Question

Every polynomial with complex coefficients can be written as the product of linear factors.
What are the linear factors of
z^5+(-9+5i)z^4+(16-21i)z^3.

Answer 1

$$z^5+(-9+5i)z^4+(16-21i)z^3=z^3(z^2+(-9+5i)z+(16-21i))$$ yes? so immediately we know three of the linear factors are identical \(z\) hence the \(z^3\)

Answer 2

I know that there are four factors in total and I already knew about the zbeing one of them, or is it z^3?

Answer 3

uh there are 5 factors total... are you familiar with the fundamental theorem of algebra? ;-)

Answer 4

the quadratic can be factored by finding its roots... are you familiar with the quadratic formula?$$\begin{align*}z&=\frac{-(-9+5i)\pm\sqrt{(-9+5i)^2-4(16-21i)}}2\\&=\frac{9-5i\pm\sqrt{56-90i-72+84i}}2\\&=\frac{9-5i\pm\sqrt{-8-6i}}2\end{align*}$$we need to simplify this still

Answer 5

Oh yes, so what you are saying is that the z's are to considered to have the 2 parts, real and imaginary?

Answer 6

uh well that's true but no I haven't said anything about that

Answer 7

the key is to simplify \(\sqrt{-8-6i}\) into the form \(a+bi\), so that:$$(a+bi)^2=-8-6i\\(a^2-b^2)+(2ab)i=-8-6i$$

Answer 8

NOW equate the real and imaginary parts:$$a^2-b^2=-8\\2ab=-6$$so \(ab=-3\) and therefore \(b=-3/a\) hence:$$a^2-b^2=-8\\a^2-(-3/a)^2=-8\\a^2-9/a^2=-8$$by inspection we see immediately \(a^2=1\) solves the above hence \(a=1,b=-3\) and \(a=-1,b=3\) appear as candidate solutions. this is why the \(\pm\) is there in the quadratic formula... anyways this gives us \(\sqrt{-8-6i}=1-3i\) hence:$$z=\frac{9-5i\pm(1-3i)}2\\z=\frac{9-5i+1-3i}2=\frac{10-8i}2=5-4i\text{ and }z=\frac{9-5i-1+3i}2=\frac{8-2i}2=4-i$$

Answer 9

hence $$
z^2+(-9+5i)z+(16-21i)=(z-(5-4i))(z-(4-i))$$and therefore$$
z^5+(-9+5i)z^4+(16-21i)z^3=z^3(z-(5-4i))(z-(4-i))$$

Answer 10

http://www.wolframalpha.com/input/?i=z%5E3%28z-%285-4i%29%29%28z-%284-i%29%29

Answer 11

Ok so does that measn I end up with the factors being
5-4i, 4-i, z,z,z ?

Answer 12

@oldrin.bataku

Answer 13

@HazelL There is 5 roots but we say that 5-4i is one root, 4 - i is the other root, and 'z' is the last root with a 'multiplicity' of 3, since it's repeated three times.

Answer 14

The roots with multiplicity of 3 is 0: 5-4i, 4-i, 0,0,0

Answer 15

@genius12 \(z\) is not a root

Answer 16

@HazelL the linear factors are:$$z,\ z,\ z,\ z-(4-i),\ z-(5-4i)$$which correspond to the roots:$$0,0,0,4-i,5-4i$$