Factor the polynomial x^4 + 2x^3 + 3x^2 + 2x + 1 completely.

Step-by-step solution

1. 1

   Notice the palindromic coefficients. Reading $1,\ 2,\ 3,\ 2,\ 1$ forwards and backwards gives the same list. A palindromic quartic is often the square of a palindromic quadratic, so try $\left(x^2 + bx + c\right)^2$.

2. 2

   Match coefficients. Expanding the trial square:

   $\left(x^2 + bx + c\right)^2 = x^4 + 2bx^3 + \left(b^2 + 2c\right)x^2 + 2bc\,x + c^2$

   Comparing with the target: $2b = 2$ gives $b = 1$; $c^2 = 1$ with the $x$ coefficient $2bc = 2$ forces $c = 1$; and the $x^2$ check is $b^2 + 2c = 1 + 2 = 3$. Every coefficient agrees.

3. 3

   Write the factorisation.

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

4. 4

   Show the factorisation is complete over the reals. The inner quadratic has discriminant $1 - 4 = -3 < 0$, so it is irreducible over $\mathbb{R}$ and never zero. Completing the square, $x^2 + x + 1 = \left(x + \tfrac12\right)^2 + \tfrac34 \ge \tfrac34$, so the quartic is bounded below by $\left(\tfrac34\right)^2 = \tfrac{9}{16}$ and has no real roots at all.

5. 5

   Note the complex factorisation. Over $\mathbb{C}$ the roots of $x^2 + x + 1$ are the primitive cube roots of unity $\omega = \tfrac{-1 \pm i\sqrt3}{2}$, each a double root of the quartic:

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

6. 6

   Check numerically. At $x = 2$: the quartic gives $16 + 16 + 12 + 4 + 1 = 49$, and $\left(4 + 2 + 1\right)^2 = 7^2 = 49$. At $x = -1.7$ both forms return $4.7961$.