Let p(x) be a monic, quartic polynomial, such that p(1)=3,p(3)=11, and p(5)=27. Find p(−2)+7p(6).
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Let q(x)=p(x)−(x2+2). Then q(1)=q(3)=q(5)=0, so q(x)=(x−1)(x−3)(x−5)(x−r)for some real number r. Then p(x)=q(x)+x2+2=(x−1)(x−3)(x−5)(x−r)=x2+2, so p(−2)p(6)=(−2−1)(−2−3)(−2−5)(−2−r)+(−2)2+2=105r+216,=(6−1)(6−3)(6−5)(6−r)+62+2=128−15r,so p(−2)+7p(6)=(105r+216)+7(128−15r)=1112.