MHB Polynomial Challenge: Find # of Int Roots of Degree 3 w/ Coeffs

[anemone]

If $P(0)=3$ and $P(1)=11$ where $P$ is a polynomial of degree 3 with integer coefficients and $P$ has only 2 integer roots, find how many such polynomials $P$ exist?

 

[kaliprasad]

Spoiler

beacuse P(0) is odd it does not have any even root and because P(1) is odd it does not have any odd root. So it cannot have any integer roots. So there is no polynomial.

reason : P (a) - P(b) is divisible by a - b

 

[kaliprasad]

Spoiler

the above is based on http://mathhelpboards.com/linear-abstract-algebra-14/polynomal-divisibility-10507.html#post48739

 

[anemone]

Hi kaliprasad,

Thanks for participating and the follow-up explanation post! I also want to thank you for your continuous support to my challenge problems!:)

 

[Albert1]

If $P(0)=3$ and $P(1)=11$ where $P$ is a polynomial of degree 3 with integer coefficients and $P$ has only 2 integer roots, find how many such polynomials $P$ exist?

Spoiler

for convenience we let the leading coefficient=1, then :
$P(x)=x^3+ax^2+bx+3$
$P(1)=1+a+b+3=11$
$\therefore a+b=7$-----(1)
if m,n are 2 intger roots of $ P(x)$ then m.n must be a factor of 3
if $P(-1)=0$ we have $-1+a-b+3=0, \,, =>a-b=-2---(2)$
from (1)(2)
$a=\dfrac {5}{2}$
does not fit (since a must be integer)
if $P(3)=0 $
we have $27+9a+3b+3=0, \,, =>3a+b=-10---(3)$
if $P(-3)=0$
we have $-27+9a-3b+3=0\,, =>3a-b=8----(4)$
from (1)(3) and (1)(4) we find both "a" are not integer
and we conclude such P(x) does not exist