Posted by **Zac** on Monday, January 10, 2011 at 12:11pm.

According to the rational root theorem, which is not a possible rational root of x3 + 8x2 x 6 = 0?

- algebra -
**Reiny**, Monday, January 10, 2011 at 12:26pm
verify that you meant

x^3 + 8x^2 + 6 = 0

- algebra -
**Zac**, Monday, January 10, 2011 at 12:30pm
x3 + 8x2-x-6 = 0?

- algebra -
**Reiny**, Monday, January 10, 2011 at 12:38pm
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?

- algebra -
**Zac**, Monday, January 10, 2011 at 12:42pm
yeh but my homework has +-2 and +-1 as answers abd i can only choose one

- algebra -
**Reiny**, Monday, January 10, 2011 at 12:46pm
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

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