To find the roots of the polynomial equation x3 + x2 + x – 18 = 0, we can utilize several methods, including synthetic division, the Rational Root Theorem, or numerical approaches if necessary.
Firstly, it’s good to start with the Rational Root Theorem, which suggests that any rational solution (root) of the polynomial equation, being in the form of p/q, should have p as a factor of the constant term (-18 in this case) and q as a factor of the leading coefficient (1 here). Therefore, the possible rational roots can be:
- ±1, ±2, ±3, ±6, ±9, ±18
Let’s test these possible roots:
- Testing x = 2:
23 + 22 + 2 – 18 = 8 + 4 + 2 – 18 = -4 (Not a root)
- Testing x = 3:
33 + 32 + 3 – 18 = 27 + 9 + 3 – 18 = 21 (Not a root)
- Testing x = -3:
(-3)3 + (-3)2 + (-3) – 18 = -27 + 9 – 3 -18 = -39 (Not a root)
- Testing x = 1:
13 + 12 + 1 – 18 = 1 + 1 + 1 – 18 = -15 (Not a root)
- Testing x = -1:
(-1)3 + (-1)2 + (-1) – 18 = -1 + 1 – 1 – 18 = -19 (Not a root)
- Testing x = 6:
63 + 62 + 6 – 18 = 216 + 36 + 6 – 18 = 240 (Not a root)
- Testing x = 0:
03 + 02 + 0 – 18 = -18 (Not a root)
After testing several rational roots with no results, we might try numerical approximations like Newton’s method or graphing the function to identify where it crosses the x-axis.
By using numerical software or a graphing calculator, we can find that the polynomial has one real root, approximately at:
- x ≈ 2.20
We can find the other roots through polynomial division, factoring, or using the cubic formula, but often the process is more complex and may produce complex roots.
Therefore, the equation has one real root around x ≈ 2.20 and potentially two other complex roots, which can be determined through further calculations or numerical methods.