SOLUTION: The solutions of x^3 – 2x^2 + 3x + 4 = 0 are a, b, and c. what is the numerical value of (a + b)(a + c)(b + c) ?
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Question 548600: The solutions of x^3 – 2x^2 + 3x + 4 = 0 are a, b, and c. what is the numerical value of (a + b)(a + c)(b + c) ?
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
By Vieta's formulas, a+b+c = 2. Hence we are evaluating (2-c)(2-b)(2-a). We could expand this entire expression (since we are guaranteed to obtain useful cyclic sums:
(2-b)(2-a) = 8 - 4a - 4b - 4c + 2ab + 2ac + 2bc - abc)
 + 2(ab + ac + bc) - abc)
However, we have a+b+c = 2, ab+ac+bc = 3, abc = 4 (you can verify these by expanding (x-a)(x-b)(x-c) and equating like terms). Hence, the expression we want is equal to
 + 2(3) - 4 = 2)
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