algebra 2 problem

Question

 Degree 5; zeros:7; -I;-9+1 what is the polynomial? F(x) = a(?)

Answer 1

Assuming that f(x) has real coefficients, complex roots occur in conjugate pairs.  So, the roots of f(x) are 7, -i, i, -9+i, -9-i.

 

f(x) = (x-7)[x-i][x-(-i)][x-(-9+i)][x-(-9-i)]

 

      = (x-7)(x-i)(x+i)[(x+9)-i][(x+9)+i]

 

      = (x-7)(x²-i²)[(x+9)²-i²]

 

      = (x-7)(x²+1)(x²+18x+81+1)

 

      = (x-7)(x²+1)(x²+18x+82)

 

      = (x-7)(x⁴+18x³+82x²+x²+18x+82)

 

      = (x-7)(x⁴+18x³+83x²+18x+82)

 

      = x⁵+18x⁴+83x³+18x²+82x-7x⁴-126x³-581x²-126x-574

 

      = x⁵+11x⁴-43x³-563x²-44x-574