Solving a cubic polynomial with complex roots

[missmerisha]

Homework Statement

For a cubic polynomial P(x), with real coefficients, P(2+i)=0, P(1)=0 and P(0)=10.
Express P(x) in the form P(x)=ax^3+bx^2+cx+d
and solve the equation P(x)=0

Homework Equations

The conjugate factor theorem

The Attempt at a Solution

Using remainder theorem

When P(2+i) = 0,

P(2+i)=a(2+i)^3+b(2+i)^2+c(2+i)+d

0=2a+3b+2c+d+11ai+4bi+ci


P(1)=0
0= a+b+c+d

P(0)=10
d=10

P(2-i)=0 <--- according to the conjugate theorem

P(2-i) =0
0= 2a+3b+2c+d-11ai-4bi-ci


I have trouble solving this through simultaneous equations. Is there another method?

 

[gabbagabbahey]

If P(1)=P(2+i)=P(2-i)=0, then (x-1),(x-2-i) and (x-2+i) are all factors of P(x). So, you can write
P(x)=A(x-1)(x-2-i)(x-2+i). Then just use the fact that P(0)=10 to solve for A, and finally expand your function to get it into the desired form.