- Thread starter xc630
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http://www.1728.com/cubic.htm

to find out what the answers are or go to

http://www.sosmath.com/algebra/factor/fac11/fac11.html

for a (very messy) longhand solution.

I don't know what they expected you to do. Are you sure you typed it correctly?

I believe they expect you to know a theorem for this problem.

In baby-talk, the theorem goes like this:I have to find the sum and product of the roots of 3x<sup>3</sup> + 6x<sup>2</sup> - 2x + 9 .= .0

.

Given: .x<sup>3</sup> + ax<sup>2</sup> + bx + c .= .0 . . . Note: the leading coefficient is 1.

If the three roots are p, q, r, then:

. . . . .p + q + r .= .-a . . . the sum of the roots is the negative of a.

. . pq + qr + pr .= .b

. . . . . . . . .pqr .= .-c . . . the product of the roots is the negative of c.

Your equation is: .3x<sup>3</sup> + 6x<sup>2</sup> - 2x + 0 .= .0

Divide by 3: . x<sup>3</sup> + 2x<sup>2</sup> - (2/3)x + 3 .= .0

. . . . . .Sum of roots: . -2

. . . Product of roots: . -3

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- Feb 4, 2004

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I'm sorry, but you've been provided the fully-worked solution. I'm not understanding what you still need.

Eliz.