To find the zeros of the cubic function f(x) = 6x³ + 35x² + 26x + 5, we need to determine the values of x for which f(x) = 0. This involves a few steps, as cubic equations can have one real root and two complex roots, or three real roots.
1. **Using the Rational Root Theorem**: This theorem helps us identify possible rational roots by examining the factors of the constant term (5) and the leading coefficient (6). The factors of 5 are ±1, ±5, and factors of 6 are ±1, ±2, ±3, ±6. So, the possible rational roots could be ±1, ±5, ±1/2, ±5/2, ±1/3, ±5/3, ±1/6, and ±5/6.
2. **Testing the possible rational roots**: We’ll test these rational roots in the function:
- For x = -1: f(-1) = 6(-1)³ + 35(-1)² + 26(-1) + 5 = -6 + 35 – 26 + 5 = 8 (not a root)
- For x = -5: f(-5) = 6(-5)³ + 35(-5)² + 26(-5) + 5 = -750 + 875 – 130 + 5 = 0 (this is a root)
3. **Factoring the polynomial**: Since we found that x = -5 is a root, we can use synthetic division to factor the polynomial:
Dividing f(x) by (x + 5):
6 35 26 5
-5
—————–
6 25 1 0
This gives us the quotient of 6x² + 25x + 1. So, we have:
f(x) = (x + 5)(6x² + 25x + 1)
4. **Finding the remaining zeros**: To find the remaining zeros, we can use the quadratic formula: x = (-b ± √(b² – 4ac)) / 2a, where a = 6, b = 25, and c = 1.
Calculating the discriminant:
b² – 4ac = 25² – 4(6)(1) = 625 – 24 = 601
Next, applying the quadratic formula:
x = (-25 ± √601) / 12
5. **Final Result**: The zeros of the function f(x) = 6x³ + 35x² + 26x + 5 are:
- x = -5 (real root)
- x = (-25 + √601) / 12 (real root)
- x = (-25 – √601) / 12 (real root)
In summary, the function has one real zero at x = -5 and two additional real zeros derived from the quadratic equation, contributing to a total of three real zeros for the polynomial.