A polynomial with integer coefficients is of the form

\(2x^4 + a_3 x^3 + a_2 x^2 + a_1 x + 1 = 0.\)

Find the number of different possible rational roots of this polynomial.

Guest May 19, 2021

#1**0 **

The possible rational roots are 1 and 1/2, so there are 2 possible rational roots.

Guest May 19, 2021

#3**+1 **

A polynomial with integer coefficients is of the form

\(2x^4 + a_3 x^3 + a_2 x^2 + a_1 x + 1 = 0.\)

Find the number of different possible rational roots of this polynomial.

I just found this HERE

The rational root test theorem says that, if rational factors of a polynomial exist, then they are always in the form of

±(factor of last coefficient) / (factor of first coefficient)

So the rational roots could be 1, -1, 0.5, -0.5 and that is it.

It can't be all of those because the product of the roots must be +0.5 so it can't have 4 rational roots

It could definitely have no real roots, so therefore no rational roots.

So could it have 1 or 2 or 3 rational roots?

It can have 3

2(x-1)(x+1)(x+0.5)(x-1)= 0 that works

I don't know about 1 or 2

I suspect it could have 2 but not 1, not sure though.

So it can't have 4 or more. It could have 3 or 0.

I don't think it can have 1 but I am not sure. I can probably have 2.

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I expect are theorem/s that would make this nice and easy ....

Melody May 19, 2021

#4**+2 **

Just use RRT, you get the possible roots are \(\pm \frac12, \pm 1\) so there are 4 possible rational roots.

ilovepizza547 May 19, 2021

#5**+1 **

We have the form

qx^n + ....... + ax + p

The possible rational roots = ± [ all factors of p / all factors of q ]

Here p = 1 and all the possible factors = 1

q = 2 and all possible factors = 1 , 2

So all the possible rational roors = ± 1 , ± 1/2 = 4 possible rational roots

CPhill May 19, 2021

#6**+1 **

**Sure, there is 4 possible rational roots. I said that too**

**But they cannot all be true at once.**

I went on to interpret the question to say how many integer solutions can there be in any specific values of a_n

I found that for any specific solution there cannot be 4.

There can be 0

There can be 3

Here is a different combination of 3 rational roots that work

I am not sure about 1 or 2.

**It cannot be 4**

Melody May 19, 2021