Let \(k\) and \(m\) be real numbers, and suppose that the roots of the equation

\(x^3 - 7x^2 + kx - m = 0\)

are three distinct positive integers. Compute \(k+m\)

Let the roots be p, q and r, and show that p + q + r = 7.

If p, q and r are distinct positive integers, they have to be 1, 2 and 4.

Ohh, I see. I haven't grasped Vietas yet.

Now that I think about it, why do p+q+r=7?

If the roots are p, q and r then:

(x-p)(x-q)(x-r) = x^3 - (p+q+r)x^2 + (pq+pr+qr)x - pqr

Compare with the x^2 term of the original cubic to see that p+q+r = 7

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ohh, ok, so would we get k+m=7?

No! Look at Guest #1's reply. Set p = 1, q = 2, r = 4, then work out k and m by comparing with the x term and the constant term.

ahh, Ok I see. Got it, k+m=22. I get how to do it.