Compute the following without the calculator 
$$4\left(5+3\sqrt2\over 2\right)^4-16\left(5+3\sqrt2\over 2\right)^3-17\left(5+3\sqrt2\over 2\right)^2+27\left(5+3\sqrt2\over 2\right)-3$$

Please solve the following equation without using calculator.
Substituting $\left(5+3\sqrt2\over 2\right)$ to x must be the first step, but then I don't know how to factor it.
 A: Let $f(x)=4x^4-16x^3-17x^2+27x-3$. Then $g(x)=f(x)-4=(4x^2 - 20x + 7)(x^2 + x - 1)$.
Since $a=(5+3\sqrt{2})/2$ satisfies $4a^2-20a+7=0$ we know that $g(a)=0$ and hence
$f(a)=4$. So the answer is $4$.
A: Let $\alpha=\displaystyle\frac{5+\sqrt{3}}2, \beta=\displaystyle\frac{5-\sqrt{3}}2$.
$$\alpha+\beta=5; \alpha\beta=\frac 74$$
$\alpha,\beta$ are roots of the quadratic 
$$x^2-(\alpha+\beta)+\alpha\beta=0\\
x^2-5x+\frac 74=0\\
4x^2-20x+7=0$$
i.e. 
$$\overbrace{4\alpha^2-20\alpha+7}^{f(\alpha)}=0$$
Now consider the following:
$$\begin{array}
&&&&&\\
\alpha^2\cdot f(\alpha):&4\alpha^4&-2 0\alpha^3   &+7\alpha^2  &         &  &=0\\
\alpha\cdot f(\alpha):  &         &\;\;\;4\alpha^3&-20\alpha^2 &+7\alpha &  &=0\\
- f(\alpha):            &         &               &-4\alpha^2  &+20\alpha&-7&=0\\
+4:                     &         &               &            &         &+4&=4\\
\hline
\text{Adding}:          &4\alpha^4&-16\alpha^3    &-17\alpha^2 &+27\alpha &-3&=\color{red}4\\
\hline
\end{array}$$
A: We know that $$4x^4-16x^3-17x^2+27x-7 =(4x^2-20x+7)(x^2+x-1) $$ $$=(x- \frac{5+3\sqrt {2}}{2})(x-\frac {5-3\sqrt {2}}{2})(x-\frac {-1 +\sqrt {5}}{2})(x+\frac {-1-\sqrt {5}}{2}) $$
Observe that $\frac {5+3\sqrt {2}}{2} $ is a root of this polynomial. Also observe that the above polynomial is $4$ less than the required polynomial in question. Thus, the answer is $4$. Hope it helps. 
