Factor a Quartic Expression

Question

I'm trying to factor $$x^4+2x^3+6x^2+2x+1$$

I compare a quartic equation to the form:
$$(x^2+ax+b)(x^2+cx+d)$$
$$x^4+(a+c)x^3+(b+d+ac)x^2+(ad+bc)x+bd$$
By comparing, I get 4 equations to solve:
$$a+c=2$$ $$b+d+ac=6$$ $$ad+bc=2$$ $$bd=1$$

I picked the solutions on the first line to get that
$$x^4+2x^3+6x^2+2x+1=(x^2+(1-\sqrt3)x+1)(x^2+(1+\sqrt3)x+1)$$

I check it with the value x=7. On the left side, I get 3396. On the right side, I get 3102. That means that I must have factored it wrongly.

What went wrong?

Answer 1

Note that $a=1-\sqrt3\,i$ and that $c=1+\sqrt3\,i$. So, you should have written$$\left(x^2+\left(1-\sqrt3\,i\right)x+1\right)\left(x^2+\left(1+\sqrt3\,i\right)x+1\right)$$instead.

Answer 2

Let $$f(x)=x^4+2x^3+6x^2+2x+1$$

then

$$f'(x)=2(2x^3+3x^2+6x+1)$$

$$f''(x)=12(x^2+x+1)$$ $f''(x)=0 $ has no real root,

So, $f(x)=0$ cannot have $ 4$ real roots.

Answer 3

\begin{align}x^4+2x^3+6x^2+2x+1&= x^2\Big((x^2+{1\over x^2})+2(\underbrace{x+{1\over x}}_{t})+6\Big)\\ &= x^2\Big((t^2-2)+2t+6\Big)\\ &= x^2(t^2+2t+4)\\ &= x^2\Big((t+1)^2+3\Big)\\ &=x^2 \Big(t+1+i\sqrt{3}\Big)\Big(t+1-i\sqrt{3}\Big)\\ &=\Big(x^2+(1+i\sqrt{3})x+1\Big)\Big(x^2+(1-i\sqrt{3}))x+1\Big) \end{align}