Consider the integral expression in $x$ $$P=x^3+x^2+ax+1,$$ where $a$ is a rational number. At $a= ?$ 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 $P=?$
I don't know how the answer came key $P= -1$, $A=-4$.
What I did is I tried to find the second equation roots but it is irrational. I don't know how to get the answer.
When $x^2 + 2x − 2 = 0,\;$ then
$\quad x^2 = 2 - 2x \;$ and
$\quad x^3 = 2x - 2x^2 = 2x - 2(2 - 2x) = 6x - 4$
So $P = x^3 + x^2 + ax + 1 = (6x-4) + (2-2x) + ax + 1 = (4+a)x - 1$
You should be able to see the rest.
We want $(x^2+2x-2)(x+b)=x^3+x^2+ax+1-P$. Expanding the lefthand side we get $x^3+(b+2)x^2+(2b-2)x-2b$. To get the correct coefficient of $x^2$ we need $b=-1$. The lefthand side then becomes $x^3+x^2-4x+2$. So we need $a=-4,P=-1$.
Alternatively, the sum of the two roots of $x^2+2x-2$ is $-2$ (minus the coefficient of $x^2$). Similarly, the sum of the three roots to $x^3+x^2+ax+1-P=0$ is $-1$. So the third root must be $1$. Hence we must have: $(x^2+2x-2)(x-1)=x^3+x^2+ax+1-P$. Hence $a=-4,P=-1$.
From the quadratic formula, the solutions of $x^2-2x-2=0$ are $x=-1\pm\sqrt{3}$
Substituting $x=-1\pm\sqrt{3}$ into $P=x^3+x^2+ax+1$ gives: $\left( \pm\sqrt{3}-1\right) \,a\pm4\,\sqrt{3}-5$. As $a$ is rational, for this to be rational requires that the $\sqrt{3}$ terms cancel, and this occurs when $a=-4$ and then $P=-1$
