Factoring: Methods and Examples

The factoring is a method through which a polynomial is expressed in the form of multiplication of factors, which can be numbers, letters or both. To factorize the factors that are common to the terms are grouped, and in this way the polynomial is decomposed into several polynomials.

Thus, when the factors multiply each other the result is the original polynomial. Factorization 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 can not be factored because there is no common factor between its terms; thus, these algebraic expressions are divisible only between 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 the following:

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 removed and the factorization is completed.

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

Example 1

Factor (b ² x) + (b ² Y).

Solution

First there is the common factor of each term, which in this case is b ² , and then divide the terms between 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

Factorize (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 elevated to a power. To factor them, first the two terms are broken down into their long form:

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

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

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

Factorization by grouping

As not in all cases the maximum 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 these 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 you get a binomial that is common for both terms. To factor it is multiplied by the remaining factors; That way you have to:

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

Factorization by inspection

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 â???? aâ???? is different from 1. This method is also used when the trinomial has the form x ² ± bx + c and the value of â???? aâ???? = 1

Example 1

Factor x ² + 5x + 6

Solution

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

2 _* 3 = 6

2 + 3 = 5.

In that way, the expression is simplified like this:

(x ² + 2x) + (3x + 6)

Each term is factored:

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

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

Thus, the expression remains:

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

As you have a common binomial, to reduce the expression is multiplied by the leftover terms and you have to:

x ² + 5x + 6 = (x + 2) _* (x + 3)

Example 2

Factor 4a ² + 12a + 9 = 0.

Solution

You have a quadratic trinomial of the form ax ² ± bx + c and to factorize it all the expression is multiplied 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 to ² + 12a (4) + 36 = 0

Now we must find two numbers that, when multiplied together, give as a result the value of"c"(which is 36) and that when added together result in 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 ² to ² = 4a _* 4a. Therefore, the distributive property for each term is applied:

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

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

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

The expression is as follows:

4a ² + 12a +9 = (2a +3) _* (2a + 3)

Factoring with notable products

There are cases in which, to fully factor the polynomials with the previous 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 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 = (a-b) _* (to ² + ab + b ² )

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

Example 1

Factor (5 ² - x ² )

Solution

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

(to 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 we have a perfect square of a sum, because we can identify two terms squared, and the remaining term is the result of multiplying two times the square root of the first term, by the square root of the second term.

to 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 separated by the sign of the operation, and the whole 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 raised to the cube. In order to factor them, the formula of the notable product of the cube difference is applied, which is:

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

Thus, to factorize, the cubic root of each term of the binomial is extracted and multiplied by the square of the first term, plus the product of the first by the second term, plus the second term by the square.

27a ³ - b ³

³â???? (27a ³ ) = 3a

³â???? (- b ³ ) = -b

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

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 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 rest 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 divider. 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 among the factors found for the division to be exact. The polynomial terms are ordered from highest to lowest exponent; in the case that a term with the degree that follows in the sequence is missing, a 0 is placed in its place.

The coefficients are located in a scheme as seen 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 is added vertically with that result that was obtained 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 grade. 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)

To finish, 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, L. H. (1996). Algebra and trigonometry with analytical geometry. Pearson Education.
2. J, V. (2014). How to Teach Kids About Factoring to Polynomial.
3. Manuel Morillo, A. S. (s.f.). Basic Mathematics With Applications.
4. Roelse, P. L. (1997). Linear methods for polynomial factorization over finite fields: theory and implementations. Universität Essen.
5. Sharpe, D. (1987). Rings and Factorization.