If $a+b=8$ and $ab+c+d = 23$ and $ad+bc=28$ and $cd=12\;,$ Then $abcd$ 
If $a+b=8$ and $ab+c+d = 23$ and $ad+bc=28$ and $cd=12\;,$ Then value of
$(1)\;\; a+b+c+d=$
$(2)\;\; ab+bc+cd+da = $
$(3)\;\; abcd=$

My attempt: Let $x=a\;,b$ be the roots of $(x-a)(x-b)=0$ and $x=c\;,d$ be the roots
of $(x-c)(x-d)=0.$
So, $(x-a)(x-b)(x-c)(x-d)=\left[x^2-(a+b)x+ab\right]\cdot \left[x^2-(c+d)x+cd\right]=0.$
Thus, $$\left[x^2-8x+ab\right]\cdot \left[x^2-(c+d)x+12\right]=0.$$
Where do I go from here? Any help would be much appreciated.

This problem actually checks your observational skills of algebraic products.
Now take this:
$(x^2+ax+c)(x^2+bx+d) $
$=x^4+(a+b)x^3+(ab+c+d)x^2+(ad+bc)x+cd $
$=x^4+8x^3+23x^2+28x+12=f(x) $
$12$ is the product of roots so by putting $ x=−1 $ we find  $f(−1)=0 $
$⟹(x+1) $ is factor of $  f(x) $
Similarly  $(x+2) $ and $(x+3)$ are factors of $f(x) $
Hence  $f(x)=(x+1)×(x+2)×(x+2)×(x+3) $
This can be written in two ways:
$(x^2+3x+2)×(x^2+5x+6) $
$⟹a=3,b=5,c=2,d=6$
$(x^2+4x+3)×(x^2+4x+4) $
$⟹a=4,b=4,c=3,d=4 $
Hence $2$ sets of solutions are possible
