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 A Vanishing Polynomial Problem (Posted on 2007-01-17)
A cubic polynomial M(x) is such that M(-2)=0, and has relative extremes at x=-1 and x=1/3.

Determine M(x) given that ∫-11M(x)dx= 14/3.

 See The Solution Submitted by K Sengupta Rating: 1.5000 (2 votes)

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 Solution Comment 2 of 2 |
`Let M(x) = Ax^3 + Bx^2 + Cx + D. then`
`    M'(x) = 3Ax^2 + 2Bx + C`
`Setting M'(x) = 0 gives`
`       B     sqrt(B^2 -3AC)    - --- - ---------------- = -1      (1)       3A         3A `
`       B     sqrt(B^2 -3AC)     1    - --- + ---------------- = ---     (2)       3A         3A            3`
`Adding (1) and (2) gives`
`    B = A`
`Subtracting (1) from (2) gives`
`    A(A+C) = 0`
`Since M(x) is to be cubic,`
`    C = -A`
`Therefore,`
`    M(x) = Ax^3 + Ax^2 - Ax + D`
`Since M(-2) = 0, D = 2A. Thus,`
`    M(x) = A(x^3 + x^2 - x + 2)`
`           |- 1     14    |    --- =  |  M(x)dx      3     |          -| -1`
`           |- 1           |        = A| (x^3 + x^2 - x + 2)dx            |          -| -1`
`              x^4   x^3   x^2        | 1        = A ( --- + --- - --- + 2x ) |                4     3     2         | -1`
`             14        = A ---             3`
`Thus, A = 1 and`
`    M(x) = x^3 + x^2 - x + 2`
` `

 Posted by Bractals on 2007-01-17 12:52:16

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