π/π = β7 Assuming a = 1 π/1 = β7 c = β7 Product of zeroes Product of zeroes = β14 (βπ)/π = β 14 π/π = 14 Assuming a = 1 π/1 = 14 d = 14 Thus, a = 1, b = β2 , c = β7, d = 14 Hence, our cubic polynomial is p(x) = ax3 + bx2 + cx + d = x3 β 2x2 β 7x + 14