show that 3x^3- x^2- 6x+ 12 has no ratioinal zeros?

Question

how do i show without graphic?

10 points to first comprehendable answer!

sorry typo- rational* and graphing*

Answer 1

You can use the rational root theorem.

This basically states that all possible rational roots can be found by dividing factors of your constant (In your case 12) by factors of your leading coefficient (in your case 3). Don't forget to to include both positive and negative factors. That is, 12 = 1 x 12 and -1 x -12, etc.

Once you have the list of possibilities, you can plug them into your function to see if they are zeros or not.

Since these are ALL of the possibilities, if none of them produce zero as a result, you have shown there are no rational zeros.

Hope this helps!

Best of luck,

~Angel

Answer 2

You can find the possible rational zeros by listing all the factors of the leading coefficient and the constant:

1, 3
1, 2, 3, 4, 5, 6 ,12

Then all the combinations you can make with Constant/Leading Coefficient

1/3, 2/3, 1, 4/3, 5/3, 2, 3, 4, 5, 6, and 12

Then, plug them into f(x) = 3x^3 - x^2 - 6x + 12. If f(x) = 0, it is a zero.

f(1/3) = 1/9 - 1/9 - 2 + 12 = 10
f(2/3) = 8/9 - 4/9 - 4 + 12 = 8 4/9
f(1) = 3 - 1 - 6 + 12 = 8
etc...

If none of them are equal to zero, then you have a polynomial with no rational zeros.

Answer 3

calculate for x values from -12 to 12 step 0,001 for it
using excel

check the results.