FACTORING: METHODS AND EXAMPLES - MATHS - 2025

The factorization is a method by which a polynomial is expressed as multiplication factors, which may be numbers or letters or both. To factor, the factors that are common to the terms are grouped together, and in this way the polynomial is decomposed into several polynomials.

Thus, when the factors are multiplied together, the result is the original polynomial. Factoring is a very useful method when you have algebraic expressions, because it can be converted into the multiplication of several simple terms; for example: 2a ² + 2ab = 2a _* (a + b).

There are cases in which a polynomial cannot be factored because there is no common factor between its terms; thus, these algebraic expressions are divisible only by themselves and by 1. For example: x + y + z.

In an algebraic expression, the common factor is the greatest common divisor of the terms that compose it.

Factoring methods

There are several factoring methods, which are applied depending on the case. Some of these are as follows:

Factoring by common factor

In this method those factors that are common are identified; that is, those that are repeated in the terms of the expression. Then the distributive property is applied, the greatest common divisor is taken, and the factoring is completed.

In other words, the common factor of the expression is identified and each term is divided by it; The resulting terms will be multiplied by the greatest common divisor to express the factorization.

Example 1

Factor (b ² x) + (b ² y).

Solution

First you find the common factor of each term, which in this case is b ², and then divide the terms by the common factor as follows:

(b ² x) / b ² = x

(b ² y) / b ² = y.

The factorization is expressed, multiplying the common factor by the resulting terms:

(b ² x) + (b ² y) = b ² (x + y).

Example 2

Factor (2a ² b ³) + (3ab ²).

Solution

In this case we have two factors that are repeated in each term that are "a" and "b", and that are raised to a power. To factor them, the two terms are first decomposed in their long form:

2 _* a _* a _* b _* b _* b + 3a _* b _* b

It can be seen that factor "a" is repeated only once in the second term, and factor "b" is repeated twice in this; so in the first term only 2 remains, a factor "a" and a factor "b"; while in the second term only 3 remains.

Therefore, the times that "a" and "b" are repeated are written and multiplied by the factors that are left over from each term, as shown in the image:

Grouping factoring

As not in all cases the greatest common divisor of a polynomial is clearly expressed, it is necessary to do other steps to be able to rewrite the polynomial and thus factor.

One of those steps is to group the terms of the polynomial into several groups, and then use the common factor method.

Example 1

Factor ac + bc + ad + bd.

Solution

There are 4 factors where two are common: in the first term it is «c» and in the second it is «d». In this way the two terms are grouped and separated:

(ac + bc) + (ad + bd).

Now it is possible to apply the common factor method, dividing each term by its common factor and then multiplying that common factor by the resulting terms, like this:

(ac + bc) / c = a + b

(ad + bd) / d = a + b

c (a + b) + d (a + b).

Now we get a binomial that is common for both terms. To factor it, it is multiplied by the remaining factors; that way you have to:

ac + bc + ad + bd = (c + d) _* (a + b).

Inspection factoring

This method is used to factor quadratic polynomials, also called trinomials; that is, those that are structured as ax ² ± bx + c, where the value of “a” is different from 1. This method is also used when the trinomial has the form x ² ± bx + c and the value of “a” = 1.

Example 1

Factor x ² + 5x + 6.

Solution

We have a quadratic trinomial of the form x ² ± bx + c. To factor it, you must first find two numbers that, when multiplied, give as a result the value of «c» (that is, 6) and that their sum is equal to the coefficient «b», which is 5. Those numbers are 2 and 3:

2 _* 3 = 6

2 + 3 = 5.

In this way, the expression is simplified like this:

(x ² + 2x) + (3x + 6)

Each term is factored:

- For (x ² + 2x) the common term is taken: x (x + 2)

- For (3x + 6) = 3 (x + 2)

Thus, the expression is:

x (x +2) + 3 (x +2).

Since we have a binomial in common, to reduce the expression we multiply this by the remaining terms and we have to:

x ² + 5x + 6 = (x + 2) _* (x + 3).

Example 2

Factor 4a ² + 12a + 9 = 0.

Solution

We have a quadratic trinomial of the form ax ² ± bx + cy to factor it, multiply the entire expression by the coefficient of x ²; in this case, 4.

4a ² + 12a +9 = 0

4a ² (4) + 12a (4) + 9 (4) = 0 (4)

16 a ² + 12a (4) + 36 = 0

4 ² a ² + 12a (4) + 36 = 0

Now we must find two numbers that, when multiplied by each other, give as a result the value of "c" (which is 36) and which, when added, give as a result the coefficient of the term "a", which is 6.

6 _* 6 = 36

6 + 6 = 12.

In this way the expression is rewritten, taking into account that 4 ² a ² = 4a _* 4a. Therefore, the distributive property applies for each term:

(4a + 6) _* (4a + 6).

Finally, the expression is divided by the coefficient of a ²; that is, 4:

(4th + 6) _* (4th + 6) / 4 = ((4th + 6) / 2) _* ((4th + 6) / 2).

The expression is as follows:

4a ² + 12a +9 = (2a +3) _* (2a + 3).

Factoring with notable products

There are cases where, to fully factor the polynomials with the above methods, it becomes a very long process.

That is why an expression can be developed with the formulas of the remarkable products and thus the process becomes simpler. Among the most widely used notable products are:

- Difference of two squares: (a ² - b ²) = (a - b) _* (a + b)

- Perfect square of a sum: a ² + 2ab + b ² = (a + b) ²

- Perfect square of a difference: a ² - 2ab + b ² = (a - b) ²

- Difference of two cubes: a ³ - b ³ = (ab) _* (a ² + ab + b ²)

- Sum of two cubes: a ³ - b ³ = (a + b) _* (a ² - ab + b ²)

Example 1

Factor (5 ² - x ²)

Solution

In this case there is a difference of two squares; therefore the remarkable product formula applies:

(a ² - b ²) = (a - b) _* (a + b)

(5 ² - x ²) = (5 - x) _* (5 + x)

Example 2

Factor 16x ² + 40x + 25 ²

Solution

In this case, you have a perfect square of a sum, because you can identify two terms squared, and the term that remains is the result of multiplying two by the square root of the first term, by the square root of the second term.

a ² + 2ab + b ² = (a + b) ²

To factor only the square roots of the first and third terms are calculated:

√ (16x ²) = 4x

√ (25 ²) = 5.

Then the two resulting terms are expressed separated by the sign of the operation, and the entire polynomial is squared:

16x ² + 40x + 25 ² = (4x + 5) ².

Example 3

Factor 27a ³ - b ³

Solution

The expression represents a subtraction in which two factors are cubed. To factor them, the formula for the notable product of the difference of cubes is applied, which is:

a ³ - b ³ = (ab) _* (a ² + ab + b ²)

Thus, to factor, the cube root of each term of the binomial is taken and multiplied by the square of the first term, plus the product of the first by the second term, plus the second term squared.

27a ³ - b ³

³√ (27a ³) = 3a

³√ (-b ³) = -b

27a ³ - b ³ = (3a - b) _*

27a ³ - b ³ = (3a - b) _* (9a ² + 3ab + b ²)

Factoring with Ruffini's rule

This method is used when you have a polynomial of degree greater than two, in order to simplify the expression to several polynomials of lesser degree.

Example 1

Factor Q (x) = x ⁴ - 9x ² + 4x + 12

Solution

First we look for the numbers that are divisors of 12, which is the independent term; These are ± 1, ± 2, ± 3, ± 4, ± 6, and ± 12.

Then the x is replaced by these values, from lowest to highest, and thus it is determined with which of the values ​​the division will be exact; that is, the remainder must be 0:

x = -1

Q (-1) = (-1) ⁴ - 9 (-1) ² + 4 (-1) + 12 = 0.

x = 1

Q (1) = 1 ⁴ - 9 (1) ² + 4 (1) + 12 = 8 ≠ 0.

x = 2

Q (2) = 2 ⁴ - 9 (2) ² + 4 (2) + 12 = 0.

And so on for each divisor. In this case, the factors found are for x = -1 and x = 2.

Now the Ruffini method is applied, according to which the coefficients of the expression will be divided by the factors found so that the division is exact. The polynomial terms are ordered from highest to lowest exponent; in the case that a term with the next degree is missing in the sequence, a 0 is placed in its place.

The coefficients are located in a scheme as shown in the following image.

The first coefficient is lowered and multiplied by the divisor. In this case, the first divisor is -1, and the result is placed in the next column. Then the value of the coefficient with that result that was obtained is added vertically and the result is placed below. In this way the process is repeated until the last column.

Then the same procedure is repeated again, but with the second divisor (which is 2) because the expression can still be simplified.

Thus, for each root obtained the polynomial will have a term (x - a), where "a" is the value of the root:

(x - (-1)) _* (x - 2) = (x + 1) _* (x - 2)

On the other hand, these terms must be multiplied by the remainder of Ruffini's rule 1: 1 and -6, which are factors that represent a degree. In this way the expression that is formed is: (x ² + x - 6).

Obtaining the result of the factorization of the polynomial by the Ruffini method is:

x ⁴ - 9x ² + 4x + 12 = (x + 1) _* (x - 2) _* (x ² + x - 6)

Finally, the polynomial of degree 2 that appears in the previous expression can be rewritten as (x + 3) (x-2). Therefore, the final factorization is:

x ⁴ - 9x ² + 4x + 12 = (x + 1) _* (x - 2) _* (x + 3) _* (x-2).

References

1. Arthur Goodman, LH (1996). Algebra and trigonometry with analytical geometry. Pearson Education.
2. J, V. (2014). How to Teach Kids About Factoring a Polynomial.
3. Manuel Morillo, AS (sf). Basic Mathematics With Applications.
4. Roelse, PL (1997). Linear methods for polynomial factorization over finite fields: theory and implementations. Universität Essen.
5. Sharpe, D. (1987). Rings and Factorization.