Why is -(2+i) also a root of a polynomial?

Question

$z^6-7z^4+31z^2-25=0$ has a root $2+i$. I figured that $2-i$ is also a root as complex roots come in conjugate pairs...But:

1) How do I prove that $-(2+i)$ is also a root

2) How do I find other roots? Assuming there are three more.

Answer 1

By the factor theorem, if $2+i$ is a root, then $z^6−7z^4+31z^2−25=0$ when $z=2+i$. So this is how you prove the first part.

For your first question, if you notice that the polynomial only involves even powers of $z$, this means that the negated version of the roots are also roots because when you square them, the $-1$ becomes $1$, they also fit the factor theorem condition.

Since you know four roots out of six ($2+i,\ 2-i,\ -2+i,\ -2-i$), you can try out other values, or use the sum or product of roots formulae to find them. So you know that the sum of roots is $0$ because the coefficient of $z^5$ is zero, this means that your two remaining roots sum to zero as well, because the four roots you found previously cancel each other out.

Answer 2

To solve the equation set $x=z^2$ to get $x^3-7x^2+31x-25=0$. The root $x=1$ is evident. Then $x^3-7x^2+31x-25=(x-1)(x^2-6x+25)=0$. Now solve the quadratic equation $x^2-6x+25=0$. You have $x_1=1$, $x_2=3+4i$ and $x_3=3-4i$. Then solve $z^2=x$ for each $x$.

Answer 3

Part 1) was answered in a comment: since the polynomial has only even terms, you can treat it as a polynomial of $z^2$: $$ (z^2)^3-7(z^2)^2+31(z^2)-25=0.$$ You can make the cubic polynomial more obvious by setting $w=z^2,$ so that $$ w^3-7w^2+31w-25=0.$$

Then each solution gives you a value of $w=z^2,$ but if and if $z$ is a root of the original polynomial then $z^2 = w$ is a root of the cubic, and since $(-z)^2 = w$ as well, $-z$ is also a root of the original polynomial. Since you know $2+i$ is a root, you know that $-(2+i)$ is one too.

For part 2), you already know that the conjugate of each root is a root (since the polynomial has real coefficients). That gives you two more roots, $2-i$ and $-(2-i)$. So the original polynomial is something of the form $$(z - (2+i))(z - (2-i))(z+(2+i))(z+(2-i)) Q(z),$$ where $Q(z)$ is a polynomial. You can multiply the known factors together and find that $$ (z - (2+i))(z - (2-i))(z+(2+i))(z+(2-i)) = z^4 - 6 z^2 + 25.$$ Therefore the original polynomial is actually $$ z^6-7z^4+31z^2-25 = (z^4 - 6 z^2 + 25) Q(z). $$ You can use synthetic division to find $Q(z).$ It's a very easy division; in fact it should be almost immediately obvious that $Q(z) = z^2 - 1$ is the only possible result, since the leading term has to be $z^6/z^4 = z^2$, the final term has to be $-25/25 = -1,$ and there can't be any non-zero term with an odd power of $z.$

The roots of $z^2 - 1$ are very easy to find, and they are the final two roots.