Respuesta :

syed514
x^4+12x^3+36x^2+x^2+12x+36 x^2(x^2+12x+36)+(x^2+12x+36) (x^2+1)(x+6)^2 
x^2+1=0 x=±i (each with multiplicity 1) 
x+6=0 x=-6 (multiplicity 2) 
the total multiplicities needs to equal the degree of the function. imaginary roots always come in pairs (conjugates) the even multiplicity of the real root means it does not cross the x-axis at that point

Answer:

Two real roots and two imaginary roots.

Step-by-step explanation:

Given : Function  [tex]f(x) = x^4+12x^3+37x^2+12x+36[/tex]

To find : State how many imaginary and real zeros the function has?

Solution :  

Function [tex]f(x) = x^4+12x^3+37x^2+12x+36[/tex]

We factor the given function,

[tex]f(x) = x^4+12x^3+36x^2+x^2+12x+36[/tex]

[tex]f(x) = x^2(x^2+12x+36)+1(x^2+12x+36)[/tex]

[tex]f(x) = (x^2+1)(x^2+2\cdot 6\cdot x+36)[/tex]

[tex]f(x) = (x^2+1)(x+6)^2[/tex]

The factors of the function is [tex]f(x)=(x^2+1)(x+6)^2[/tex] equating the function to zero to find roots.

[tex](x^2+1)(x+6)^2=0[/tex]

[tex](x^2+1)=0,(x+6)^2=0[/tex]

[tex]x^2=-1,x+6=0[/tex]

[tex]x=\sqrt{-1},x=-6[/tex]

[tex]x=i,-i,x=-6[/tex]

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.

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