6.2: Finding the factors of a Monomial
- Page ID
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Products of Polynomials
Previously, we studied the multiplication of polynomials. We were given factors and asked to find their product, as shown below.
Given the factors 4 and 8, find the product. \(4 \cdot 8=32\). The product is 32.
Given the factors \(6x^2\) and \(2x−7\), find the product.
\(6x^2(2x−7)=12x^3−42x^2\)
The product is \(12x^3−42x^2\).
Given the factors \(x - 2y\) and \(3x + y\), find the product.
\(\begin{array}
(x-2y)(3x+y)&=&3x^2+xy-6xy-2y^2\\
&=&3x^3-5xy-2y^2
\end{array}\)
The product is \(3x^2-5xy-2y^2\).
Given the factors \(a+8\) and \(a+8\), find the product.
\((a+8)^2=a^2+16a+64\)
The product is \(a^2+16a+64\).
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 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 \(( )\) such that \(6 \cdot ( )=24\). We know from experience that \(( )=4\). 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 \(6 \cdot ( )=24\) as a guide.
To find the number \(( )\), we would divide \(24\) by \(6\).
\(\dfrac{24}{6} = 4\)
The other factor is \(4\).
The product is \(18x^3y^4z^2\) and one factor is \(9xy^2z\). What is the other factor?
We know that since \(9xy^2z\) is a factor of \(18x^3y^4z^2\), there must be some quantity \( )\) such that \(9xy^2z \cdot ( ) = 18x^3y^4z^2\).
Dividing \(18x^3y^4z^2\) by \(9xy^2z\), we get:
\(\dfrac{18x^3y^4z^2}{9xy^2z} = 2x^2y^2z\)
Thus, the other factor is \(2x^2y^2z\).
Checking will convince us that \(2x^2y^2z\) is indeed the proper factor.
\(\begin{array}
(2x^2y^2z)(9xy^2z)&=&18x^{2+1}y^{2+2}z^{1+1}\\
&=&18x^3y^4z^2
\end{array}\)
We should try to find the quotient mentally and avoid actually writing the division problem.
The product of \(-21a^5b^n\) and \(3ab^4\) is a factor. Find the other factor.
Mentally dividing \(-21a^5b^n\) by \(3ab^4\), we get
\(\dfrac{-21a^5b^n}{3ab^4} = -7a^{5-1}b^{n-4} = -7a^4b^{n-4}\)
Thus, the other factor is \(-7a^4b^{n-4}\).
Practice Set A
The product is 84 and one factor is 6. What is the other factor?
- Answer
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14
The product is \(14x^3y^2z^5\) and one factor is \(7xyz\). What is the other factor?
- Answer
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\(2x^2yz^4\)
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.
\(30, 6\)
- Answer
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5
\(45, 9\)
\(10a, 5\)
- Answer
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\(2a\)
\(16a, 8\)
\(21b, 7b\)
- Answer
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\(3\)
\(15a, 5a\)
\(20x^3, 4\)
- Answer
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\(5x^3\)
\(30y^4, 6\)
\(8x^4, 4x\)
- Answer
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\(2x^3\)
\(16y^5, 2y\)
\(6x^2y, 3x\)
- Answer
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\(2xy\)
\(9a^4b^5, 9a^4\)
\(15x^2b^4c^7, 5x^2bc^6\)
- Answer
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\(3b^3c\)
\(25a^3b^2c, 5ac\)
\(18x^2b^5, -2xb^4\)
- Answer
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\(−9xb\)
\(22b^8c^6d^3, -11b^8c^4\)
\(-60x^5b^3f^9, -15x^2b^2f^2\)
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\(4x^3bf^7\)
\(39x^4y^5z^{11}, 3xy^3z^{10}\)
\(147a^{20}b^6c^{18}d^2, 21a^3bd\)
- Answer
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\(7a^{17}b^5c^{18}d\)
\(-121a^6b^8c^{10}, 11b^2c^5\)
\(\dfrac{1}{8}x^4y^3, \dfrac{1}{2}xy^3\)
- Answer
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\(\dfrac{1}{4}x^3\)
\(7x^2y^3z^2, 7x^2y^3z\)
\(5a^4b^7c^3d^2, 5a^4b^7c^3d\)
- Answer
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\(d\)
\(14x^4y^3z^7, 14x^4y^3z^7\)
\(12a^3b^2c^8, 12a^3b^2c^8\)
- Answer
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\(1\)
\(6(a+1)^2(a+5), 3(a+1)^2\)
\(8(x+y)^3(x-2y), 2(x-2y)\)
- Answer
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\(4(x+y)^3\)
\(14(a-3)^6(a+4)^2, 2(a-3)^2(a+4)\)
\(26(x-5y)^{10}(x-3y)^{12}, -2(x-5y)^7(x-3y)^7\)
- Answer
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\(-13(x-5y)^3(x-3y)^5\)
\(34(1-a)^4(1+a)^8, -17(1-a)^4(1+a)^2\)
\((x+y)(x−y), x−y\)
- Answer
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\((x+y)\)
\((a+3)(a−3), a−3\)
\(48x^{n+3}y^{2n-1}, 8x^3y^{n+5}\)
- Answer
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\(6x^ny^{n-6}\)
\(0.0024x^{4n}y^{3n+5}z^2, 0.03x^{3n}y^5\)
Exercises for Review
Simplify \((x^4y^0z^2)^3\)
- Answer
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\(x^{12}z^6\)
Simplify \(−{−[−(−|6|)]}\)
Find the product \((2x-4)^2\)
- Answer
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\(4x^2-16x+16\)