According to the rational root theorem, which is not a possible rational root of x3 + 8x2 x 6 = 0?
verify that you meant
x^3 + 8x^2 + 6 = 0
x3 + 8x2-x-6 = 0?
x^3 + 8x^2 - x - 6 = 0
Let f(x) = x^3 + 8x^2 - x - 6
f(1) = 1+8-1-6 ≠ 0
f(-1) = -1 + 8 + 1 - 6 ≠ 0
f(2) = 8 + 32 - 2 - 6 ≠ 0
f(-2) = -8 + 32 + 2 - 6 ≠ 0
f(3) = 27 + .... ≠ 0
f(-3) = -27 + 72 ... ≠ 0
numbers which are NOT possible rational roots are
±1 , ±2 , ± 3
Is that what you wanted?
yeh but my homework has +-2 and +-1 as answers abd i can only choose one
I used the 1 , 2, and 3 since they were factors of the 6 at the end.
There are an infinite number of choices of rational numbers which are NOT possible roots.
This is a poorly worded question.
go with the ±1 and ±2
To determine which values are not possible rational roots of the given polynomial equation, we can use the Rational Root Theorem. The Rational Root Theorem states that if a polynomial has a rational root (x = p/q), then p must be a factor of the constant term (6 in this case), and q must be a factor of the leading coefficient (1 in this case).
The polynomial equation is x³ + 8x² - x + 6 = 0.
To find the factors of the constant term (6), we can list all the possible values:
Factors of 6: ±1, ±2, ±3, ±6
Now, let's find the factors of the leading coefficient (1):
Factors of 1: ±1
According to the Rational Root Theorem, the possible rational roots are fractions where the numerator is a factor of 6 and the denominator is a factor of 1. So, the possible rational roots are:
±1/1, ±2/1, ±3/1, ±6/1
Simplifying these fractions, we get:
±1, ±2, ±3, ±6
Therefore, the possible rational roots of the equation x³ + 8x² - x + 6 = 0 are ±1, ±2, ±3, and ±6.
To determine which value is not a possible rational root, you can compare the options given and see if any of them do not match ±1, ±2, ±3, and ±6.