Doctor otter needs your help

Set $x^4-2x^3-5x^2+10x-3 = (x^2+\alpha x+\beta)(x^2+\gamma x+\delta)=x^4+(\alpha+\gamma)x^3+(\beta+\delta+\alpha\gamma)x^2+(\alpha\delta+\beta\gamma)x+\beta\delta$ . Equating coefficients produces the following system of equations:

$\alpha+\gamma=-2$ (1)

$\beta+\delta+\alpha\gamma=-5$ (2)

$\alpha\delta+\beta\gamma=10$ (3)

$\beta\delta=-3$ (4)

Assume $\alpha,\beta,\gamma,\delta$ are all integers. Looking at equation (4), there are only two possibilities for the integer pair, $\beta$ and $\delta$ , that are not ultimately equivalent: $\beta =1, \delta=-3$ and $\beta=-1, \delta=3$ .

First, I'll try $\beta =-1, \delta=3$ to see if it satisfies the other equations. Substituting these values into equation (3) gives $3\alpha-\gamma=10$ . Combining this equation with equation (1) gives $\alpha=2,\gamma=-4$ . Next, substitute these found values into equation (2) to see if it is satisfied. The result is $-1+3+(2)(-4)=-6\neq -5$ . So, the system is not satisfied with $\beta =-1, \delta=3$ .

Finally, I try $\beta =1, \delta=-3$ . Substituting these values into equation (3) gives $-3\alpha+\gamma=10$ . Combining this equation with equation (1) gives $\alpha=-3,\gamma=1$ . Then we test these values in equation (2) and see that it is satisfied. Thus, the system is solved by $\alpha=-3,\beta=1,\gamma=1,\delta=-3$ . These values lead to the following factorization for $x^4-2x^3-5x^2+10x-3$ :

$(x^2-3x+1)(x^2+x-3)$

From there, I just used the quadratic formula and a calculator to find the ordering of the roots and then, of course, used a calculator again to calculate $a^cb^d\approx 0.785$ .

$x^4-2x^3-5x^2+10x-3=0$

$(x^2-3x+1)(x^2+x-3)=0$

$x^2-3x+1=0$

$x^2+x-3=0$

The solutions for both equations are $\frac {3 \pm \sqrt{5}}{2}$ and $\frac {-1 \pm \sqrt{13}}{2}$ Then the value of $a^cb^d=\boxed{.785}$ .

Sorry if my solution is not elegant i will post another one in a few days.