Multiple choice

The cubic equation whose roots are each 2 more than the roots of the equation ${x^3} + 4{x^2} - 3x + 12 = 0$ is

  1. ${x^3} + 2{x^2} - 31x - 2 = 0$
  2. ${x^3} + 6{x^2} - x + 14 = 0$
  3. ${x^3} - 2{x^2} - 7x + 26 = 0$
  4. ${x^3} + 2{x^2} - 5x + 10 = 0$
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C Correct answer
Explanation

If the roots of the new equation are x = y + 2, then y = x - 2. Substituting (x - 2) for x in the original equation (x^3 + 4x^2 - 3x + 12 = 0) gives (x - 2)^3 + 4(x - 2)^2 - 3(x - 2) + 12 = 0. Expanding this yields x^3 - 6x^2 + 12x - 8 + 4(x^2 - 4x + 4) - 3x + 6 + 12 = x^3 - 2x^2 - 7x + 26 = 0.

AI explanation

To increase the roots of the original polynomial by 2, we substitute (x - 2) for every x in the given equation. This gives the new equation (x - 2)^3 + 4(x - 2)^2 - 3(x - 2) + 12 = 0. Expanding the terms results in x^3 - 6x^2 + 12x - 8 + 4x^2 - 16x + 16 - 3x + 6 + 12 = 0. Combining like terms yields the final cubic equation x^3 - 2x^2 - 7x + 26 = 0.

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