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# Polynomial Roots

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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.

May 19, 2021

#1
0

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

May 19, 2021
#3
+113698
+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 ....

May 19, 2021
#4
+177
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Just use RRT, you get the possible roots are $$\pm \frac12, \pm 1$$ so there are 4 possible rational roots.

May 19, 2021
#5
+120075
+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

May 19, 2021
edited by CPhill  May 19, 2021
#6
+113698
+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

May 19, 2021
edited by Melody  May 19, 2021
#7
+113698
0

I would be interested in seeing other graphs that work with this.

See what combinations of rational roots you can find.

Accept it as a challenge if you want.

May 19, 2021