# Math Help - A 3th degree polynome problem

1. ## A 3th degree polynome problem

hello I don't know how to solve a problem:

P= x^3+x^2+Ax+1

at A the value of P is a rational number for any x that satisfies the equation x^2+2x-2=0, and in this case P=B

They ask me to find A and B

I do have the answer, but what I need is to show me the method.

Thank you very much!

2. Originally Posted by Viou
hello I don't know how to solve a problem:

P= x^3+x^2+Ax+1

at A the value of P is a rational number for any x that satisfies the equation x^2+2x-2=0, and in this case P=B

They ask me to find A and B

I do have the answer, but what I need is to show me the method.

Thank you very much!
$x^2+ 2x- 2= x^2+ 2x+ 1- 3= (x+1)^2- 3= 0$
so "any x that satisfies the equation $x^2+ 2x- 2= 0$" simply means $x= -1+\sqrt{3}$ and $-1-\sqrt{3}$.

If you put $x= -1+ \sqrt{3}$ into the formula $x^3+ x^2+ Ax+ 1$ you get [tex](-1+\sqrt{3})^3+ (-1+ \sqrt{3})^2+ A(-1+\sqrt{3})+ 1= 6\sqrt{3}- 10+ 4- 2\sqrt{3}- A- A\sqrt{3}+ 1= -5- A+ (4+A)\sqrt{3}[/itex].

Do you see what A must be in order that P be a rational number?

3. Originally Posted by HallsofIvy
$x^2+ 2x- 2= x^2+ 2x+ 1- 3= (x+1)^2- 3= 0$
so "any x that satisfies the equation $x^2+ 2x- 2= 0$" simply means $x= -1+\sqrt{3}$ and $-1-\sqrt{3}$.

If you put $x= -1+ \sqrt{3}$ into the formula $x^3+ x^2+ Ax+ 1$ you get $(-1+\sqrt{3})^3+ (-1+ \sqrt{3})^2+ A(-1+\sqrt{3})+ 1= 6\sqrt{3}- 10+ 4- 2\sqrt{3}- A- A\sqrt{3}+ 1= -5- A+ (4+A)\sqrt{3}$.

Do you see what A must be in order that P be a rational number?
No I don't. What you did is exactly what I did before. The problem is what should I do from here.
After I found that f( $-1+ \sqrt{3}$) = $-5 - A +(4+A)\sqrt{3}$, what should I do?

Wh

4. If $x^2+2x-2=0$ then $x^2 = -2x+2$. Also (multiply by x) $x^3 = -2x^2+2x = -2(-2x+2)+2x = 6x-4$. Therefore

$P= x^3+x^2+Ax+1 = (6x-4) + (-2x+2) + Ax + 1 = (A+4)x -1.$

Now you can put $x = -1+\sqrt3$, which is much easier than doing so at an earlier stage.

5. Originally Posted by Opalg
If $x^2+2x-2=0$ then $x^2 = -2x+2$. Also (multiply by x) $x^3 = -2x^2+2x = -2(-2x+2)+2x = 6x-4$. Therefore

$P= x^3+x^2+Ax+1 = (6x-4) + (-2x+2) + Ax + 1 = (A+4)x -1.$

Now you can put $x = -1+\sqrt3$, which is much easier than doing so at an earlier stage.

Thanks, Opalg, that really makes the counting easier.
But it doesn't solve the problem, we still have the unknown A. Can you tell me what should I do next? How can I find A , knowing that
$f( \sqrt3 - 1)= - A - 5 + (A+4)\sqrt3$

6. Originally Posted by Viou
Thanks, Opalg, that really makes the counting easier.
But it doesn't solve the problem, we still have the unknown A. Can you tell me what should I do next? How can I find A , knowing that
$f( \sqrt3 - 1)= - A - 5 + (A+4)\sqrt3$
P = B for this value of x. That is one equation in terms of A and B. Now use your other x value. That will get you a second equation in terms of A and B...

-Dan

7. Originally Posted by Viou
How can I find A , knowing that
$f( \sqrt3 - 1)= - A - 5 + (A+4)\sqrt3$
You are told that P is rational, so you must ensure that the coefficient of $\sqrt3$ is zero.

8. Originally Posted by topsquark
P = B for this value of x. That is one equation in terms of A and B. Now use your other x value. That will get you a second equation in terms of A and B...

-Dan

I did so before, but what I got is only the statement that $4\sqrt(3)= -4\sqrt(3)$ so its not very helpful

Originally Posted by Opalg
You are told that P is rational, so you must ensure that the coefficient of $\sqrt3$ is zero.
Oh thank you, now I get it!!!!!

B being a rational number AND
$f( \sqrt3 - 1)= B$
$f( \sqrt3 - 1) = - A - 5 + (A+4)\sqrt3$

$(A+4)\sqrt3=0$
$A+4 = 0$
$A = -4$
$B = - A - 5$
$B = -(-4) - 5$
$B = -1$

yay! happy ^w^