# 6.1 Finding the factors of a monomial

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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: be reminded of products of polynomials, be able to determine a second factor of a polynomial given a first factor.

## Overview

• Products of Polynomials
• Factoring

## Products of polynomials

Previously, we studied multiplication of polynomials (Section [link] ). We were given factors and asked to find their product , as shown below.

Given the factors 4and 8, find the product. $4\cdot 8=32$ . The product is 32.

Given the factors $6{x}^{2}$ and $2x-7$ , find the product.

$6x^{2}\left(2x-7\right)=12x^{3}-42x^{2}$

The product is $12{x}^{3}-42{x}^{2}$ .

Given the factors $x-2y$ and $3x+y$ , find the product.

$\begin{array}{lll}\left(x-2y\right)\left(3x+y\right)\hfill & =\hfill & 3{x}^{2}+xy-6xy-2{y}^{2}\hfill \\ \hfill & =\hfill & 3{x}^{2}-5xy-2{y}^{2}\hfill \end{array}$

The product is $3{x}^{2}-5xy-2{y}^{2}$ .

Given the factors $a+8$ and $a+8$ , find the product.

${\left(a+8\right)}^{2}={a}^{2}+16a+64$

The product is ${a}^{2}+16a+64$ .

## Factoring

Now, let’s reverse the situation. We will be given the product, and we will try to find the factors. This process, which is the reverse of multiplication, is called factoring .

## Factoring

Factoring is the process of determining the factors of a given product.

## Sample set a

The number 24 is the product, and one factor is 6. What is the other factor?

We’re looking for a number $\left(\begin{array}{cc}& \end{array}\right)$ such that $6\cdot \left(\begin{array}{cc}& \end{array}\right)=24$ . We know from experience that $\left(\begin{array}{cc}& \end{array}\right)=4$ . As problems become progressively more complex, our experience may not give us the solution directly. We need a method for finding factors. To develop this method we can use the relatively simple problem $6\cdot \left(\begin{array}{cc}& \end{array}\right)=24$ as a guide.
To find the number $\left(\begin{array}{cc}& \end{array}\right)$ , we would divide 24 by 6.

$\frac{24}{6}=4$

The other factor is 4.

The product is $18{x}^{3}{y}^{4}{z}^{2}$ and one factor is $9x{y}^{2}z$ . What is the other factor?

We know that since $9x{y}^{2}z$ is a factor of $18{x}^{3}{y}^{4}{z}^{2}$ , there must be some quantity $\left(\begin{array}{cc}& \end{array}\right)$ such that $9x{y}^{2}z\cdot \left(\begin{array}{cc}& \end{array}\right)=18{x}^{3}{y}^{4}{z}^{2}$ . Dividing $18{x}^{3}{y}^{4}{z}^{2}$ by $9x{y}^{2}z$ , we get

$\frac{18{x}^{3}{y}^{4}{z}^{2}}{9x{y}^{2}z}=2{x}^{2}{y}^{2}z$

Thus, the other factor is $2{x}^{2}{y}^{2}z$ .

Checking will convince us that $2{x}^{2}{y}^{2}z$ is indeed the proper factor.

$\begin{array}{lll}\left(2{x}^{2}{y}^{2}z\right)\left(9x{y}^{2}z\right)\hfill & =\hfill & 18{x}^{2+1}{y}^{2+2}{z}^{1+1}\hfill \\ \hfill & =\hfill & 18{x}^{3}{y}^{4}{z}^{2}\hfill \end{array}$

We should try to find the quotient mentally and avoid actually writing the division problem.

The product is $-21{a}^{5}{b}^{n}$ and $3a{b}^{4}$ is a factor. Find the other factor.

Mentally dividing $-21{a}^{5}{b}^{n}$ by $3a{b}^{4}$ , we get

$\frac{-21{a}^{5}{b}^{n}}{3a{b}^{4}}=-7{a}^{5-1}{b}^{n-4}=-7{a}^{4}{b}^{n-4}$

Thus, the other factor is $-7{a}^{4}{b}^{n-4}$ .

## Practice set a

The product is 84 and one factor is 6. What is the other factor?

14

The product is $14{x}^{3}{y}^{2}{z}^{5}$ and one factor is $7xyz$ . What is the other factor?

$2{x}^{2}y{z}^{4}$

## Exercises

In the following problems, the first quantity represents the product and the second quantity represents a factor of that product. Find the other factor.

$30,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}6$

5

$45,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}9$

$10a,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}5$

$2a$

$16a,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}8$

$21b,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}7b$

3

$15a,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}5a$

$20{x}^{3},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}4$

$5{x}^{3}$

$30{y}^{4},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}6$

$8{x}^{4},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}4x$

$2{x}^{3}$

$16{y}^{5},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2y$

$6{x}^{2}y,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3x$

$2xy$

$9{a}^{4}{b}^{5},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}9{a}^{4}$

$15{x}^{2}{b}^{4}{c}^{7},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}5{x}^{2}b{c}^{6}$

$3{b}^{3}c$

$25{a}^{3}{b}^{2}c,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}5ac\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}$

$18{x}^{2}{b}^{5},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-2x{b}^{4}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}$

$-9xb$

$22{b}^{8}{c}^{6}{d}^{3},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-11{b}^{8}{c}^{4}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}$

$-60{x}^{5}{b}^{3}{f}^{9},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-15{x}^{2}{b}^{2}{f}^{2}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}$

$4{x}^{3}b{f}^{7}$

$39{x}^{4}{y}^{5}{z}^{11},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3x{y}^{3}{z}^{10}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}$

$147{a}^{20}{b}^{6}{c}^{18}{d}^{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}21{a}^{3}bd\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}$

$7{a}^{17}{b}^{5}{c}^{18}d$

$-121{a}^{6}{b}^{8}{c}^{10},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}11{b}^{2}{c}^{5}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}$

$\frac{1}{8}{x}^{4}{y}^{3},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{1}{2}x{y}^{3}$

$\frac{1}{4}{x}^{3}$

$7{x}^{2}{y}^{3}{z}^{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}7{x}^{2}{y}^{3}z$

$5{a}^{4}{b}^{7}{c}^{3}{d}^{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}5{a}^{4}{b}^{7}{c}^{3}d$

$d$

$14{x}^{4}{y}^{3}{z}^{7},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}14{x}^{4}{y}^{3}{z}^{7}$

$12{a}^{3}{b}^{2}{c}^{8},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}12{a}^{3}{b}^{2}{c}^{8}$

1

$6{\left(a+1\right)}^{2}\left(a+5\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3{\left(a+1\right)}^{2}$

$8{\left(x+y\right)}^{3}\left(x-2y\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2\left(x-2y\right)$

$4{\left(x+y\right)}^{3}$

$14{\left(a-3\right)}^{6}{\left(a+4\right)}^{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2{\left(a-3\right)}^{2}\left(a+4\right)$

$26{\left(x-5y\right)}^{10}{\left(x-3y\right)}^{12},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-2{\left(x-5y\right)}^{7}{\left(x-3y\right)}^{7}$

$-13{\left(x-5y\right)}^{3}{\left(x-3y\right)}^{5}$

$34{\left(1-a\right)}^{4}{\left(1+a\right)}^{8},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-17{\left(1-a\right)}^{4}{\left(1+a\right)}^{2}$

$\left(x+y\right)\left(x-y\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x-y$

$\left(x+y\right)$

$\left(a+3\right)\left(a-3\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}a-3$

$48{x}^{n+3}{y}^{2n-1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}8{x}^{3}{y}^{n+5}$

$6{x}^{n}{y}^{n-6}$

$0.0024{x}^{4n}{y}^{3n+5}{z}^{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0.03{x}^{3n}{y}^{5}$

## Exercises for review

( [link] ) Simplify ${\left({x}^{4}{y}^{0}{z}^{2}\right)}^{3}$ .

${x}^{12}{z}^{6}$

( [link] ) Simplify $-\left\{-\left[-\left(-|6|\right)\right]\right\}$ .

( [link] ) Find the product. ${\left(2x-4\right)}^{2}$ .

$4{x}^{2}-16x+16$

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