Factor (x^2 - 6x + 1)(x^2 + x + 1) + 12x^2 completely.

Step-by-step solution

1. 1

   Expand the product term by term. Distributing each part of the first bracket:

   $x^{2}\left(x^{2}+x+1\right)=x^{4}+x^{3}+x^{2},\quad -6x\left(x^{2}+x+1\right)=-6x^{3}-6x^{2}-6x,\quad 1\left(x^{2}+x+1\right)=x^{2}+x+1.$

   Adding and collecting like terms gives $x^{4}-5x^{3}-4x^{2}-5x+1$.

2. 2

   Add the $12x^{2}$. Only the quadratic coefficient changes:

   $x^{4}-5x^{3}-4x^{2}-5x+1+12x^{2}=x^{4}-5x^{3}+8x^{2}-5x+1.$

   The extra $12x^{2}$ is clearly engineered — it is what turns an unremarkable quartic into a factorable one.

3. 3

   Notice the palindromic coefficients. The coefficients read $1,-5,8,-5,1$, the same forwards and backwards. Such a reciprocal polynomial always factors (when it factors at all) in the shape

   $\left(x^{2}+ax+1\right)\left(x^{2}+bx+1\right)=x^{4}+(a+b)x^{3}+(ab+2)x^{2}+(a+b)x+1,$

   whose own coefficient list is automatically palindromic. Spotting this saves a blind search over all possible factorisations.

4. 4

   Match coefficients and solve. Comparing with $1,-5,8,-5,1$:

   $a+b=-5,\qquad ab+2=8\ \Rightarrow\ ab=6.$

   The numbers with sum $-5$ and product $6$ are $a=-2$ and $b=-3$, so

   $x^{4}-5x^{3}+8x^{2}-5x+1=\left(x^{2}-2x+1\right)\left(x^{2}-3x+1\right).$

5. 5

   Factor the remaining quadratic and finish. The first factor is a perfect square, $x^{2}-2x+1=(x-1)^{2}$, while $x^{2}-3x+1$ has discriminant $9-4=5$, so it has no rational factorisation (its roots are $\frac{3\pm\sqrt5}{2}$, the golden-ratio pair). The complete factorisation over the rationals is

   $\left(x^{2}-6x+1\right)\left(x^{2}+x+1\right)+12x^{2}=(x-1)^{2}\left(x^{2}-3x+1\right).$

   Numerical check at $x=2.1$: the original gives $-1.0769$ and the factored form gives $-1.0769$ ✓, with two further random values agreeing to nine digits.