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# Polynomials

Asked by hemant2020 25th March 2010, 6:21 PM

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 28th March 2010, 7:46 AM
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