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# Polynomials. Question from RD Sharma

Asked by AkshayPathak12 14th April 2010, 1:22 PM

The given expression can't be a polynomial, Please use proper + or - signs instead of =.

However you can refer to a similarly answered question, reproduced below,

 Query: Polynomials Query Description: if the zeros of the polynomial f(x)= ax^3 + 3bx^2 +3cx + d are in AP prove that 2b^3 -3abc + da^2 = 0 Answer: Let α, β and γ be the roots or zeros, of the polynomial, ax3 +3bx2 +3cx + d then we have α + β + γ = -3b/a αβ+ βγ + αγ = 3c/a αβγ = -d/a, Since the roots are in AP, let they be, A-D, A and A+D, where A is a term and D is common difference, then α + β + γ = -3b/a A-D + A + A+D = -3b/a 3A = -3b/a A = -b/a  ... (1) Next, αβ+ βγ + αγ = 3c/a (A-D)A + A(A+D) + (A-D)(A+D) = 3c/a A2 - AD + A2 + AD + A2 - D2 = 3c/a 3A2 - D2 = 3c/a D2 = 3A2 - c/a = 3b2/a2 - 3c/a            ... from (1) αβγ = -d/a,              (A-D)A(A+D) = -d/a A(A2-D2) = -d/a Put for A and D2, (-b/a)((b2/a2 - (3b2/a2 - 3c/a)) = -d/a (b2/a2 - (3b2/a2 - 3c/a)) = d/b (b2 - (3b2 - 3ac)) = a2d/b (b3 - (3b3 - 3abc)) = a2d (-2b3 + 3abc) = a2d 2b3 - 3abc + a2d = 0 Regards, Team, TopperLearning.
Answered by Expert 14th April 2010, 3:39 PM
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