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Polynomial Puzzle (Posted on 2007-03-16)

Let p(x)=x^4+ax^3+bx^2+cx+d, where a, b, c, and d are constants.

If p(1)=-9, p(2)=-18, and p(3)=-27, find the value of ¼(p(10)+p(-6)).

See The Solution Submitted by Dennis

Rating: 4.0000 (5 votes)

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answer

Comment 7 of 7 |

the answer is 2007 and it is unique.

plug in 1,2,and3 for x and get system

a+b+c=-10-d

8a+4b+2c=-34-d

27a+9b+3c=-108-d

get a=-6-d/6

b=11+d

c=-15-(11/6)d

evaluate (1/4)(p(10)+p(-6)) with the above values of a,b andc and get 2007

Posted by kevin raponi on 2007-03-18 12:00:52

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