Find the cubic polynomial with integer coefficients whose roots are twice the roots of x^3 - 2x^2 + 3x - 1 = 0.

Answer and explanation

Correct answer: A. x^3 - 4x^2 + 12x - 8 = 0

  1. Step 1: If original polynomial has roots r, the new polynomial has roots 2r. Substitute y = 2x => x = y/2. Step 2. (y/2)^3 - 2(y/2)^2 + 3(y/2) - 1 = 0 => y^3/8 - 2y^2/4 + 3y/2 - 1 = 0. Multiply by 8: y^3 - 4y^2 + 12y - 8 = 0. Answer: x^3 - 4x^2 + 12x - 8 = 0. Why not the others: Transforming roots by a factor k requires replacing x with x/k.

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