Answer:
Two real roots and two imaginary roots.
Step-by-step explanation:
Given : Function \(f(x) = x^4+12x^3+37x^2+12x+36\)
To find : State how many imaginary and real zeros the function has?
Solution :
Function \(f(x) = x^4+12x^3+37x^2+12x+36\)
We factor the given function,
\(f(x) = x^4+12x^3+36x^2+x^2+12x+36\)
\(f(x) = x^2(x^2+12x+36)+1(x^2+12x+36)\)
\(f(x) = (x^2+1)(x^2+2\cdot 6\cdot x+36)\)
\(f(x) = (x^2+1)(x+6)^2\)
The factors of the function is \(f(x)=(x^2+1)(x+6)^2\) equating the function to zero to find roots.
\((x^2+1)(x+6)^2=0\)
\((x^2+1)=0,(x+6)^2=0\)
\(x^2=-1,x+6=0\)
\(x=\sqrt{-1},x=-6\)
\(x=i,-i,x=-6\)
As the number of roots are 4 because of 4 degree polynomial.
So, The roots are x=i,-i,-6,-6.
Therefore, Two real roots and two imaginary roots.
