To find the zeros of the function defined by f(x) = 4x^4 + 4x^3 + 3x^2, we first set the function equal to zero:
4x^4 + 4x^3 + 3x^2 = 0
Next, we can factor out the common term, which is x^2:
x^2(4x^2 + 4x + 3) = 0
This gives us one zero immediately, which is x = 0. The multiplicity of this zero is 2 because we factored out x^2.
Next, we need to solve the quadratic equation 4x^2 + 4x + 3 = 0. To determine the nature of the roots, we use the discriminant:
D = b^2 – 4ac = (4)^2 – 4(4)(3) = 16 – 48 = -32
Since the discriminant is negative (D < 0), this quadratic has no real solutions, meaning no additional real zeros exist for our function.
In summary, the function has one real zero at x = 0 with a multiplicity of 2. There are no other real zeros for this function.