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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.
$6x^{2}(2x-7)=12x^{3}-42x^{2}$
The product is $12{x}^{3}-42{x}^{2}$ .
Given the factors $x-2y$ and $3x+y$ , find the product.
$\begin{array}{lll}(x-2y)(3x+y)\hfill & =\hfill & 3{x}^{2}+xy-6xy-2{y}^{2}\hfill \\ \hfill & =\hfill & 3{x}^{2}-5xy-2{y}^{2}\hfill \end{array}$
The product is $3{x}^{2}-5xy-2{y}^{2}$ .
Given the factors $a+8$ and $a+8$ , find the product.
${(a+8)}^{2}={a}^{2}+16a+64$
The product is ${a}^{2}+16a+64$ .
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 .
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.
$\begin{array}{lll}(2{x}^{2}{y}^{2}z)(9x{y}^{2}z)\hfill & =\hfill & 18{x}^{2+1}{y}^{2+2}{z}^{1+1}\hfill \\ \hfill & =\hfill & 18{x}^{3}{y}^{4}{z}^{2}\hfill \end{array}$
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 $-7{a}^{4}{b}^{n-4}$ .
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?
$2{x}^{2}y{z}^{4}$
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
$6{(a+1)}^{2}(a+5),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3{(a+1)}^{2}$
$8{(x+y)}^{3}(x-2y),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2(x-2y)$
$4{\left(x+y\right)}^{3}$
$14{(a-3)}^{6}{(a+4)}^{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2{(a-3)}^{2}(a+4)$
$26{(x-5y)}^{10}{(x-3y)}^{12},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-2{(x-5y)}^{7}{(x-3y)}^{7}$
$-13{\left(x-5y\right)}^{3}{\left(x-3y\right)}^{5}$
$34{(1-a)}^{4}{(1+a)}^{8},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-17{(1-a)}^{4}{(1+a)}^{2}$
$(x+y)(x-y),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x-y$
$\left(x+y\right)$
$(a+3)(a-3),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}a-3$
$48{x}^{n+3}{y}^{2n-1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}8{x}^{3}{y}^{n+5}$
$6{x}^{n}{y}^{n-6}$
$0.0024{x}^{4n}{y}^{3n+5}{z}^{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0.03{x}^{3n}{y}^{5}$
( [link] ) Simplify ${\left({x}^{4}{y}^{0}{z}^{2}\right)}^{3}$ .
${x}^{12}{z}^{6}$
( [link] ) Simplify $-\left\{-\left[-\left(-\left|6\right|\right)\right]\right\}$ .
( [link] ) Find the product. ${\left(2x-4\right)}^{2}$ .
$4{x}^{2}-16x+16$
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