How to solve the following system of equations
$$ \begin{cases} a+c=12\\ b+ac+d=86\\ bc+ad=300\\ bd=625\\ \end{cases} $$
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Sign up to join this communityHow to solve the following system of equations
$$ \begin{cases} a+c=12\\ b+ac+d=86\\ bc+ad=300\\ bd=625\\ \end{cases} $$
This system of equations corresponds to a factorization of $$f(x)=x^4+12x^3+86x^2+300x+625$$ into two quadratics as $$f(x)=(x^2+ax+b)(x^2+cx+d)$$ The roots of $f(x)$ are $-3-4i$ and $-3+4i$, each having multiplicity $2$. So what should the quadratic factors of $f(x)$ look like?
Well, $b=\cfrac {625}{d}$ from your last equation, and $a=-c+12$ from your first equation.
Substitute $b$ and $a$ into your second equation to get $\cfrac{625}{d}+(-c+12)(c)+d=86$
Do the same for your third equation.