To determine the potential roots of the polynomial function 6x³ + 19x² + 15x – 28, we can utilize a few mathematical tools and techniques. One common approach is the Rational Root Theorem, which suggests that any rational solution, or root, of a polynomial equation can be expressed as \\frac{p}{q}\, where:
- p is a factor of the constant term (the last number in the polynomial), and
- q is a factor of the leading coefficient (the coefficient of the highest degree term).
In this case:
- The constant term is -28. The factors of -28 are ±1, ±2, ±4, ±7, ±14, ±28.
- The leading coefficient is 6. The factors of 6 are ±1, ±2, ±3, ±6.
We can list the potential rational roots by taking all combinations of these factors:
- Potential roots: ±1, ±2, ±4, ±7, ±14, ±28, ±1/2, ±3/2, ±7/2, ±14/3, ±28/3, ±1/6, ±7/6.
Next, to identify which of these potential roots are actual roots of the polynomial, we can substitute these values back into the polynomial and see if the result equals zero.
Let’s check a few substitutions:
x = 1: 6(1)³ + 19(1)² + 15(1) - 28 = 6 + 19 + 15 - 28 = 12 (not a root)
x = -1: 6(-1)³ + 19(-1)² + 15(-1) - 28 = -6 + 19 - 15 - 28 = -30 (not a root)
x = 2: 6(2)³ + 19(2)² + 15(2) - 28 = 6(8) + 19(4) + 30 - 28 = 48 + 76 + 30 - 28 = 126 (not a root)
x = -2: 6(-2)³ + 19(-2)² + 15(-2) - 28 = -48 + 76 - 30 - 28 = -30 (not a root)
This process can continue for each potential root until we find those that yield zero. For complicated roots, numerical methods or graphing might also help identify the roots.
In conclusion, the potential roots of 6x³ + 19x² + 15x – 28 could be any of the values obtained through the Rational Root Theorem, and to find the actual roots, we must evaluate these values within the polynomial itself.