Chapter 4: Polynomial Factorization

In the previous chapter, we introduced polynomials as a fundamental class of functions that can be written in the general form:

$f (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0}$

A polynomial consists of terms involving a variable (here, $x$ ) raised to non-negative integer powers and multiplied by constant coefficients. Formally, $n \in N_{0}$ denotes the degree of the polynomial, and $a_{n}, a_{n - 1}, \dots, a_{0} \in R$ are its coefficients, with $a_{n}$ being the leading coefficient.

In this chapter, we will learn how to manipulate and simplify polynomials in order to better understand their behavior, find their zeros, and analyze their graphs.
A key step in this process is factorization, which allows us to rewrite a polynomial as a product of simpler factors.

Before exploring general methods, recall that certain algebraic identities, introduced in Chapter 2, can be applied directly to polynomials. These identities often enable quick factorizations of specific expressions without the need for more elaborate techniques.

Example: The Square of a Difference

Suppose we want to factor the following polynomial:

$x^{2} - 4$

This expression simply matches the difference of squares identity, so we apply it directly as follow:

$x^{2} - 4 = x^{2} - 2^{2} = (x - 2) (x + 2)$

While simple cases like this can be solved using known identities, most polynomial expressions, particularly trinomials, require a more systematic approach. Let us now turn to the process of factoring trinomials.

Factoring Trinomials

One of the most common and useful techniques in algebra is factoring a trinomial, i.e., an expression with three terms, typically of the form:

$a x^{2} + b x + c$

The goal of factoring is to rewrite the trinomial as a product of two binomials:

$a x^{2} + b x + c = (p x + q) (r x + s)$

Here $p$ , $q$ , $r$ , and $s$ are real coefficients chosen so that the product on the right expands back to the original expression on the left-hand side.

Algorithm 1: Factoring Trinomials

This method assumes the coefficients $a$ , $b$ , and $c$ of the trinomial are integers.

To factor a trinomial of the form:

$a x^{2} + b x + c$

Step 1: Identify the coefficients:

$a$ is the coefficient of $x^{2}$
$b$ is the coefficient of $x$
$c$ is the constant term

Step 2: Find two integers $m, n \in Z$ such that:

$m \cdot n = a \cdot c$
$m + n = b$

Step 3: Rewrite the middle term $b x$ as $m x + n x$ , giving a four-term polynomial: $a x^{2} + m x + n x + c$

Step 4: Proceed to factor by grouping (described in the next Algorithm 2).

Once the middle term has been split, the trinomial becomes a four-term polynomial. The next step is to apply the factorization by grouping method, a general strategy for breaking down such polynomials into products of simpler factors.

Algorithm 2: Factorization by Grouping

This method assumes the coefficients $a$ , $m$ , $n$ , and $c$ of the four-term polynomial are integers, where $m$ and $n$ are the integers identified in the previous algorithm.

To factor the four-term polynomial:

$a x^{2} + m x + n x + c$

Step 1: Group the terms into two pairs: $(a x^{2} + m x) + (n x + c)$

Step 2: Factor out the greatest common factor (GCF) from each group:

For $(a x^{2} + m x)$ , factor out $r x$ : $r x (p x + q) such that r x \cdot p x = a x^{2} and r x \cdot q = m x$
For $(n x + c)$ , factor out $s$ : $s (p x + q) such that s \cdot p x = n x and s \cdot q = c$ Here, $r$ and $s$ are constants obtained from factoring each group, and $p$ , and $q$ are the coefficients in the common binomial.

Step 3: Check for a common binomial factor:

If both groups contain the same binomial $(p x + q)$ , factor it out: $(p x + q) (r x + s)$
Otherwise, if no common binomial appears, try a different grouping or another factoring technique.

Warning: Limitations of the Grouping Method

While factoring by grouping is a useful technique, it does not always work.

If no common factor (such as a binomial) appears after grouping, the expression cannot be simplified by this method, and other factoring techniques should be considered instead.

In some cases, particularly when the coefficients are irrational, complex, or when no integer factorization exists, no factoring method may succeed. When this occurs, the polynomial is said to be prime, meaning that it cannot be factored further over the number set under consideration (for example, the integers or the real numbers).

Example 1: Factorization by Grouping

Factor the following polynomial using factorization by grouping (Algorithm 2):

$x^{2} - 5 x + 2 x - 10$

Step 1: Group the terms into two pairs to prepare for factoring.

$(x^{2} - 5 x) + (2 x - 10)$

Step 2: Factor out the greatest common factor from each group.

$x (x - 5) + 2 (x - 5)$

Step 3: Factor out the common binomial.

$(x + 2) (x - 5)$

Example 2: Factorization by Grouping

Factor the following polynomial using factorization by grouping (Algorithm 2):

$x^{3} + 3 x^{2} - x - 3$

Step 1: Group the terms into two pairs to prepare for factoring.

$(x^{3} + 3 x^{2}) + (- x - 3)$

Step 2: Factor out the greatest common factor from each group.

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

Step 3: Factor out the common binomial.

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

The remaining quadratic $x^{2} - 1$ can be factored further using the difference of squares identity:

$x^{2} - 1 = (x - 1) (x + 1)$

If we substitute this back into the expression, then we get:

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

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Mathematics Brush-up for Data Science

Chapter 4: Polynomial Factorization

Factoring Trinomials