Find an nth-degree polynomial function with real coefficients satisfying the given conditions.
n=4;
2i and 3i are zeros;
f(−1)equals=100

f(x) = 2x⁴ + 26x² + 72

Step-by-step explanation:

Imaginary and complex roots come in conjugate pairs.  So if 2i and 3i are zeros, then -2i and -3i are also zeros.

f(x) = a (x − 2i) (x + 2i) (x − 3i) (x + 3i)

f(x) = a (x² − 4i²) (x² − 9i²)

f(x) = a (x² + 4) (x² + 9)

f(x) = a (x⁴ + 13x² + 36)

f(-1) = 100

100 = a ((-1)⁴ + 13(-1)² + 36)

100 = a (1 + 13 + 36)

100 = 50a

a = 2

f(x) = 2 (x⁴ + 13x² + 36)

f(x) = 2x⁴ + 26x² + 72

65

step-by-step explanation:

the quotient is 4x^2 - 27x + 167 with a remainder of -996

step-by-step explanation:

in synthetic division, use -6 as the divisor and use the coefficients {4, -3, 5, 6}:

-6     )   4     -3     5     6

                -24   162 -1002

          4   -27   167   -996

thus, the quotient is 4x^2 - 27x + 167 with a remainder of -996.

note:   your "4x°-3x+5x+6" is ambiguous / incomplete.   i had to assume that you meant '4x^3 - 3x^2 + 5x + 6."