Question

(1 point) Let pp be the quartic (degree 4) polynomial that satisfies p(i)=2i,i=0,1,2,3,4. p(i)=2i,i=0,1,2,3,4. Then p(x)=p(x)= . Hint: You may have a better idea, but a brute force approach is to write p(x)=ax4+bx3+cx2+dx+e p(x)=ax4+bx3+cx2+dx+e where aa, bb, cc, dd, and ee, are the unknown coefficients, and then solve the linear system p(0)=1p(0)=1, p(1)=2p(1)=2, p(2)=4p(2)=4, p(3)=8p(3)=8, and p(4)=16p(4)=16 for aa, bb, cc, dd, and ee. Preview My AnswersSubmit Answers

Answer 1

Answer:

a = 1/3

b = -3

c = 26/3

d = -6

e = 0

Step-by-step explanation:

Given the quartic polynomial

p(x)=ax⁴+bx³+cx²+dx+e and

p(i) =2i when i=0,1,2,3,4

If i = 0:

p(0) = 2(0)

p(0) = 0

0 = 0+0+0+0+0++e

e = 0

When i = 1

p(1) = 2(1) = 2

2 = a(1)⁴+b(1)³+c(1)²+d(1)+e

2 = a+b+c+d+0

a+b+c+d = 0... (1)

When i = 2, p(2) = 2(2)

p(2) = 4

4 = a(2)⁴+b(2)³+c(2)²+d(2)+e

4 = 16a+8b+4c+2d+0

16a+8b+4c+2d = 4

8a+4b+2c+d = 2... (2)

When i = 3

p(3) = 8

8 = a(3)⁴+b(3)³+c(3)²+d(3)+0

8 = 81a+27b+9c+3d..(3)

When i = 4

p(4) =16

16 = a(4)⁴+b(4)³+c(4)²+d(4)+0

16 = 256a+64b+16c+4d

64a+16b+4c+d = 4...(4)

Solving equation 1 to 4 simultaneously.

Check the attachment for solution.