Let a, b, c, d be four real numbers satisfying the system of equation a+b=8, ab+c+d = 23, ad +bc = 28 and cd = 12 .If $\mathrm{A}=\left[\begin{array}{l}\mathrm{a}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{b}\\ \mathrm{c}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{d}\end{array}\right]$ then the sum of all trace of A is

$\begin{array}{l}=\left({\mathrm{x}}^2+\mathrm{ax}+\mathrm{c}\right)\left({\mathrm{x}}^2+\mathrm{bx}+\mathrm{d}\right)\\ ={\mathrm{x}}^4+(\mathrm{a}+\mathrm{b}){\mathrm{x}}^3+(\mathrm{ab}+\mathrm{c}+\mathrm{d}){\mathrm{x}}^2+(\mathrm{ad}+\mathrm{bc})\mathrm{x}+\mathrm{cd}\\ ={\mathrm{x}}^4+8{\mathrm{x}}^3+23{\mathrm{x}}^2+28\mathrm{x}+12\\ =(\mathrm{x}+1)(\mathrm{x}+2{)}^2(\mathrm{x}+3)\\ \left({\mathrm{x}}^2+\mathrm{ax}+\mathrm{c}\right)\left({\mathrm{x}}^2+\mathrm{bx}+\mathrm{d}\right)=\left[(\mathrm{x}+1)(\mathrm{x}+3)\right]\left[\left(\mathrm{x}+2\right)\left(\mathrm{x}+2\right)\right]\\ \left({\mathrm{x}}^2+\mathrm{ax}+\mathrm{c}\right)\left({\mathrm{x}}^2+\mathrm{bx}+\mathrm{d}\right)=\left(x^2+4x+3\right)\left(x^2+4x+4\right)\\ Comparing Equations on both sides then\\ a=4 ,b=4, c=3 ,d=4\end{array}$