Using the Rational Root Theorem, state all the possible rational roots for the function.
Then, solve the function using synthetic division to check and find the actual roots of the function.
f (x) = 2x2 + 3x + 1
you missed x = -1/2
f(-1/2) = 2(1/4) + 3(-1/2) + 1 = 1/2 - 3/2 + 1 = 0
Apologies for the oversight. You are correct.
The actual roots of the function f(x) = 2x^2 + 3x + 1 are x = -1 and x = -1/2.
Thank you for pointing that out.
The Rational Root Theorem states that any rational roots of the function f(x) = 2x^2 + 3x + 1 must be of the form p/q, where p is a factor of the constant term (1) and q is a factor of the leading coefficient (2). Therefore, the possible rational roots are ±1, ±1/2.
To check these roots, we will use synthetic division:
1. For x = 1:
2 | 2 3 1
| 2 5
------------
2 5 6
Since the remainder is not 0, x = 1 is not a root.
2. For x = -1:
2 | 2 3 1
| -2 -1
------------
2 1 0
Since the remainder is 0, x = -1 is a root.
3. For x = 1/2:
2 | 2 3 1
| 1/2 2
------------
2 3/2 3
Since the remainder is not 0, x = 1/2 is not a root.
4. For x = -1/2:
2 | 2 3 1
| -1 1/2
------------
2 2 1/2
Since the remainder is not 0, x = -1/2 is not a root.
Therefore, the actual roots of the function f(x) = 2x^2 + 3x + 1 are x = -1.