use the rational theorem to select the values that are possible zeroes of the function f(x)=6x^3-2x^2+x+3 select all that apply
-3
-2/3
3/2
6
the answers are -3 and 3/2
c
b
c
a and c
Well, let's bring in the Rational Root Theorem to help us out. According to the theorem, the possible rational roots of a polynomial function can be found by taking the factors of the constant term (in this case, 3) and dividing them by the factors of the leading coefficient (in this case, 6).
So, let's list the factors of 3: ±1, ±3. And the factors of 6: ±1, ±2, ±3, ±6.
Now, dividing the factors of 3 by the factors of 6, we get the following possible rational roots: ±1/6, ±3/6, ±1/2, ±3/2.
Simplifying these fractions, we have: ±1/6, ±1/2, ±3/2.
Now, let's check which of these values are possible zeroes by substituting them into the function:
For f(-3): 6(-3)^3 - 2(-3)^2 + (-3) + 3 = -162 + 18 - 3 + 3 = -144. Not zero.
For f(-2/3): 6(-2/3)^3 - 2(-2/3)^2 + (-2/3) + 3 = -2. Not zero.
For f(3/2): 6(3/2)^3 - 2(3/2)^2 + (3/2) + 3 = 54 - 9 + 3 + 3 = 54 - 6 = 48. Not zero.
For f(6): 6(6)^3 - 2(6)^2 + (6) + 3 = 2595. Not zero.
So, none of the values selected -3, -2/3, 3/2, and 6 - are possible zeroes based on the Rational Root Theorem.
Boy, this Rational Root Theorem can be quite picky sometimes!
To use the rational root theorem to identify possible zeroes of a polynomial function, you need to consider the factors of the constant term divided by the factors of the leading coefficient.
In this case, the constant term is 3 and the leading coefficient is 6.
Factors of 3: ±1, ±3
Factors of 6: ±1, ±2, ±3, ±6
Therefore, the possible rational zeroes are the combinations of these factors:
±1, ±(1/2), ±(1/3), ±(1/6), ±3, ±(3/2), ±(3/4), ±(3/6), ±(3/12), ±6
From the provided options, the possible zeroes that apply to the function f(x) = 6x^3 - 2x^2 + x + 3 are:
-3
-2/3
3/2
6
To find zeros, I would have tried x = ±3, and ± 2/3
for f(x)
However, this is a poor equation to use, since it doesn't have any rational factors at all.