Question

Problem 4. Find a degree three polynomial p(n) := ax3 + bx2 + cx + d such that p(n) = $k2 by solving the system of equations p(1) = a +b+c+d = 1 p(2) = 8a + 4b + 2c + d = 5 p(3) = 27a + 96 + 3c + d = 14 p(4) = 64a + 16b + 4c + d = 30 with the reduction algorithm. Also find the reduced echelon form of the augmented matrix corresponding to the above.

Answer 1

System of equations: p(n) = ax^3 + bx^2 + cx + d p(1) = a + b + c + d = 1 ---(1) p(2) = 8a + 4b + 2c + d = 5 ---(2) p(3) = 27a + 9b + 3c + d = 14 ---(3) p(4) = 64a + 16b + 4c + d = 30 ---(4) from equation(1) d = 1-a-b-c substitute in equations (2) , (3) , and (4) , 8a + 4b + 2c + d + 1-a-b-c = 5 + d 7a + 3b + c = 4 --- (5) 27a + 9b + 3c + d + 1-a-b-c = 14 + d 26a + 16b + 4c + d + 1-a-b-c = 30 + d 63a + 15b + 3c = 29 ---(7) from equation 5 we get c = 4-7a-3b substitute in equation (6) and (7) 26a + 8b + 2(4-7a-3b) = 13 26a + 8b + 8-14a-6b = 13 12a + 2b = 5 --- (8) 63a + 15b + 3(4-7a-3b) = 29 63a + 15b + 12-21a-9b = 29 42a + 6b = 17 --- (9) from equation (8) b = 5-12a/2 substitute in equation (9) 42a + 6(5-12a/2) = 17 42a + 15-36a = 17 6a = 2 a = 1/3 b = s-12a/2 = 5-4/2 = 1/2 c = 4-7a-3b = 4-7/3-3/2 = 24-14-9/6 = 1/6 d = 1-a-b-c = 1-1/3 - 1/2 - 1/6 = 6-2-3-1/6 = 0 d = 0 therefore a = 1/3