# Find the value of $A$ and $B$ such that $P$ is a rational number

Well, I got stuck in that question:

Consider the integral expression in $x$: $P = x^3+x^2+ax+1$ where $a$ is a rational number. At $a=A$ the value of $P$ is a rational number for any $x$ which satisfies the equation $x^2+2x-2=0$, and in this case the value of $P$ is $B$. Find $A$ and $B$.

I've tried to use the last equation in the first one, I multiplied it by $x$ and then I got $x^3 = 2x-2x^2$ and I put that in place of $x^3$ in $P$ and this led me to $P = -x^2+(2+a)x+1$. I thought that as $P$ is a rational number, I would choose a rational number to put in place of $P$, but I don't know if this expression includes all rational numbers. Anyway, I tried the value $P=1$ and this led me to $x=2+a$, well, this didn't help, I was trying to find $a$ and I found an $x$ that makes $P$ be $1$.

Any idea?

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The procedure you began works after not much more calculation. We are considering $x$ such that $x^2=2-2x$. So $x^3=2x-2x^2$.

Thus $x^3+x^2=2x-x^2=2x-(2-2x)=4x-2$.

We conclude that if $x$ is either root of the quadratic $x^2+2x-2=0$, then our original expression $x^3+x^2+ax+1$ is equal to $4x-2+ax+1$. But the roots of the quadratic are irrational, and $a$ is rational. So the only way for $4x-2+ax+1$ to be rational is if the coefficient of $x$ is $0$. Thus $A=-4$. It follows that $B=-2+1=-1$.

Remark: Or else we could do it the boring way, by solving the quadratic and substituting. The cubes and squares of the roots can be easily calculated explicitly. It would have been a better problem if it had involved powers of $x$ say up to $6$, and the quadratic had roots more complicated than $-1\pm\sqrt{3}$. Then the method we used above is not much more stressful to carry out, while substituting and expanding gets quite unpleasant.

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Well, if $a=-4$, then the last equation you wrote down becomes $4x-2-4x+1=0$ then $-1=0$. I think I hasn't really got your idea or it is not yet okay as answer. – Ataias Reis Sep 8 '12 at 20:55
I had a typo earlier. Did not mean $=0$, meant $4x-2+ax+1$. – André Nicolas Sep 8 '12 at 20:57
Thanks, I understood it now. – Ataias Reis Sep 8 '12 at 21:00
@AndréNicolas I can't seem to get $x^3+x^2=4x-2$, since $x^2=2-2x$. Can you please expand on that? – Secret Apr 18 '14 at 7:15
@Secret: It is all in the second and third line of the answer. From $x^2=2-2x$ we get by multiplying by $x$ that $x^3=2x-2x^2$, so $x^3+x^2=2x-x^2=2x-(2-2x)=4x-2$. – André Nicolas Apr 18 '14 at 13:14