Factoring the quartic $\left(x^{2}+x-1\right)\left(x^{2}+2x-1\right)-2sx\left(2x-1\right)^{2}$ Define $Q{\left(s;x\right)}$ to be the quartic function of $x$ with real parameter $s$ such that $0\le s\le1$ given as
$$Q{\left(s;x\right)}=\left(x^{2}+x-1\right)\left(x^{2}+2x-1\right)-2sx\left(2x-1\right)^{2}.\tag{1}$$

Question: Is there a (relatively) straightforward way of factoring $Q{\left(s;x\right)}$ into a product of two quadratics with real parameters?

In principle, this problem could be solved by brute force using the general methods for finding roots of quartic functions such as the Ferrari method described in the quartic function wiki.
However, I'm having a terrible time actually carrying out the process for my particular quartic $Q$. I'd really like to know if there is a more efficient approach or any cute shortcuts for avoiding some of the cumbersome intermediate expressions.
Note: for reference,the discriminant of $Q$ is
$$\begin{align}
\Delta_{Q}
&=-8\left(s-1\right)\left(128s^{4}-602s^{3}+982s^{2}-508s+5\right).\tag{2}\\
\end{align}$$

Only the start of some thoughts...
$Q{\left(s;x\right)}=\left(x^{2}+x-1\right)\left(x^{2}+2x-1\right)-2sx\left(2x-1\right)^{2}$
$Q{\left(s;x\right)}=x^4 + 2x^3-x^2+x^3+2x^2-x-x^2-2x+1-8sx^3+8sx^2-2sx$
$Q{\left(s;x\right)}=x^4 + (3-8s)x^3+8sx^2+(-3-2s)x+1$
Consider $Q{\left(s;x\right)}=(x^2+ax+p)(x^2+bx+\frac 1p)$
$Q{\left(s;x\right)}=x^4+bx^3+\frac 1p x^2 + ax^3+abx^2+ \frac ap x+px^2+bpx+ 1$
$Q{\left(s;x\right)}=x^4+(a+b)x^3+(p+\frac 1p+ab)x^2+ (\frac ap +bp)x+ 1$
Comparing coefficients:
$(1): a+b=3-8s$
$(2): p+\frac 1p+ab=8s$
$(3): \frac ap +bp=-3-2s$
$(2) + 4 \times (3) \Rightarrow p+\frac 1p+ab+4(\frac ap +bp)=-12 \Rightarrow (1+4b)p^2+(ab+12)p+ (4a+1)=0$
$(1) + (2) \Rightarrow a+b+p+\frac 1p+ab=3 \Rightarrow p^2+(a+ab+b-3)p+1=0$
$\Rightarrow (1+4b)p^2+(1+4b)(a+ab+b-3)p+(1+4b)=0$
Then $1+4b=1+4a \Rightarrow a=b$
And $(1+4b)(a+ab+b-3)=ab+12$
$(1+4a)(2a+a^2-3)=a^2+12$
$2a+a^2-3+8a^2+4a^3-12a=a^2+12$
$4a^3+8a^2+10a-15=0$
