Factoring Polynomials
Factoring a polynomial means rewriting a polynomial P(x) as a product of two or more lower-degree polynomials. $$ P_n(x) = A_s(x) \cdot B_t(x) \ \ \ \ \ where \ \ n = s + t $$
A polynomial P(x), in one or more variables, is said to be reducible if it can be expressed as the product of polynomials of lower degree.
If it cannot be factored, it is called an irreducible polynomial.
For irreducible polynomials, the factorization is unique - up to multiplication by constants.
Why does it matter? Factoring is a key technique when working with equations of degree higher than two, as it allows us to find solutions by applying the zero-product property. It also has a wide range of practical applications in solving algebraic problems, whether simple or advanced.
Methods for Factoring Polynomials
There are several techniques for factoring a reducible polynomial, but no one-size-fits-all method.
So, each case needs to be analyzed to determine the most appropriate strategy.
Here are the main factoring techniques:
- Factoring out the greatest common factor (GCF)
If every term in the polynomial contains the same factor (numeric or variable), you can factor it out using the distributive property: $$ xa + xb = x \cdot (a + b) $$Example. All terms in this polynomial share the factor x: $$ 2x^3 - 5x^2 + x $$ Factoring out x yields: $$ x \cdot (2x^2 - 5x + 1) $$ This rewrites the original expression as the product of a first-degree and a second-degree polynomial. In this form, it’s much easier to identify the values of x that make the expression equal zero, thanks to the zero-product rule.
- Partial factoring
If only some of the terms have a common factor, you can group and factor them accordingly: $$ xa + xb - yc = x \cdot (a + b) - yc $$Example. In the polynomial $$ 3x^2 - 4x + 5 $$ the first two terms share the factor x. We can group and factor those: $$ x \cdot (3x - 4) + 5 $$ Here, only part of the expression is factored - this is known as partial factoring.
- Special products
Recognizing special product identities can greatly speed up the factoring process. Some common identities include:- Square of a binomial: $$ (a + b)^2 = a^2 + 2ab + b^2 $$ $$ (a - b)^2 = a^2 - 2ab + b^2 $$
- Cube of a binomial: $$ (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 $$ $$ (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 $$
- Difference of squares: $$ a^2 - b^2 = (a + b)(a - b) $$
- Sum and difference of cubes: $$ x^3 - a^3 = (x - a)(x^2 + ax + a^2) $$ $$ x^3 + a^3 = (x + a)(x^2 - ax + a^2) $$
- Square of a trinomial: $$ (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc $$
- Factoring a special quadratic: $$ x^2 + (a + b)x + ab = (x + a)(x + b) $$
Example. This polynomial is clearly a perfect square: $$ x^2 - 4x + 2 $$ We reverse the square-of-a-binomial formula: $$ x^2 - 4x + 2 = (x - 2)^2 $$ Since squaring a binomial means multiplying it by itself, we can also write: $$ x^2 - 4x + 2 = (x - 2)^2 = (x - 2)(x - 2) $$ This gives the polynomial in its fully factored and irreducible form.
- Ruffini’s Rule
A polynomial P(x) can be factored as (x − k)·Q(x) if k is a root of the polynomial - that is, if P(k) = 0. $$ P(x) = (x - k) \cdot Q(x) $$Example. Consider the polynomial: $$ x^2 - 4x + 3 $$ We look for a root, or zero, such that: $$ x^2 - 4x + 3 = 0 $$ Since the leading coefficient is 1, we test the integer factors of the constant term, which is 3: ±1 and ±3. Trying x = 1 gives: $$ P(1) = 1 - 4 + 3 = 0 $$ So x = 1 is a root. We now write: $$ x^2 - 4x + 3 = (x - 1) \cdot Q(x) $$ To find Q(x), we perform polynomial division (using synthetic division): $$ \begin{array}{c|lc|r} & 1 & -4 & 3 \\ 1 & & 1 & -3 \\ \hline & 1 & -3 & 0 \end{array} $$ The quotient is Q(x) = x - 3, so the complete factorization is: $$ x^2 - 4x + 3 = (x - 1)(x - 3) $$
Verification. To check the result, multiply the two factors: $$ (x - 1)(x - 3) = x^2 - 3x - x + 3 = x^2 - 4x + 3 $$ Since we get back the original polynomial, the factorization is correct.
A Practical Example
Example 1
Let’s consider a fourth-degree polynomial (n = 4):
$$ x^4 - 1 $$
This expression can be factored into the product of two quadratics:
$$ x^4 - 1 = (x^2 - 1)(x^2 + 1) $$
Each factor, (x2 − 1) and (x2 + 1), is of degree 2, which is lower than the degree of the original polynomial.
The sum of the degrees of the two factors (2 + 2 = 4) matches the degree of the original polynomial.
Verification. To confirm the factorization, we multiply the two quadratics: $$ (x^2 - 1)(x^2 + 1) = x^4 + x^2 - x^2 - 1 = x^4 - 1 $$ The result matches the original expression, so the factorization is correct.
But the process doesn’t end here. The binomial (x2 − 1) can itself be factored further as:
$$(x - 1)(x + 1)$$
Verification. A quick check: $$ (x - 1)(x + 1) = x^2 + x - x - 1 = x^2 - 1 $$
We can now rewrite the original polynomial as the product of three factors:
$$ x^4 - 1 = (x^2 - 1)(x^2 + 1) = (x - 1)(x + 1)(x^2 + 1) $$
This is the complete factorization, since all remaining factors are irreducible over the real numbers.
The factored form consists of three polynomials: two binomials of degree 1, and one binomial of degree 2.
$$ x^4 - 1 = (x - 1)(x + 1)(x^2 + 1) $$
As expected, the sum of the degrees (1 + 1 + 2 = 4) equals the degree of the original polynomial.
Why factor the polynomial? In its factored form, it’s much easier to determine when the original expression x4 − 1 is equal to zero: $$ x^4 - 1 = (x - 1)(x + 1)(x^2 + 1) $$ Using the zero-product property, we find the solutions by setting each factor equal to zero. The factor (x − 1) is zero when x = 1. The factor (x + 1) is zero when x = −1. The factor (x2 + 1) has no real roots, since x2 + 1 = 0 has no real solutions. Therefore, the original polynomial is equal to zero only when x = 1 or x = −1 (in ℝ).
Example 2
Let’s now examine whether the polynomial x2 + 1 is reducible:
$$ x^2 + 1 $$
This is an example of an irreducible polynomial over the real numbers, meaning it cannot be expressed as the product of two or more lower-degree real polynomials.
Verification. Let’s test some possible factorizations: $$ (x + 1)(x + 1) = x^2 + x + x + 1 = x^2 + 2x + 1 \ne x^2 + 1 $$ $$ (x + 1)(x - 1) = x^2 + x - x - 1 = x^2 - 1 \ne x^2 + 1 $$ None of these products yield the original polynomial, so we conclude that x2 + 1 is irreducible over ℝ.
And so on.
