# Complex (non-real) factors of a polynomial

• Feb 18th 2011, 11:37 PM
iFuuZe
Complex (non-real) factors of a polynomial
Hi the question says If z-3i is a factor of 2z^4-4z^3+21z^2-36z+27

I tried doing long division by dividing it by z-3i but i get a perfect zero as a answer so i cant do much with that, is their another way i can approach this.? and im working in the chapter factorasation of polynomials of C.

thanks
• Feb 18th 2011, 11:58 PM
DrSteve
Complex roots come in conjugate pairs. Thus, since \$\displaystyle 3i\$ is a root, so is \$\displaystyle -3i\$. Thus \$\displaystyle z-3 \$ and \$\displaystyle z+3 \$ are factors. So the polynomial is divisible by \$\displaystyle (z-3i)(z+3i)=z^2+9\$.

You can now perform long division to get a quadratic polynomial. You can then find the roots of this quadratic polynomial by completing the square or using the quadratic formula.

Remark: Instead of doing long division with \$\displaystyle z^2+9\$, you can perform synthetic division twice with \$\displaystyle 3i\$ and \$\displaystyle -3i\$.
• Feb 18th 2011, 11:58 PM
Prove It
Since all the coefficients of your polynomial are real, you know that complex roots occur as conjugates.

Since \$\displaystyle \displaystyle z - 3i\$ is a factor, so is \$\displaystyle \displaystyle z + 3i\$, which means \$\displaystyle \displaystyle (z - 3i)(z + 3i) = z^2 + 9\$ is a factor.

So divide your polynomial by \$\displaystyle \displaystyle z^2 + 9\$ to get the remaining quadratic factor, which you can then factorise further into two linear factors.
• Feb 19th 2011, 02:56 PM
SammyS
Hello iFuuZe,

If you mean that you get zero as the remainder, that's good.