Question 1209706
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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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<pre>
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.


<U>ANSWER</U>.  In polynomial  x^3 - bx^2 - c, coefficients  are b = 4,  c = -9.
</pre>

At this point, the problem is solved completely.