The equation whose roots are $2, 1\pm 3i$ is
- $x^{3}-4x^{2}+14x-20=0$
- $x^{3}-x^{2}+5x-14=0$
- $x^{3}+2x^{2}-3x-10=0$
- $x^{3}-3x^{2}-14x+10=0$
Reveal answer
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A
Correct answer
Explanation
Roots are 2, 1+3i, 1-3i. The quadratic factor for the complex roots is (x - (1+3i))(x - (1-3i)) = (x-1)^2 + 9 = x^2 - 2x + 10. The cubic is (x-2)(x^2 - 2x + 10) = x^3 - 2x^2 + 10x - 2x^2 + 4x - 20 = x^3 - 4x^2 + 14x - 20 = 0.
AI explanation
The required equation is given by the product of factors (x-2)(x-(1+3i))(x-(1-3i))=0. Multiply the complex factors using the difference of squares identity to get ((x-1)-3i)((x-1)+3i) = (x-1)^2 + 9 = x^2 - 2x + 10. Now multiply this quadratic by (x-2) to obtain (x-2)(x^2 - 2x + 10) = 0. Expanding this yields x^3 - 2x^2 + 10x - 2x^2 + 4x - 20 = 0, which simplifies to x^3 - 4x^2 + 14x - 20 = 0.