SOLUTION: There are integers b, c for which both roots of the polynomial x^2 - x - 3 are also roots of the polynomial x^3 - bx^2 - c. Determine the ordered pair (b,c).

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Question 1209706: There are integers b, c for which both roots of the polynomial x^2 - x - 3 are also roots of the polynomial x^3 - bx^2 - c. Determine the ordered pair (b,c).
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1987)   (Show Source): You can put this solution on YOUR website!
Let $P(x) = x^2 - x - 3$. The roots of $P(x)$ are given by the quadratic formula:
$$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-3)}}{2(1)} = \frac{1 \pm \sqrt{1+12}}{2} = \frac{1 \pm \sqrt{13}}{2}$$
Let $r_1 = \frac{1 + \sqrt{13}}{2}$ and $r_2 = \frac{1 - \sqrt{13}}{2}$.
Since $r_1$ and $r_2$ are roots of $x^2 - x - 3 = 0$, we have
$r_1^2 - r_1 - 3 = 0$ and $r_2^2 - r_2 - 3 = 0$.
Thus, $r_1^2 = r_1 + 3$ and $r_2^2 = r_2 + 3$.
Also, $r_1 + r_2 = 1$ and $r_1 r_2 = -3$.
Let $Q(x) = x^3 - bx^2 - c$. Since $r_1$ and $r_2$ are roots of $Q(x)$, we have
$r_1^3 - br_1^2 - c = 0$ and $r_2^3 - br_2^2 - c = 0$.
$r_1^3 = r_1 \cdot r_1^2 = r_1(r_1+3) = r_1^2 + 3r_1 = (r_1+3) + 3r_1 = 4r_1 + 3$
$r_2^3 = r_2 \cdot r_2^2 = r_2(r_2+3) = r_2^2 + 3r_2 = (r_2+3) + 3r_2 = 4r_2 + 3$
Substituting these into the equations for $Q(x)$, we get
$4r_1 + 3 - b(r_1 + 3) - c = 0$
$4r_2 + 3 - b(r_2 + 3) - c = 0$
Subtracting the second equation from the first, we get
$4(r_1 - r_2) - b(r_1 - r_2) = 0$
$r_1 - r_2 \ne 0$, so $4 - b = 0$, which gives $b = 4$.
Substituting $b=4$ into the first equation, we get
$4r_1 + 3 - 4(r_1 + 3) - c = 0$
$4r_1 + 3 - 4r_1 - 12 - c = 0$
$-9 - c = 0$
$c = -9$
Thus, $(b, c) = (4, -9)$.
Final Answer: The final answer is $\boxed{(4,-9)}$
Answer by ikleyn(52932)   (Show Source): You can put this solution on YOUR website!
.
There are integers b, c for which both roots of the polynomial x^2 - x - 3
are also roots of the polynomial x^3 - bx^2 - c. Determine the ordered pair (b,c).
~~~~~~~~~~~~~~~~~~~~~~~~~ 

Let p and q be the roots of the polynomial x^2 - x - 3.

Due to Vieta's theorem, 

    p + q = 1,        (1)

    pq = -3.          (2)


According to the problem, p and q are also the roots of the polynomial x^3 - bx^2 - c. 
Let r be the third root of this polynomial.

Then, due to Vieta's theorem for polynomial x^3 - bx^2 - c

    p + q + r = b,          (3)

    p*q + p*r + q*r = 0,    (4)   (coefficient at x in polynomial x^3 - bx^2 - c)

    p*q*r = c.              (5)


In (3), replace p+q by 1, based on (3).  In (5), replace p*q by -3, based on (2).
Then from (3) an (4) you will have

    1 + r = b,              (6)

    -3r =  c,               (7) 


In equation (4), replace p*q  by -3, based on (2).  Then equation (4) takes the form

    -3 + pr + qr = 0,  

or

    p*r + q*r  = 3,

    (p + q)*r  = 3.


In the last equation, replace (p + q) by 1,  based on (1).  You will get

    1*r = 3,  i.e.  r = 3.


Now from  (6)  b = 1 + r = 1 + 3 = 4;

    from  (7)  c = -3r = -3*3 = -9.


ANSWER.  In polynomial  x^3 - bx^2 - c, coefficients  are b = 4,  c = -9.

At this point, the problem is solved completely.



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