To determine the number of real and imaginary zeros of the function f(x) = x³ + 5x² + x + 5, we can use methods such as the Rational Root Theorem, Synthetic Division, and the Descartes’ Rule of Signs.
1. **Familiarize with the function:** The given polynomial is a cubic function, which means it can have either 1 or 3 real roots. The remaining roots will be imaginary or complex.
2. **Evaluate the number of sign changes:** We apply the Descartes’ Rule of Signs to find the possible number of positive and negative real roots.
– For f(x), the signs of the coefficients are +, +, +, +. There are no sign changes for positive roots, which indicates that there are 0 positive real roots.
– For f(-x): substituting -x into the function gives us:
f(-x) = (-x)³ + 5(-x)² + (-x) + 5 = -x³ + 5x² - x + 5– The coefficients of f(-x) are -1, +5, -1, +5, which gives us 3 sign changes. By the Descartes’ Rule, this means there can be 3 or 1 negative real roots.
3. **Finding the roots:** We should look for rational roots using the Rational Root Theorem. The possible rational roots are the factors of the constant term (5) over the leading coefficient (1), which are ±1, ±5. Testing these values:
– f(1) = 1 + 5 + 1 + 5 = 12 (not a root)
– f(-1) = -1 + 5 – 1 + 5 = 8 (not a root)
– f(5) = 125 + 125 + 5 + 5 = 260 (not a root)
– f(-5) = -125 + 125 – 5 + 5 = 0 (is a root)
Having found that -5 is a root, we can perform synthetic division to factor the cubic polynomial:
-5 | 1 5 1 5
| -5 0 -5
--------------------
1 0 1 0
This gives us a quadratic factor: x² + 1. To find the zeros of this quadratic, we set it to zero:
x² + 1 = 0This gives us x² = -1 or x = ±i, where i is the imaginary unit.
4. **Conclusion:** Summarizing our findings:
– Real Zeros: 1 (the real root is -5)
– Imaginary Zeros: 2 (the imaginary roots are +i and -i)
Thus, the function f(x) = x³ + 5x² + x + 5 has 1 real zero and 2 imaginary zeros.