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I have the following simple problem:

Consider the polynom P(x) = x³ + ax² + bx + c of real coefficients. Knowing that...

  • -1 and (1 + $\alpha$) are roots of P(x) = 0
  • $\alpha > 0$
  • the rest of the division of p(x) by (x-1) is 8

find the value of $\alpha$.

Now, I understand (1 - $\alpha$) is a root too and I can solve it by writing: (x+1)(x - 1 - $\alpha$)(x - 1 + $\alpha$) = 0

Since P(1) = 8,

$ (1+1)(1 - 1 - \alpha)(1 - 1 + \alpha) = 8 \Leftrightarrow 2(\alpha)(\alpha) = 8$

$\alpha$ = 2

My real doubt reside on why the following method does not work:

  • Since -1 is root, a - b + c - 1 = 0
  • Since (1 + $\alpha$) is root, a(1 + $\alpha$)² + b(1 + $\alpha$) + c + (1 + $\alpha$)³ = 0
  • From both equations, a - b + c - 1 = a(1 + $\alpha$)² + b(1 + $\alpha$) + c + (1 + $\alpha$)³

Comparing the coefficients, (1 + $\alpha$) = -1 and $\alpha$ = 2i follows.

Vittorio D'Angelo
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1 Answers1

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The method of equating coefficient with a polynomial is based on the fact that the equation works for all values of x.

Now the "comparing coefficient" you did on

"From both equations, $a - b + c - 1 = a(1 + α)² + b(1 + α) + c + (1 + α)³$"

doesn't work because $a, b, c$ are fixed (becuase of the given constraints), you can not plug in different values to it like you would do with a polynomial in $x$

More water plz
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