State how many imaginary and real zeros the function has. f(x) = x³ + 5x² + x + 5

Solution:

Given: Function f(x) = x³ + 5x² + x + 5

Let us factorize by grouping

f(x) = (x³ + x) + (5x² + 5)

Taking out the common

f(x) = x(x² + 1) + 5(x² + 1)

f(x) = (x + 5)(x² + 1)

Now let us equate it to zero to find the imaginary and real zeros

⇒ x + 5 = 0

x = -5

⇒ x² + 1 = 0

x² = -1

x = ±√(-1)

We know that i² = -1

x = ± i

Therefore, the function has one real and two imaginary zeros.

State how many imaginary and real zeros the function has. f(x) = x³ + 5x² + x + 5

Summary:

The function f(x) = x³ + 5x² + x + 5 has two imaginary and one real zeros.