por The common factor factoring of an algebraic expression consists of determining two or more factors whose product is equal to the proposed expression. In this way, looking for the common factor, the factorization process always begins.
To do this, it is observed if there is a common term, which can be both letters and numbers. In the case of letters, the literals common to all the terms that have the smallest exponent are taken as a common factor, and for numbers, the greatest common divisor (GCD) of all the coefficients is calculated.
The product of both common factors, as long as it is different from 1, will be the common factor of the expression. Once found, by dividing each term by said factor, the definitive factorization is established.
Here is an example of how to do it, by factoring this trinomial:
4×5-12×3+8×2
It is seen that all the terms contain the literal “x”, whose lowest power is x2. As for the numerical coefficients: 4, -12 and 8 are all multiples of 4. Therefore the common factor is 4×2.
Once the factor is found, each term of the original expression is divided between it:
4×5 / 4×2 = x3
-12×3 / 4×2 = -3x
8×2/ 4×2 = 2
Finally, the expression is rewritten as the product of the common factor and the sum of the results of the previous operations, as follows:
4×5-12×3+8×2 = 4×2 (x3 – 3x +2)
[toc]
How to factor when there is no common factor
If the common factor is not obvious as in the previous example, it is still possible to factor by looking carefully at the expression, to see if one of the following methods can be implemented:
Difference of two perfect squares
It is a binomial expression of the form:
a2–b2
Which can be factored by applying the notable product:
a2 – b2 = (a+b)⋅(ab)
The procedure is the next:
-First extract the square root of each of the perfect squares.
-Then form the product between the sum of said roots and their difference, as indicated.
perfect square trinomial
The trinomials of the form:
x2 ± 2a⋅x + a2
They are factored by the notable product:
(x+a)2 = x2 ± 2a⋅x + a2
To apply this factorization, it is necessary to verify that the trinomial does indeed have two perfect squares, and that the remaining term is twice the product of the square roots of said values.
Trinomial of the form x2 + mx + n
If the trinomial to be factored does not have two perfect squares, try to write it as the product of two terms:
x2 + mx + n = x2 + (a + b)x + ab = (x + a)(x + b)
Where it must always be fulfilled that:
n = a⋅b
m = a+b
Factorization by grouping of terms
Sometimes the expression to be factored does not have a common factor, nor does it correspond to any of the cases described above. But if the number of your terms is even, you can try this procedure:
-Group pairs that have a common factor.
-Factorize each pair using a common factor, in such a way that the terms between parentheses are equal, that is, so that the parentheses are in turn a common factor. If the chosen grouping does not work, you have to try another combination to find it.
-The factorization sought is the product of the terms inside the parentheses by the common factors of each pair.
The examples that follow will help to clarify the cases discussed.
examples
Factor the following algebraic expressions:
a) 6ab2 – 18a2b3
This is an example of a common factor. Starting with the literal part, the letters a and b are present in both terms. For the variable «a», the smallest exponent is 1 and it is in the term 6ab2, while for the letter «b» the smallest exponent is b2.
So ab2 is a common factor in the original expression.
As for the numbers, there are 6 and -18, the latter is a multiple of 6, since -18 = -(6 × 3). Therefore, 6 is the numerical coefficient of the common factor, which multiplied with the literal part results in:
6ab2
Now divide each original term by this common factor:
6ab2 ÷ 6ab2 = 1
(-18a2b3)÷ 6ab2 = -3ab
Finally, the original expression is rewritten as a product between the common factor and the algebraic sum of the terms found in the previous step:
6ab2 – 18a2b3 = 6ab2 ⋅(1–3ab)
b) 16×2 – 9
This expression is a difference of perfect squares, therefore, by extracting the square root of both terms, we obtain, respectively:
√(16×2) = 4x
√9 = 3
The original expression is written as the product of the sum of these square roots by their difference:
16×2 – 9 = (4x+3)(4x-3)
c) z2 + 6z + 8
It is a trinomial of the form x2 + mx + n, since 8 is not the perfect square of another integer, so it is necessary to find two numbers a and b such that they satisfy simultaneously:
By trial and error, that is, trying, the numbers sought are 4 and 2, since:
4×2 = 8 and 4 + 2 = 6
So:
z2 + 6z + 8 = (z+4)⋅(z+2)
The reader can verify, by applying the distributive property on the right hand side of the equality, that both expressions are equivalent.
d) 2×2 – 3xy – 4x + 6y
This expression is a candidate for factorization by grouping of terms, since there is no common factor evident to the naked eye and it also has an even number of terms.
It is grouped as follows, knowing that the order of the addends does not alter the sum:
2×2 – 3xy + 4x – 6y = (2×2 –3xy) + (4x–6y)
Each parenthesis has its own common factor:
(2×2 – 3xy) + (4x–6y) = x(2x–3y) + 2(2x–3y)
The definitive common factor has already been revealed: it is the parenthesis that is repeated in both terms (2x -3y).
Now it can be factored again:
x(2x–3y)÷ (2x–3y) = x
2(2x–3y) ÷ (2x–3y) = 2
Therefore:
2×2 – 3xy + 4x – 6y = (2x–3y)(x+2)
Again, the reader can apply the distributive property to the right of the equality, to check the equality.
solved exercises
Factorize:
a) y2 – 10y + 25
b) 4×2 + 12xy + 9y2
c) x2 + 5x – 14
d) 3a4 + a3 + 15a + 5
Solution to
It is a perfect square trinomial, it begins by finding the square root of the perfect square terms:
√ (y2) = y
√25 = 5
It is verified that the center term is the double product of these two:
10y = 2. 5. y
And the factorization sought is:
y2 – 10y + 25 = (y-5)2
solution b
The expression is also a perfect square trinomial:
√ (4×2) = 2x
√ (9y2) = 3y
The central term is verified:
12xy = 2⋅2x⋅3y
Finally:
4×2 + 12xy + 9y2 = (2x+3y)2
solution c
The problem is a trinomial of the type x2 + mx + n:
n = a⋅b = -14 = 7 x (- 2)
m = a+b = 5 = 7 + (- 2) = 5
The appropriate numbers are 7 and -2:
x2 + 5x – 14 = (x +7)(x – 2)
solution d
3a4 + a3 + 15a + 5 = (3a4 + a3) + (15a + 5)
The common factor of (3a4 + a3) is a3 and that of (15a + 5) is 5, being grouped like this:
(3a4 + a3) + (15a + 5) = a3 (3a+1) + 5(3a+1) = (3a+1)(a3 + 5)
References
Baldor, A. 2005. Algebra. Cultural Homeland Group.
Larson, R. 2012. Precalculus. 8th Edition. Cengage Learning.
MathWorld. Factorization. Retrieved from: mathworld.wolfram.com.
MathWorld. Polynomial factorization. Retrieved from: mathworld.wolfram.com.
Stewart, J. 2007. Precalculus: Mathematics for Calculus. 5th. Edition. Cengage Learning.
Zill, D. 1984. Algebra and Trigonometry. McGraw Hill.