1 pt The zeros of Px x5 1823 81x are x1 with multiplicit

(1 pt) The zeros of P(x) = x5 + 182.3 + 81x are x1 = with multiplicity 2 with negative imaginary part, ts multiplicity is ; and with positive imaginary part, its multiplicity is

Solution

1)P(x)=x⁵+18x³+81x=0

x(x⁴+18x²+81)=0

x((x²)²+2*9x²+9²)=0

x(x²+9)²=0

x =0,(x²+9)²=0

x =0,(x²+9)=0

x =0,x²=-9

x =0,x =-3i,x =3i

x₁=0 , multiplicity 1

x₂ =0+3i, multiplicity 2

x₃=0-3i ,multiplicity 2

2)-2,3i ,-3i are zeroes

polynomial is (x-(-2))(x-3i)(x-(-3i))

(x+2)(x-3i)(x+3i)

(x+2)(x²-(3i)²)

(x+2)(x²-(-9))

(x+2)(x²+9)

(x+2)(x²) + (x+2)(9)

(x³+2x²+9x +18) is the polynomial