Problem:
The polynomial P(x) is cubic. What is the largest value of k for which the polynomials Q1(x)=x2+(k−29)x−k and Q2(x)=2x2+(2k−43)x+k are both factors of P(x) ?
Solution:
Because P(x) has three roots, if Q1(x)=x2+(k−29)x−k and Q2(x)= 2x2+(2k−43)x+k are both factors of P(x), then they must have a common root r. Then Q1(r)=Q2(r)=0, and mQ1(r)+nQ2(r)=0, for any two constants m and n. Taking m=2 and n=−1 yields the equation 15r+3k=0, so r=5−k. Thus Q1(r)=25k2−(k−29)(5k)−k=0, which is equivalent to 4k2−120k=0, whose roots are k=30 and 0. When k=30,Q1(x)= x2+x−30 and Q2(x)=2x2+17x+30, and both polynomials are factors of P(x)=(x+6)(x−5)(2x+5). Thus the requested value of k is 30.
The problems on this page are the property of the MAA's American Mathematics Competitions