Solve x⁸ − 2x⁷ + 3x⁶ − 4x⁵ + 5x⁴ − 4x³ + 3x² − 2x + 2 = 0.

Step-by-step solution

1. 1

   Read the coefficient pattern. The coefficients are $1, -2, 3, -4, 5, -4, 3, -2, 2$ — palindromic apart from the last entry, which is $2$ where symmetry would want $1$. A palindromic degree-$8$ pattern like $1, -2, 3, -4, 5, -4, 3, -2, 1$ is the signature of a squared quartic, so isolate that piece first.

2. 2

   Identify the square. Squaring the alternating quartic gives exactly the symmetric pattern:

   $\left(x^4 - x^3 + x^2 - x + 1\right)^2 = x^8 - 2x^7 + 3x^6 - 4x^5 + 5x^4 - 4x^3 + 3x^2 - 2x + 1$

   Comparing term by term with the given polynomial, every coefficient matches except the constant, which is larger by exactly $1$.

3. 3

   Rewrite the whole polynomial. Therefore

   $P(x) = \left(x^4 - x^3 + x^2 - x + 1\right)^2 + 1$

   This identity is the entire solution — it makes the question about real roots trivial and gives the complex ones cleanly.

4. 4

   Conclude that there are no real solutions. For real $x$ the square is non-negative, so

   $P(x) \ge 0 + 1 = 1 > 0$

   for every real $x$. The equation $P(x) = 0$ therefore has no real solutions. Numerically the minimum of $P$ on the reals is about $1.4537$, attained near $x \approx 0.6058$ — comfortably above zero, as the identity predicts.

5. 5

   Describe the complex roots. Over $\mathbb{C}$ the equation becomes

   $\left(x^4 - x^3 + x^2 - x + 1\right)^2 = -1 \ \Longleftrightarrow \ x^4 - x^3 + x^2 - x + 1 = \pm i$

   so the eight roots split into two quartics. (The quartic on the left is the cyclotomic polynomial $\Phi_{10}$, equal to $\frac{x^5+1}{x+1}$.) Numerically the roots are

   $x \approx -0.4661 \pm 0.7670i, \quad -0.2913 \pm 1.1399i, \quad 0.6631 \pm 0.8872i, \quad 1.0943 \pm 0.5143i$

   all genuinely non-real, in four conjugate pairs as required for a real polynomial.

6. 6

   Note what does not work. The polynomial has no rational roots, and an exhaustive search shows it does not factor over the integers into a quadratic times a sextic or into two quartics. So the square-plus-one identity is not one route among many — it is the route.