The greatest common factor

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This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. Factoring is an essential skill for success in algebra and higher level mathematics courses. Therefore, we have taken great care in developing the student's understanding of the factorization process. The technique is consistently illustrated by displaying an empty set of parentheses and describing the thought process used to discover the terms that are to be placed inside the parentheses.The factoring scheme for special products is presented with both verbal and symbolic descriptions, since not all students can interpret symbolic descriptions alone. Two techniques, the standard "trial and error" method, and the "collect and discard" method (a method similar to the "ac" method), are presented for factoring trinomials with leading coefficients different from 1. Objectives of this module: understand more clearly the factorization process, be able to determine the greatest common factor of two or more terms.

Overview

Factoring Method
Greatest Common Factor

Factoring method

In the last two types of problems (Sections [link] and [link] ), we knew one of the factors and were able to determine the other factor through division. Suppose, now, we’re given the product without any factors. Our problem is to find the factors, if possible. This procedure and the previous two procedures are based on the distributive property.

An equation showing the product of a and the sum of b and c equal to ab plus ac. The product on the left are identified as factors and the expression on the right of the equal sign is identified as the product.

We will use the distributive property in reverse.

$\underset{product}{\underset{︸}{a b + a c}} = \underset{factors}{\underset{︸}{a (b + c)}}$

We notice that in the product, $a$ is common to both terms. (In fact, $a$ is a common factor of both terms.) Since $a$ is common to both terms, we will factor it out and write

$a ()$

Now we need to determine what to place inside the parentheses. This is the procedure of the previous section. Divide each term of the product by the known factor $a .$

$\begin{matrix} \frac{a b}{a} = b & and & \frac{a c}{a} = c \end{matrix}$

Thus, $b$ and $c$ are the required terms of the other factor. Hence,

$a b + a c = a (b + c)$

When factoring a monomial from a polynomial, we seek out factors that are not only common to each term of the polynomial, but factors that have these properties:

The numerical coefficients are the largest common numerical coefficients.
The variables possess the largest exponents common to all the variables.

Greatest common factor

A monomial factor that meets the above two requirements is called the greatest common factor of the polynomial.

Sample set a

Factor $3 x - 18.$

The greatest common factor is 3.

$\begin{array}{l} 3 x - 18 = 3 \cdot x - 3 \cdot 6 & Factor out 3. \\ 3 x - 18 = 3 () & Divide each term of the product by 3. \\ \frac{3 x}{3} = x and \frac{- 18}{3} = - 6 \\ (Try to perform this division mentally .) \\ 3 x - 18 = 3 (x - 6) \end{array}$

Factor $9 x^{3} + 18 x^{2} + 27 x .$

Notice that $9 x$ is the greatest common factor.

$\begin{array}{l} 9 x^{3} + 18 x^{2} + 27 x = 9 x \cdot x^{2} + 9 x \cdot 2 x + 9 x \cdot 3. & Factor out 9 x . \\ 9 x^{3} + 18 x^{2} + 27 x = 9 x () & Mentally divide 9 x into each term of the product . \\ 9 x^{3} + 18 x^{2} + 27 x = 9 x (x^{2} + 2 x + 3) \end{array}$

Factor $10 x^{2} y^{3} - 20 x y^{4} - 35 y^{5} .$

Notice that $5 y^{3}$ is the greatest common factor. Factor out $5 y^{3} .$

$10 x^{2} y^{3} - 20 x y^{4} - 35 y^{5} = 5 y^{3} ()$

Mentally divide $5 y^{3}$ into each term of the product and place the resulting quotients inside the $() .$

$10 x^{2} y^{3} - 20 x y^{4} - 35 y^{5} = 5 y^{3} (2 x^{2} - 4 x y - 7 y^{2})$

Factor $- 12 x^{5} + 8 x^{3} - 4 x^{2} .$

We see that the greatest common factor is $- 4 x^{2} .$

$- 12 x^{5} + 8 x^{3} - 4 x^{2} = - 4 x^{2} ()$

Mentally dividing $- 4 x^{2}$ into each term of the product, we get

$- 12 x^{5} + 8 x^{3} - 4 x^{2} = - 4 x^{2} (3 x^{3} - 2 x + 1)$

Practice set a

Factor $4 x - 48.$

$4 (x - 12)$

Factor $6 y^{3} + 24 y^{2} + 36 y .$

$6 y (y^{2} + 4 y + 6)$

Factor $10 a^{5} b^{4} - 14 a^{4} b^{5} - 8 b^{6} .$

$2 b^{4} (5 a^{5} - 7 a^{4} b - 4 b^{2})$

Factor $- 14 m^{4} + 28 m^{2} - 7 m .$

$- 7 m (2 m^{3} - 4 m + 1)$

Consider this problem: factor $A x + A y .$ Surely, $A x + A y = A (x + y) .$ We know from the very beginning of our study of algebra that letters represent single quantities. We also know that a quantity occurring within a set of parentheses is to be considered as a single quantity. Suppose that the letter $A$ is representing the quantity $(a + b) .$ Then we have

$A x + A y = A (x + y)$

$(a + b) x + (a + b) y = (a + b) (x + y)$

When we observe the expression

$(a + b) x + (a + b) y$

we notice that $(a + b)$ is common to both terms. Since it is common, we factor it out.

$(a + b) ()$

As usual, we determine what to place inside the parentheses by dividing each term of the product by $(a + b) .$

$\begin{matrix} \frac{(a + b) x}{(a + b)} = x & and & \frac{(a + b) y}{(a + b)} = y \end{matrix}$

Thus, we get

$(a + b) x + (a + b) y = (a + b) (x + y)$

This is a forerunner of the factoring that will be done in Section $5.4.$

Sample set b

Factor $(x - 7) a + (x - 7) b .$

Notice that $(x - 7)$ is the greatest common factor. Factor out $(x - 7) .$

$\begin{matrix} (x - 7) a + (x - 7) b & = & (x - 7) (\begin{matrix} \end{matrix}) \\ Then, \frac{(x - 7) a}{(x - 7)} & = & a and \frac{(x - 7) b}{(x - 7)} = b . \\ (x - 7) a + (x - 7) b & = & (x - 7) (a + b) \end{matrix}$

Factor $3 x^{2} (x + 1) - 5 x (x + 1)$ .

Notice that $x$ and $(x + 1)$ are common to both terms. Factor them out. We’ll perform this factorization by letting $A = x (x + 1) .$ Then we have

$\begin{matrix} 3 x A - 5 A & = & A (3 x - 5) \\ But A & = & x (x + 1), so \\ 3 x^{2} (x + 1) - 5 x (x + 1) & = & x (x + 1) (3 x - 5) \end{matrix}$

Practice set b

Factor $(y + 4) a + (y + 4) b .$

$(y + 4) (a + b)$

Factor $8 m^{3} (n - 4) - 6 m^{2} (n - 4) .$

$2 m^{2} (n - 4) (4 m - 3)$

Exercises

For the following problems, factor the polynomials.

$9 a + 18$

$9 (a + 2)$

$6 a + 24$

$8 b + 12$

$4 (2 b + 3)$

$16 x + 12$

$4 x - 6$

$2 (2 x - 3)$

$8 x - 14$

$21 y - 28$

$7 (3 y - 4)$

$16 f - 36$

$12 x^{2} + 18 x$

$6 x (2 x + 3)$

$10 y^{2} + 15 y$

$8 y^{2} + 18$

$2 (4 y^{2} + 9)$

$7 x^{2} - 21$

$3 y^{2} - 6$

$3 (y^{2} - 2)$

$2 x^{2} - 2$

$6 y^{2} - 6 y$

$6 y (y - 1)$

$a x^{2} - a$

$b y^{2} + b$

$b (y^{2} + 1)$

$7 b y^{2} + 14 b$

$5 a^{2} x^{2} + 10 x$

$5 x (a^{2} x + 2)$

$24 a x^{2} + 28 a$

$10 x^{2} + 5 x - 15$

$5 (2 x^{2} + x - 3)$

$12 x^{2} - 8 x - 16$

$15 y^{3} - 24 y + 9$

$3 (5 y^{3} - 8 y + 3)$

$a x^{2} + a x + a$

$b y^{3} + b y^{2} + b y + b$

$b (y^{3} + y^{2} + y + 1)$

$2 y^{2} + 6 y + 4 x y$

$9 x^{2} + 6 x y + 4 x$

$x (9 x + 6 y + 4)$

$30 a^{2} b^{2} + 40 a^{2} b^{2} + 50 a^{2} b^{2}$

$13 x^{2} y^{5} c - 26 x^{2} y^{5} c - 39 x^{2} y^{5}$

$13 x^{2} y^{5} (- c - 3)$

$- 4 x^{2} - 12 x - 8$

$- 6 y^{3} - 8 y^{2} - 14 y + 10$

$- 2 (3 y^{3} + 4 y^{2} + 7 y - 5)$

$A b + A c$

$N x + N y$

$N (x + y)$

$Q x + Q y$

$A x - A y$

$A (x - y)$

$(x + 4) b + (x + 4) c$

$(x - 9) a + (x - 9) b$

$(x - 9) (a + b)$

$(2 x + 7) a + (2 x + 7) b$

$(9 a - b) w - (9 a - b) x$

$(9 a - b) (w - x)$

$(5 - v) X + (5 - v) Y$

$3 x^{5} y^{4} - 12 x^{3} y^{4} + 27 x^{5} y^{3} - 6 x^{2} y^{6}$

$3 x^{2} y^{3} (x^{3} y - 4 x y + 9 x^{3} - 2 y^{3})$

$8 a^{3} b^{15} + 24 a^{2} b^{14} + 48 a^{3} b^{6} - 20 a^{3} b^{7} + 80 a^{4} b^{6} - 4 a^{3} b^{7} + 4 a^{2} b$

$- 8 x^{3} y^{2} - 3 x^{3} y^{2} + 16 x^{4} y^{3} + 2 x^{2} y$

$- x^{2} y (11 x y - 16 x^{2} y^{2} - 2)$

Exercises for review

( [link] ) A quantity plus $21 %$ more of that quantity is $26.25.$ What is the original quantity?

( [link] ) Solve the equation $6 (t - 1) = 4 (5 - s) if s = 2.$

$t = 3$

( [link] ) Given that $4 a^{3}$ is a factor of $8 a^{3} - 12 a^{2},$ find the other factor.

Questions & Answers

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Source: OpenStax, Algebra ii for the community college. OpenStax CNX. Jul 03, 2014 Download for free at http://cnx.org/content/col11671/1.1

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