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Write an equation each of whose roots are 2 less than 3 times the roots of 3x^3 + 10x^2 + 7x - 10 = 0
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This problem's solution is to apply the Vieta's theorem several times.
Let a, b and c be the roots of the given equation;
let u, v and w be the roots of the projected equation.
According to the condition, we have
u = 3a-2, v = 3b-2, w = 3c-2.
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| Let the projected equation be px^3 + qx^2 + rx + s = 0. |
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| Our goal is to determine the coefficients p, q, r and s. |
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According to Vieta's theorem, a + b + c = .
Hence, u + v + w = (3a-2) + (3b-2) + (3c-2) = 3*(a+b+c) - 6 = = -10 - 6 = -16.
Thus = 16, according to Vieta's theorem.
Next, according to Vieta's theorem, ab + ac + bc = .
Hence, uv + uw + vw = (3a-2)*(3b-2) + (3a-2)*(3c-2) + (3b-2)*(3c-2) = 9(ab + ac + bc) - (6a + 6b + 6a + 6c + 6b + 6c) + (4+4+4) =
= 9(ab + ac + bc) - 12(a+b+c) + 12 = = 21 + 40 + 12 = 73.
Thus = 73, according to Vieta's theorem.
Finally, according to Vieta's theorem, abc = {{10/3}}}.
Hence, uvw = (3a-2)*(3b-2)*(3c-2) = 27abc - 18(ab + ac + bc) + 12(a + b + c) - 8 =
= = 9*10 - 6*7 - 4*10 - 8 = 0.
Thus = 0, according to Vieta's theorem.
Since the projected polynomial coefficients ratios are integer numbers, we can take p = 1.
It gives q = 16, r = 73, s = 0.
Thus the projected equation is x^3 + 16x^2 + 73x = 0. ANSWER
Solved.
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The post-solution note
The solution by @MathLover1 works due to that happy occasion, that the given equation has a rational root.
My approach / (my solution) works independently of this occasion.
With my approach, there is no need in finding the roots of the given polynomial.