Find p so that the roots of x^3 + 2px^2 - px + 10 = 0 ......
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Find p so that the roots of x^3 + 2px^2 - px + 10 = 0 ......

[From: ] [author: ] [Date: 12-09-11] [Hit: ]
the desired equation is,[x - (-1)](x - 2)(x - 5) = 0 ==> x^3 - 6x^2 + 3x + 10 = 0,and p = -3.I hope this helps!......
Find p so that the roots of x^3 + 2px^2 - px + 10 = 0 are integers and are in arithmetic progression.

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We are given that the roots are integers and are, hence, rational.

By the Rational Root Theorem, any rational root to x^3 + 2px^2 - px + 10 = 0 must be a factor of 10 (±1, ±2, ±5, ±10). The only three-root combination of these eight whose roots are in AP is:
x = -1, x = 2, and x = 5 (AP with a first term of -1 and common difference of 3).

So, the desired equation is, by the Factor Theorem:
[x - (-1)](x - 2)(x - 5) = 0 ==> x^3 - 6x^2 + 3x + 10 = 0,

and p = -3.

I hope this helps!
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