Question

A polynomial P is given. P(x) = x3 + 3x2 + 6x(a) Find all zeros of P, real and complex.x = (b) Factor P completely.P(x) =

Answer 1

P(x) is defined by the expression

`P(x)=x^3+3x^2+6x`

Note

`x^3+3x^2+6x=x(x^2+3x+6)\text{ }`

Therefore, one solution is 0.

The other two solutions come from

`x^2+3x+6=0`

Apply the general solution in order to find complex solutions

`(-3\pm√(3^2-4(1)(6)))/(2(1))=(-3\pm√(-15))/(2)`

The solutions are

`0,\frac{-3+i√(15)\text{ }}{2},(-3-i√(15))/(2)`

We calculate the factor from the solutions, like this

`x=(-3+i√(15))/(2)\Rightarrow x+(3)/(2)-(i√(15))/(2)=0`
`x=(-3-i√(15))/(2)\Rightarrow x+(3)/(2)+(i√(15))/(2)=0`

The factor is

`P(x)=x(x+(3)/(2)-(i√(15))/(2))(x+(3)/(2)+(i√(15))/(2))`