# Thread: How to calculate zeros withour graphing technology?

1. ## How to calculate zeros withour graphing technology?

The function f(x) = 3x^4 + 2x^3 - 15x^2 + 12x -2 is not able to factor without graphing technology because you can only find that the zero is x = 1 without graphing technology.

Is there a method that I am missing for using test points on the cubic function. I used all plus/minus 0.5,1,2.

2. Originally Posted by Barthayn
The function f(x) = 3x^4 + 2x^3 - 15x^2 + 12x -2 is not able to factor without graphing technology because you can only find that the zero is x = 1 without graphing technology.

Is there a method that I am missing for using test points on the cubic function. I used all plus/minus 0.5,1,2.
I suppose you could use Descartes' rule of signs to find an approximation for the number of roots.

But you are really asking about the rational roots test, I think, which is a small expansion of the factor theorem.

You have to consider that some factors might be of the form $(3x\pm k)$, where 'k' refers to any of the factors of $2$, by testing $f(\frac{-k}{3})$, as you would with the factor theorem. You should find that there are indeed more factors to this polynomial.

3. Originally Posted by Quacky
I suppose you could use Descartes' rule of signs to find an approximation for the number of roots.

But you are really asking about the rational roots test, I think, which is a small expansion of the factor theorem.

You have to consider that some factors might be of the form $(3x\pm k)$, where 'k' refers to any of the factors of $2$, by testing $f(\frac{-k}{3})$, as you would with the factor theorem. You should find that there are indeed more factors to this polynomial.
I know that there are more factors, however, the zeros are x = -2.89681, x = 0.23013, and x = 1.

The -2.89 and 0.23 zeros cannot come from any test points that one can find. Therefore, one must use graphing technology, correct?

EDIT: Nevermind, I forgot that there was a (x-1)^2 on this function. I feel like a mathematical fool.