Polynomial Factoring Methods Explained
Case I: Common Factor Factoring
Removing the common factor involves identifying the greatest common factor (GCF) shared by all terms in a polynomial (binomial, trinomial, etc.). The GCF consists of the greatest common divisor of the numerical coefficients and the common variables raised to their smallest exponents. Factoring out the GCF simplifies the polynomial.
Common Monomial Factor
This involves finding a single term (monomial) that is a factor of every term in the polynomial.
Common Factor by Grouping Terms
This technique is often used when not all terms share a common factor, but groups of terms do. It is detailed further in Case II.
…and this holds if the polynomial equals 0, resulting in x. This can be relevant for tetranomials (four-term polynomials).
Common Polynomial Factor
First, we must determine the common factor, which might be a binomial or polynomial itself, not just a monomial. Consider the variables (with the lowest exponent) and coefficients.
For example:
It’s clear that the polynomial (x – y) is repeated, so this will be the common factor. The other factor consists of the remaining parts of the original polynomial:
The factored answer is:
In some cases, you might need to factor out a 1, for example:
This can be treated as:
So the factored answer is:
Case II: Factoring by Grouping Terms
To factor a polynomial by grouping terms, look for shared characteristics among pairs or groups of terms. This method is often applicable when a polynomial has an even number of terms (typically four or more) and no single factor is common to all terms.
A numerical example could be:
You can group them as follows:
Applying the common factor method to each group:
Now, factor out the common binomial factor (a + b):
Case III: Perfect Square Trinomial Factoring
A perfect square trinomial is identified by having three terms where:
- Two terms are perfect squares (e.g., a², b²).
- The remaining term is twice the product of the square roots of the perfect square terms (e.g., 2ab or -2ab).
To factor a perfect square trinomial:
- Arrange the terms if necessary, placing the perfect squares first and third.
- Extract the square root of the first and third terms.
- Write these roots in parentheses, separated by the sign of the middle term.
- Square the entire binomial expression.
The forms are:
Example 1:
Example 2:
Example 3:
Example 4:
Arranging the terms, we have:
Extracting the square root of the first (5x) and last (2y) terms, grouping them in parentheses separated by the sign of the second term (-), and squaring the binomial, we get:
To verify, check if twice the product of the roots (2 * 5x * -2y = -20xy) matches the middle term. If it does, the factorization is correct.
Case IV: Difference of Squares Factoring
This pattern is identified by having two terms that are perfect squares, separated by a minus sign (subtraction).
It is factored into two binomials, one with a minus sign and one with a plus sign (conjugate binomials):
Or, for even exponents in general:
Using a product notation (productoria), we can define a factorization for any exponent, yielding r+1 factors (Note: This advanced notation might refer to repeated factoring or specific contexts):
Example 1:
Example 2: Assume r=2 for this example (likely referring to the power in the general form).
Factoring a difference of squares involves obtaining the square root of each term and representing these as the product of conjugate binomials.
Case V: Perfect Square by Adding & Subtracting
This technique, often related to ‘completing the square’, applies to trinomials that are not perfect squares but can be transformed. It involves adding and subtracting the same term to manipulate the expression into a difference of squares format.
Identify by three terms where the first and third might be perfect squares, but the middle term isn’t correct for a perfect square trinomial. Add and subtract a term to make the first three terms a perfect square trinomial, then factor the resulting difference of squares.
The value added is the same as the value subtracted, so the overall value of the expression does not change.
Note: The parentheses “(xy – xy)” likely represent adding and subtracting the term needed to complete the square, shown here conceptually.