To find the roots of the function f(x) = x^3 + 3x^2 + x + 3 using the Remainder Theorem, we first verify that f(1) = 0, which indicates that x = 1 is one of the roots of the function.
Let’s evaluate f(1):
f(1) = 1^3 + 3(1^2) + 1 + 3 = 1 + 3 + 1 + 3 = 8.
Since f(1)
eq 0, we conclude that x = 1 is actually not a root. Hence, we need to find the correct roots.
The Remainder Theorem states that if a polynomial f(x) is divided by x – r, the remainder is f(r). To find the other roots, we can perform polynomial long division using synthetic division.
Next, let’s try to find a different root by testing some rational roots. We can test x = -1 (since r should be among the factors of the constant term and lead coefficient):
f(-1) = (-1)^3 + 3(-1)^2 + (-1) + 3 = -1 + 3 – 1 + 3 = 4.
Now testing x = -3:
f(-3) = (-3)^3 + 3(-3)^2 + (-3) + 3 = -27 + 27 – 3 + 3 = 0. Thus, x = -3 is a root.
Now we can divide the polynomial f(x) by (x + 3) using synthetic division:
| 3 | 1 3 1 3 |
| -3 0 1 |
| —————- |
| 1 0 1 0 |
After the division, we get a quotient of x^2 + 1. To find the remaining roots, we can set x^2 + 1 = 0, which gives:
x^2 = -1 or x = i and x = -i.
In summary, the roots of the function f(x) = x^3 + 3x^2 + x + 3 are x = -3, x = i, and x = -i.