Factorization of complex polynomials

Complex numbers: Complex polynomials

Factorization of complex polynomials

Integers can be factored into primes. Real polynomials can be factured into irreducible factors. The same applies to complex polynomials. On this page we treat the already known notions and theorems from the real case, but now for complex polynomials.

Irreducible polynomials

A polynomial $f(x)$ in $x$ of degree $m\ge1$ can be factored in factors $g(x)$ and $h(x)$ if $f(x) = g(x)\cdot h(x)$ and the degree of both $g(x)$ and $h(x)$ is smaller than $m$ . A factor like $g(x)$ is called a divisor of $f$ . In this case, $f(x)$ is said to be reducible.

If $f(x)$ cannot be factored, then $f(x)$ is called irreducible.

For example: $x^2+4x+3= (x+1)\cdot(x+3)=\dfrac{1}{2}(2x+2)\cdot(x+3)$ . The first factorization and the second are very similar: they differ by a constant.

Constant factors are unequal to $0$ , for otherwise the whole polynomial would be equal to $0$ . Therefore, we can always remove them by division by a constant. The most common standard is to give all factors the leading coefficient $1$ , as was done in the first factorization.

Linear factors (of degree $1$ ) cannot be factored further. Later on we will show that, when it comes to complex polynomials, these are the only ones with that property.

$x^2+4$ is not the product of two linear factors.

The complex polynomial $z^2+4$ is the product of the two linear factors $z-2\ii$ en $z+2\ii$ .

Regarding the real polynomial: let us assume it is a product of linear factors: $x^2+4 = f(x)\cdot g(x)$ , where $f(x)$ and $g(x)$ are polynomials of degree smaller than $2$ . Then both must have degree $1$ . We can choose the leading coefficients of $f(x)$ and $g(x)$ both equal to $1$ . Then there are two numbers $a$ and $b$ , such that $f(x)= x-a$ and $g(x) = x-b$ . Hence, the factorization can be written as $x^2+4=(x-a)\cdot(x-b)$ . Since $(x-a)\cdot(x-b)=x^2-(a+b)\,x+a\cdot b$ , it follows that $a+b=0$ and $a\cdot b=4$ . The first equation gives $b=-a$ and entering this value for $b$ in $a\cdot b=4$ gives $-a^2 = 4$ , a contradiction with the fact that squares are never negative. Hence, there is no factorization of $x^2+4$ .

Regarding the complex polynomial: the factors $z-2\ii$ and $z+2\ii$ indeed are linear and satisfy $(z-2\ii)\cdot(z+2\ii)=z^2-(-2)\cdot2+(2-2)\cdot\ii = z^2+4\tiny.$

Division with remainder for polynomials

Let $f(z)$ be a polynomial of degree $m\ge1$ and $g(z)$ a polynomial of degree $n$ . We can test whether $g(z)$ is a divisor of $f$ as follows. Use $b$ to indicate the leading coefficient of $g(z)$ .

Unique polynomials $q(z)$ and $r(z)$ exist, such that $f(z) = q(z)\cdot g(z) + r(z)$ and the degree of $r(z)$ is smaller than $n$ .

The polynomial $q(z)$ and $r(z)$ can be found as follows:

start with with $q(z) =0$ and $r(z) = f(z)$ ;
subsequently, change and in the following manner as long as the degree of is at least , in which is the leading coeffienct of :
- add $\dfrac{a}{b}z^{k-n}$ with $q(z)$ ,
- subtract the multiple $\dfrac{a}{b}z^{k-n}g(z)$ from $r(z)$ .

If the degree of $r(z)$ is smaller than $n$ , then $q(z)$ and $r(z)$ are found.

The polynomial $g(z)$ is a divisor of $f(z)$ if and only if $r(z) = 0$ .

The procedure for finding the quotient $q(z)$ and remainder $r(z)$ is chosen in such a way, that at the end of each step where $q(z)$ and $r(z)$ change, $f(z) = q(z) g(z) + r(z)$ holds. At each step the degree of $r(z)$ decreases. Since the degree of $r(z)$ in the beginning is the degree of $f(z)$ and during the course is lowered by at least $1$ at each step, the procedure will never take more than $m$ steps.

Assume that $q_1(z)$ and $r_1(z)$ are two polynomials with $f(z) = q_1(z) g(z) + r_1(z)$ such that the degree of $r_1(z)$ is smaller than the degree of $g(z)$ . Then we have $q(z)\cdot g(z)+r(z) = f(z) = q_1(z)\cdot g(z) + r_1(z)$ , from which it follows that $(q(z)-q_1(z))\cdot g(z) = r_1(z) - r(z)$ . However the right hand side contains a polynomial of degree smaller than the degree of $g(z)$ . Therefore the polynomial at the left hand side has degree smaller than $g(z)$ . Since it is a multiple of $g(z)$ , this can only be the case if the multiple is $0$ , that is, $q(z)-q_1(z)= 0$ . This means that $q_1(z) = q(z)$ , and implies $0= (q(z)-q_1(z))\cdot g(z) = r_1(z) - r(z)$ , so $r_1(z) = r(z)$ . We conclude that the quotient $q(z)$ and the remainder $r(z)$ are indeed unique.

If $r(z) = 0$ , then $f(z) = q(z)\cdot g(z)$ is a factorization of $f(z)$ , therefore $g(z)$ is a divisor of $f(z)$ .

Conversely, if $g(z)$ is a divisor of $f(z)$ , then there will be a polynomial $s(z)$ with $f(z) = s(z)\cdot g(z) + 0$ . Because the degree of $0$ is smaller than the degree of $g(z)$ , the uniqueness implies that $s(z) =q(z)$ and $r(z) = 0$ .

Determine the quotient $q(x)$ and the remainder $r(x)$ when dividing with remainder $f(x) = 3 x^3+10 x^2+19 x+8$ by $g(x) = x^2+3 x+5$ .

Do this by entering pairs $\rv{q(x),r(x)}$ such that $f(x) = q(x)\cdot g(x) + r(x)$ and the degree of $r(x)$ keeps getting smaller.

$\rv{q(x),r(x)} =$ $\rv{3 x+1,x+3}$

To see this, we follow the theory and start with $q(x) =0$ and $r(x) = f(x) = 3 x^3+10 x^2+19 x+8$ . The degree of $r(x)$ is bigger than $2$ , the degree of $g(x)=x^2+3 x+5$ . Therefore, we subtract $3 x \cdot g(x)$ from $r(x)$ and add $3 x$ to $q(x)$ . This gives $\rv{q(x),r(x)} = \rv{3 x,x^2+4 x+8}$ The degree of $r(x)$ is not yet smaller than $2$ , so we repeat this process: we subtract $1\cdot g(x)$ from $r(x)$ and add $1$ to $q(x)$ . This gives $\rv{q(x),r(x)} = \rv{3 x+1,x+3}$ Now the degree of $r(x)$ is smaller than $2$ . The conclusion is that we have found the required $q(x)$ and $r(x)$ .

Since $r(x)\ne0$ , the polynomial $g(x)$ is not a divisor of $f(x)$ .

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