23 fx 4x4 3x3 20x2 x k 24 fx 2x4 x3 15x2 3x k 3 25 fx 3x4 4x

23. f(x) 4x4 3x3 20x2 x; k 24. f(x) 2x4 x3 15x2 3x: k -3 25. f(x) 3x4 4x3 10x2 15; k 26. f(x) 5x4 x3 2x2 3x 1: k 1 For each polynomial function, use the remainder theorem and synthetic division to find f(k). See Example 2. 27. f(x) 28. f(x) x2 5x 6: k 30. f(x) x3 8x2 63; K 4 3 3x 32, f(x) 2x 4x2 2 x 1: k 1. f(x) 3. f(r) 2x5 10x 19x 50: k 3 x4 6x3 9x 2 3x 3; 4 A. f(x) 3 8x2 5 x 6; kei f(x 15x; k 3 31x 38. f(x) 2 5 x 1; ke 2 i f(x)

Solution

Your question is not clear
I think you just need to find f(k)

30)
f(x) = -x^3 + 8*x^2 + 63
f(k) = f(4)
= -4^3 + 8*4^2 + 63
= -64 + 128 + 63
= 127

Answer: 127