7.5: Factoring, a general strategy
- Page ID
- 45070
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Step 1. Factor out the greatest common factor, if possible. This includes factoring a negative if the leading coefficient is negative.
Step 2. Determine the number of terms in the polynomial.
Step 3.
- Two Terms
- Difference of two squares: \(a^2 − b^2 = (a + b)(a − b)\)
- Difference of two cubes: \(a^3 − b^3 = (a − b)(a^2 + ab + b^2)\)
- Sum of two cubes: \(a^3 + b^3 = (a + b)(a^2 − ab + b^2)\)
- Difference of two fourth powers: \(a^4 − b^4 = (a^2 + b^2)(a + b)(a − b)\)
- Three Terms
- Perfect square trinomial: \(a^2 + 2ab + b^2 = (a + b)^2\) or \(a^2 − 2ab + b^2 = (a − b)^2\)
- Old fashion way:
- \(x^2 + (p + q)x + p\cdot q = (x + p)(x + q)\)
- \(ax^2 + bx + c\) → Factor by grouping or by trial-and-error.
- Four Terms
- Factor by grouping, rearranging terms, if needed.
Step 4. Check your work by FOIL or multiplying out the product of factors
Factor completely: \(4x^2+56xy+196y^2\)
Solution
Let’s first factor the GCF. Recall, there are three terms. So we can use either the perfect square trinomial formula or factor as usual by grouping or trial-and-error.
\[\begin{array}{rl}4x^2+56xy+196y^2&\text{Factor a GCF }4 \\ \color{blue}{4}\color{black}{}(x^2+14xy+49y^2)&\text{Perfect square trinomial} \\ \color{blue}{4}\color{black}{}((x)^2+2(x)(7y)+(7y)^2)&\text{Factor, where }a=x\text{ and }b=7y \\ \color{blue}{4}\color{black}{}(x+7y)^2&\text{Factored form}\end{array}\nonumber\]
Factor completely: \(5x^2y+15xy-35x^2-105x\)
Solution
Let’s first factor the GCF. Recall, there are four terms. So we can use factor by grouping.
\[\begin{array}{rl}5x^2y+15xy-35x^2-105x&\text{Factor a GCF }5x \\ \color{blue}{5x}\color{black}{}(xy+3y-7x-21)&\text{Factor by grouping} \\ \color{blue}{5x}\color{black}{}((xy+3y)+(-7x-21))&\text{Factor the GCF from each group} \\ \color{blue}{5x}\color{black}{}(y(x+3)-7(x+3))&\text{Factor the GCF }(x+3) \\ \color{blue}{5x}\color{black}{}(x+3)(y-7)&\text{Factored form}\end{array}\nonumber\]
Factor completely: \(100x^2-400\)
Solution
Let’s first factor the GCF. Recall, there is a difference of two terms. Since the variable \(x\) is squared, let’s see if we can use the difference of two squares formula.
\[\begin{array}{rl}100x^2-400&\text{Factor a GCF }100 \\ \color{blue}{100}\color{black}{}(x^2-4)&\text{Difference of two squares} \\ \color{blue}{100}\color{black}{}((x)^2-(2)^2)&\text{Factor} \\ \color{blue}{100}\color{black}{}(x+2)(x-2)&\text{Factored form}\end{array}\nonumber\]
Factor: \(108x^3y^2-39x^2y^2+3xy^2\)
Solution
Notice all three terms have a common factor of \(3y^2\). We factor \(3y^2\) first, then factor as usual or by using a special product.
\[\begin{array}{rl}108x^3y^2-39x^2y^2+3xy^2&\text{Factor the GCF} \\ \color{blue}{3xy^2}\color{black}{}(36x^2-13x+1)\end{array}\nonumber\]
Next, we only concentrate on the expression in the parenthesis. Let’s factor by trial-and-error. We know the first’s product is \(36x^2\) and the last’s product is \(1\). Since the signs of the last two terms are negative and positive, respectively, then the binomial factors will have negative signs.
| binomials | FOIL | Yes or no? |
|---|---|---|
| \((6x-1)(6x-1)\) | \(36x^2-12x+1\) | NO |
| \((18x-1)(2x-1)\) | \(36x^2-20x+1\) | NO |
| \((9x-1)(4x-1)\) | \(36x^2-13x+1\) | YES |
We have found the factored form of the original expression:
\[\color{blue}{3xy^2}\color{black}{}(9x-1)(4x-1)\nonumber\]
Variables originated in ancient Greece where Aristotle would use a single capital letter to represent a number.
Factor completely: \(5+625y^3\)
Solution
Let’s factor the GCF. Recall, there is a sum of two terms. Since the only formula with a sum of two terms is the sum of two cubes, then we most likely will be using this special product formula.
\[\begin{array}{rl}5+625y^3&\text{Rewrite in standard form} \\ 625y^3+5&\text{Factor a GCF }5 \\ \color{blue}{5}\color{black}{}(125y^3+1)&\text{Sum of two cubes} \\ \color{blue}{5}\color{black}{}((5y)^3+(1)^3)&\text{Factor, where }a=5y\text{ and }b=1 \\ \color{blue}{5}\color{black}{}(5y+1)((5y)^2-(5y)(1)+(1)^2)&\text{Simplify} \\ \color{blue}{5}\color{black}{}(5y+1)(25y^2-5y+1)&\text{Factored form}\end{array}\nonumber\]
Factoring, A General Strategy Homework
Factor completely.
\(24az − 18ah + 60yz − 45yh\)
\(5u^2 − 9uv + 4v^2\)
\(−2x^3 + 128y^3\)
\(5n^3 + 7n^2 − 6n\)
\(54u^3 − 16\)
\(n^2-n\)
\(x^2 − 4xy + 3y^2\)
\(9x^2 − 25y^2\)
\(m^2 − 4n^2\)
\(36b^2 c − 16xd − 24b^2d +24xc\)
\(128 + 54x^3\)
\(2x^3 + 6x^2y − 20y^2x\)
\(n^3 + 7n^2 + 10n\)
\(27x^3 − 64\)
\(5x^2+2x\)
\(3k^3 − 27k^2 + 60k\)
\(mn − 12x + 3m − 4xn\)
\(16x^2 − 8xy + y^2\)
\(27m^2 − 48n^2\)
\(9x^3 + 21x^2y − 60y^2x\)
\(2m^2 + 6mn − 20n^2\)
\(2x^2 − 11x + 15\)
\(16x^2 + 48xy + 36y^2\)
\(20uv − 60u^3 − 5xv + 15xu^2\)
\(2x^3 + 5x^2y + 3y^2x\)
\(54 − 128x^3\)
\(5x^2 − 22x − 15\)
\(45u^2 − 150uv + 125v^2\)
\(x^3 − 27y^3\)
\(12ab − 18a + 6nb − 9n\)
\(3m^3 − 6m^2n − 24n^2m\)
\(64m^3 + 27n^3\)
\(3ac + 15ad^2 + x^2c + 5x^2d^2\)
\(64m^3 − n^3\)
\(16a^2 − 9b^2\)
\(2x^2 − 10x + 12\)
\(32x^2 − 18y^2\)
\(2k^2 + k − 10\)
\(v^2+v\)
\(x^3+4x^2\)
\(9n^3 − 3n^2\)
\(2u^2v^2 − 11uv^3 + 15v^4\)