Unit roots/Factorization and Solving Equations

In this chapter, we will find some use of the roots of unity in factorization and solving equations.

Factorization and solving equations

To solve an equation is to find the set of values of the unknowns satisfying the equation. While we can easily solve equations of lower degrees, it is not easy to solve equations of higher degrees. However, by factorization, we can rewrite a high degree polynomial into a product of low degree polynomials. The method of solving equations be factorization is based on the following theorem:

Zero product property Let $A$ and $B$ be real or complex valued. If $A\cdot B=0$, then $A=0$ or $B=0$.
Proof Either $A=0$ or $A\ne 0$ holds true. If $A=0$, then "$A=0$ or $B=0$" holds true. If $A\ne 0$, then $A^{-1}$ exists. Multiply it to both sides of $A\cdot B=0$:

$A^{-1}\cdot A\cdot B=A^{-1}\cdot 0$

$B=0$.

Then "$A=0$ or $B=0$" also holds true.QED

Therefore, when we solve a polynomial equation, we can first factorize the polynomial and form smaller equations equating the factors with zero.

Example 1 Solve $x^3-x^2-x+1=0$.
Solution

$x^3-x^2-x+1=0$

$x^2(x-1)-(x-1)=0$

$(x-1)(x^2-1)=0$

$(x-1)(x-1)(x+1)=0$

$x-1=0$ or $x-1=0$ or $x+1=0$

$x=1$ or $x=1$ or $x=-1$

$x=1$ or $x=-1$

In the previous chapter, we have seen the use of unit roots in determining the divisibility of a polynomial. In fact we can similarly factorize some polynomials by considering the properties of unit roots.

Example 2 Factorize $x^8+x^6+x^4+x^2+1$.
Anaylsis The indices in the expression are 8, 6, 4, 2, 0. When they are divided by 5, the remainders are 3, 1, 4, 2, 0. Therefore, when $x$ is replaced by a non-real fifth root of unity, the expression equals zero. So the expression is divisible by $x^4+x^3+x^2+x+1$.
Answer $x^8+x^6+x^4+x^2+1=(x^4+x^3+x^2+x+1)(x^4-x^3+x^2-x+1)$

If we allow complex coefficients in the factors, any polynomials can be factorized as a product of linear factors; if we allow any real coefficients, then any polynomials with real coefficients can be factorized as a product of factors of at most the second degree.

Example 3 Factorize
(a) $x^8+x^4+1$ (three factors)
(b) $x^5+x+1$ (two factors)
Analysis We can check that both (a) and (b) is zero when $x$ is replaced by a non-real cube root of unity. However, we have to further find another factor for (a). For this purpose, we first let $y=x^2$ so the expression becomes $y^4+y^2+1$, which also equals zero when $x$ is replaced by a non-real cube root of unity.
Solution (a) $y^4+y^2+1=(y^2+y+1)(y^2-y+1)$. So,

$x^8+x^4+1=(x^4+x^2+1)(x^4-x^2+1)=(x^2+x+1)(x^2-x+1)(x^4-x^2+1)$

(b) $x^5+x+1=(x^5-x^2)+(x^2+x+1)$

$x^2(x-1)(x^2+x+1)+(x^2+x+1)$

$(x^3-x^2+1)(x^2+x+1)$

The cubic equation

Using the cube root of unity, we can derive the formula for cubic equations.

Let $\omega$ be a non-real cube root of unity, then:

$\omega +{\omega }^2=-1$, ${\omega }^3=1$.

Then we can show that:

$(x+y+z)(x+\omega y+{\omega }^{2}z)(x+{\omega }^{2}y+\omega z)=x^3+y^3+z^3-3xyz$.

Therefore, the roots of the cubic equation in $x$:

$x^3-3yz\cdot x+(y^3+z^3)=0$

are:

$x_1=-(y+z)$, $x_2=-(\omega y+{\omega }^{2}z)$, $x_3=-({\omega }^{2}y+\omega z)$.

Now consider the cubic equation:

$x^3+px+q=0$.

We may let $q=y^3+z^3$ and $p=-3yz$. So $\frac{p^3}{27}=-y^3\cdot z^3$. Therefore, $y^3$ and $z^3$ are the roots of the following equation:

$X^2-qX-\frac{p^3}{27}=0$.

The roots of this equation are:

$X=\frac{q}{2}\pm \sqrt{\frac{q^2}{4}+\frac{p^3}{27}}$.

We let $y^3$ be any of these roots, and $z^3$ be the other root. Then:

$y=\sqrt[3]{\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$,

$z=\sqrt[3]{\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$.

(In fact we may also take other non-real cube roots, as long as the relation $p=-3yz$ holds. However, we need not consider the non-real cube roots because we have considered them through the specification of $x_2$ and $x_3$.)

Example 4 Solve $x^3-12x+16=0$.
Solution In the example, $p=-12$, $q=16$. So:

$\frac{q^2}{4}+\frac{p^3}{27}=(\frac{q}{2}{)}^2+(\frac{p}{3}{)}^3=8^2+(-4{)}^3=0$,

$y=z=\sqrt[3]{\frac{q}{2}}=\sqrt[3]{8}=2$.

Therefore, the roots are:

$x_1=-(2+2)=-4$, $x_2=-(2\omega +2{\omega }^2)=2$, $x_3=-(2{\omega }^2+2\omega )=2$.

We can verify that $x^3-12x+16=(x+4)(x-2{)}^2$.

Higher degree equations

We can also use roots of unity to solve some higher degree equations in some special forms.

Example 5 Solve $x^8+x^6+x^4+x^2+1=0$
Solution From the previous example, $x^8+x^6+x^4+x^2+1=P(x)\cdot Q(x)$, where:

$P(x)=x^4+x^3+x^2+x+1$

$Q(x)=x^4-x^3+x^2-x+1$

From $P(x)=0$, we can obtain four roots:

$x^4+x^3+x^2+x+1=0\Rightarrow x_j=\cos \frac{2j\pi }{5}+i\sin \frac{2j\pi }{5},(j=1,2,3,4)$ (i.e. the non-real fifth roots of unity.)

Note that $Q(-x)=(-x{)}^4-(-x{)}^3+(-x{)}^2-(-x)+1=x^4+x^3+x^2+x+1=P(x)$, Therefore, $Q(-x_j)=P(x_j)=0$. So, the other four roots are:

$x{\text{'}}_j=-x_j,(j=1,2,3,4)$.

The original equation has a total of eight roots:

$x=\pm (\cos \frac{2j\pi }{5}+i\sin \frac{2j\pi }{5}),(j=1,2,3,4)$.

Alternative solution Let $y=x^2$, the equation becomes:

$y^4+y^3+y^2+y+1=0$.

The four roots of this equation are:

$y_j=\cos \frac{2j\pi }{5}+i\sin \frac{2j\pi }{5}),(j=1,2,3,4)$

To obtain the corresponding values of $x$, we take the square roots of each of these values of $y$. So, the roots are

$x=\pm (\cos \frac{j\pi }{5}+i\sin \frac{j\pi }{5}),(j=1,2,3,4)$.

Although the form of the roots given by the two methods are different, they are identical set of numbers. Both can be written as:

$x=\cos \frac{j\pi }{10}+i\sin \frac{j\pi }{10},(j=1,2,3,4,6,7,8,9)$.

They are the eight non-real tenth roots of unity.

Example 6 Find a equations with rational coefficients such that its roots equal ${\alpha }^4+{\alpha }^6+{\alpha }^7+{\alpha }^9$, where $\alpha$ is a root of the equation $x^{13}-1=0$.