Multiple choice

If there is a multiple root of order $3$ for the equation ${x}^{4}-2{x}^{3}+2x-a=0$, then the other root is

  1. $-1$
  2. $0$
  3. $1$
  4. $2$
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A Correct answer
Explanation

Let f(x) = x^4 - 2x^3 + 2x - a. If there is a root of order 3, then f(x) = (x-r)^3 * (x-k). Expanding this: (x^3 - 3x^2r + 3xr^2 - r^3)(x-k) = x^4 - (k+3r)x^3 + (3r^2 + 3rk)x^2 - (r^3 + 3r^2k)x + kr^3. Comparing coefficients with x^4 - 2x^3 + 0x^2 + 2x - a: k+3r = 2, 3r^2 + 3rk = 0, r^3 + 3r^2k = -2. From 3r^2 + 3rk = 0, r(r+k) = 0. If r=0, then k=2, but then the x term is 0, not 2. If r+k=0, k=-r. Then k+3r = 2 becomes -r+3r=2, so 2r=2, r=1. Then k=-1. The root of order 3 is 1, and the other root is -1.

AI explanation

If x = a is a root of multiplicity 3, then f(x) = x^4 - 2x^3 + 2x - a can be written as (x-a)^3(x-b). Equating the x^3 coefficients gives -3a - b = -2, and equating the x^2 coefficients gives 3a^2 = 0. Since a must be 0, substituting it into the first equation yields b = 2, meaning the other root is 2.

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