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)).

Setting up simultaneous equations:

1 + a + b + c + d = -9
16 + 8a + 4b + 2c + d = -18
81 + 27a + 9b + 3c + d = -27

results in each of b, c, d's dependence on a as:

  b = -6a - 25
  c = 11a + 51
  d = -36 - 6a
 
This program accepts any value for a, and produces the various values for b, c and d, and verifies the given values of the polynomial for x=1, 2 and 3, and the value of the given expression:

DO
  INPUT a
  b = -6 * a - 25
  c = 11 * a + 51
  d = -36 - 6 * a
  PRINT a; b; c; d
  x = 1
  PRINT x ^ 4 + a * x ^ 3 + b * x ^ 2 + c * x + d
  x = 2
  PRINT x ^ 4 + a * x ^ 3 + b * x ^ 2 + c * x + d
  x = 3
  PRINT x ^ 4 + a * x ^ 3 + b * x ^ 2 + c * x + d
  x = 10
  t = x ^ 4 + a * x ^ 3 + b * x ^ 2 + c * x + d
  x = -6
  t = t + x ^ 4 + a * x ^ 3 + b * x ^ 2 + c * x + d
  PRINT t / 4
LOOP

For various values it finds:

? 0
 0 -25  51 -36
-9 -18 -27
 2007
? 1
 1 -31  62 -42
-9 -18 -27
 2007
? 2
 2 -37  73 -48
-9 -18 -27
 2007
? 3
 3 -43  84 -54
-9 -18 -27
 2007
? 4
 4 -49  95 -60
-9 -18 -27
 2007
? 5
 5 -55  106 -66
-9 -18 -27
 2007
? 6
 6 -61  117 -72
-9 -18 -27
 2007
? 7
 7 -67  128 -78
-9 -18 -27
 2007
? 8
 8 -73  139 -84
-9 -18 -27
 2007
? 9
 9 -79  150 -90
-9 -18 -27
 2007
? 10
 10 -85  161 -96
-9 -18 -27
 2007
? 11
 11 -91  172 -102
-9 -18 -27
 2007
 
 but perhaps the most satisfying is when a=-5:
 
 ? -5
 -5  5 -4 -6
 -9 -18 -27
 2007
 
 indicating that then: b=5, c=-4, d=-6.
 The values for x=1, 2 and 3 are verified, and as in all of these occasions, the requested value is the current year number, 2007.

Posted by Charlie on 2007-03-16 13:59:30