# Application of synthetic division by second degree polynomials. [closed]

Exercise: Use synthetic division rather than DeMoivre's Theorem to find $(2-i)^7$. Then given $z=2-i$ use synthetic division to find $2z^3-7z^2+5z-3$.

If one finds the remainder $R(z)$ when $F(z)=z^7$ is divided by

$$[z-(2-i)]\cdot[z-(2+i)]=z^2-4z+5$$

one finds that

$$F(z)=(z^2-4z+5)Q(z)+R(z)$$

so

$$F(2-i)=(2-i)^7=0\cdot Q(2-i)+R(2-i)=-278+29i$$

For $$F(z)=(2z^3-7z^2+5z-3)=(z^2-4z+5)Q(z)+R(z)\\$$ and $$R(z)=-i-8\\$$ so $$F(z)=(z^2-4z+5)Q(z)+R(z)\\$$ and $$F(2-i)=0\cdot Q(2-i)-(2-i)-8=-10+i$$

• After $F(z)$ in second line: $-4z$ or $-5z$? – imranfat Feb 29 '16 at 3:47
• Yes, you are correct, everywhere I wrote $z^2-5x+5$ it should be $z^2-4z+5$. Thank you for catching that mistake. The remainders are correct, however since I actually used $z^2-4z+5$ in the calculation. – John Wayland Bales Mar 4 '16 at 18:22