Chapter 6: Polynomial Factorization

In Chapter 4, polynomials were introduced 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, $a_{n}, a_{n - 1}, \dots, a_{0} \in R$ are its coefficients. For a nonzero polynomial, $a_{n} \neq = 0$ , the number $n \in N_{0}$ is its degree, and $a_{n}$ is its leading coefficient. The zero polynomial, whose coefficients are all zero, is assigned no degree in this book.

In this chapter, we will learn how to manipulate and simplify polynomials in order to better understand their behavior, find their roots, 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.

Basic Factorization

Before exploring general methods, recall that certain algebraic identities, introduced in Chapter 3, can be applied directly to polynomials. These identities often enable quick factorizations of specific expressions without the need for more elaborate techniques.

Definition: Factorization

Factorization is the process of rewriting an expression as a product of simpler factors.

For example, rewriting

$x^{2} - 4$

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

is a factorization because the two factors multiply back to the original polynomial.

Example: A Difference of Squares

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 follows:

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

While simple cases like this can be solved using known identities, most polynomial expressions, particularly trinomials, require a more systematic approach. Before applying any other method, first check whether all terms share a greatest common factor and factor it out. 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.

Method: 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 method.

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.

Method: Factorization by Grouping

Factorization by grouping applies when the terms can be divided into groups that share a common polynomial factor. In the trinomial method above, we apply it to 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:

When grouping is successful, the result has the form

$(a x^{2} + m x) + (n x + c) = G_{1} (x) H (x) + G_{2} (x) H (x),$

where the same factor $H (x)$ appears in both groups.

Step 3: Factor out the common polynomial factor:

$G_{1} (x) H (x) + G_{2} (x) H (x) = H (x) (G_{1} (x) + G_{2} (x)) .$

If the two groups do not produce a common factor, try changing the order or grouping of the terms. If this still does not work, another factoring technique may be needed.

Example: Factoring a Trinomial

To see the trinomial method in a simple case, we will factor the trinomial

$x^{2} + 5 x + 6.$

First identify the coefficients $a = 1$ , $b = 5$ , and $c = 6$ . We need two integers $m$ and $n$ such that

$m \cdot n = 6 and m + n = 5.$

The integers $m = 2$ and $n = 3$ satisfy both conditions, so we rewrite the middle term:

$x^{2} + 5 x + 6 = x^{2} + 2 x + 3 x + 6.$

Now factor by grouping:

$x^{2} + 2 x + 3 x + 6 = x (x + 2) + 3 (x + 2) = (x + 3) (x + 2) .$

The final product shows the factorized form of the original trinomial.

Example: Factoring with a Nonunit Leading Coefficient

Factor

$2 x^{2} + 7 x + 3.$

Here $a \cdot c = 2 \cdot 3 = 6$ . The integers $6$ and $1$ have product $6$ and sum $7$ , so we split the middle term and group:

$2 x^{2} + 7 x + 3 = 2 x^{2} + 6 x + x + 3 = 2 x (x + 3) + 1 (x + 3) = (2 x + 1) (x + 3) .$

Multiplying the factors back together gives $2 x^{2} + 7 x + 3$ , which checks the factorization.

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, another factoring technique may be needed. Failure of the grouping method does not by itself show that no factorization exists.

Example: Grouping a Four-Term Polynomial

Here is a four-term polynomial where grouping immediately reveals a common binomial factor. We want to factor the polynomial

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

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: Grouping and Difference of Squares

In this example, grouping produces a factor that can be simplified once more using the difference of squares identity:

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

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 6: Polynomial Factorization

Basic Factorization

Factoring Trinomials