6.6: Factoring Two Special Products
- Page ID
- 49377
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)The Difference of Two Squares
Recall that when we multiplied together the two binomials \((a+b)\) and \((a−b)\), we obtained the product \(a^2-b^2\).
\((a+b)(a−b)=a^2−b^2\)
Perfect Square
Notice that the terms \(a^2\) and \(b^2\) in the product can be produced by squaring \(a\) and \(b\), respectively. A term that is the square of another term is called a perfect square. Thus, both \(a^2\) and \(b^2\) are perfect squares. The minus sign between \(a^2\) and \(b^2\) means that we are taking the difference of the two squares.
Since we know that \((a+b)(a−b)=a^2−b^2\), we need only turn the equation around to find the factorization form.
\(a^2-b^2 = (a+b)(a-b)\)
The factorization form says that we can factor \(a^2−b^2\), the difference of two squares, by finding the terms that produce the perfect squares and substituting these quantities into the factorization form.
When using real numbers (as we are), there is no factored form for the sum of two squares. That is, using real numbers,
\(a^2 + b^2\) cannot be factored
Sample Set A
Factor \(x^2 - 16\). Both \(x^2\) and \(16\) are perfect squares. The terms that, when squared, produce \(x^2\) and \(16\) are \(x\) and \(4\), respectively.
Thus,
\(x^2 - 16 = (x+4)(x-4)\)
We can check our factorization simply by multiplying.
\(\begin{array}{flushleft}
(x+4)(x-4)&=&x^2 - 4x + 4x - 16\\
&=&x^2 - 16
\end{array}\)
\(49a^2b^4 - 121\). Both \(49a^2b^4\) and \(121\) are perfect squares. The terms that, when squared, produce \(49a^2b^4\) and \(121\) are \(7ab^2\) and \(11\), respectively. Substituting these terms into the factorization form we get
\(49a^2b^4 - 121 = (7ab^2 + 121)(7ab^2 - 11)\)
We can check our facorization by multiplying
\(\begin{array}{flushleft}
(7ab^2 + 121)(7ab^2 - 11)&=&49a^2b^4 - 11ab^2 + 11ab^2 - 121\\
&=&49a^2b^4 - 121
\end{array}\)
\(3x^2 - 27\). This doesn't look like the difference of two squares since we don't readily know the terms that produce \(3x^2\) and \(27\)(. However, notice that \(3\) is common to both the terms. Factor out 3.
\(3(x^2-9)\).
Now we see that \(x^2 - 9\) is the difference of two squares. Facotring the \(x^2 - 9\) we get
\(\begin{array}{flushleft}
3x^2 - 27&=&3(x^2-9)\\
&=&3(x+3)(x-3)
\end{array}\)
Practice Set A
If possible, factor the following binomials completely.
\(m^2 - 25\)
- Answer
-
\((m+5)(m−5)\)
\(36p^2 - 81q^2\)
- Answer
-
\(9(2p−3q)(2p+3q)\)
\(49a^2 = b^2c^2\)
- Answer
-
\((7a^2 + bc)(7a^2 - bc)\)
\(x^8y^4 - 100x^{12}\)
- Answer
-
\((x^4y^2+10w^6)(x^4y^2−10w^6)\)
\(3x^2−75\)
- Answer
-
\(3(x+5)(x−5)\)
\(a^3b^4m−am^3n^2\)
- Answer
-
\(am(ab^2+mn)(ab^2−mn)\)
Fundamental Rules of Factoring
There are two fundamental rules that we follow when factoring:
Fundamental Rules of Factoring
- Factor out all common monomials first.
- Factor completely.
Sample Set B
Factor each binomial completely.
\(4a^8b - 36b^5\). Factor out the common factor \(4b\).
\(4b(a^8 - 9b^4)\).
Now we can see a difference of two squares, whereas in the original polynomial we could not. We'll complete our factorization by factoring the difference of two squares.
\(\begin{array}{flushleft}
4a^8-36b^5&=&4b(a^8-9b^4)\\
&=&4b(a^4+3b^2)(a^4-3b^2)
\end{array}\)
\(x^{16}-y^8\). Factor this difference of two squares.
\(x^{16} - y^8 = (x^8 + y^4) (x^8 - y^4)\)
We see a sum of two squares, and then a difference of two squares.
\(= (x^8+y^4)(x^4+y^2)(x^4-y^2)\)
Factor again!
\(= (x^8+y^4)(x^4+y^2)(x^2+y)(x^2-y)\)
Finally, the factorization is complete.
These types of products appear from time to time, so be aware that you may have to factor more than once.
Practice Set B
Factor each binomial completely.
\(m^4-n^4\)
- Answer
-
\((m^2+n^2)(m−n)(m+n)\)
\(16y^8−1\)
- Answer
-
\((4y^4+1)(2y^2+1)(2y^2−1)\)
Perfect Square Trinomials
Recall the process of squaring a binomial.
\((a+b)^2=a^2+2ab+b^2(a−b)^2=a^2−2ab+b^2\)
| Our Method Is | We Notice |
| Square the first term | The first term of the product should be a perfect square. |
| Take the product of the two terms and double it. | The middle term of the product should be divisible by 2 (since it’s multiplied by 2). |
| Square the last term. | The last term of the product should be a perfect square. |
Perfect square trinomials always factor as the square of a binomial.
To recognize a perfect square trinomial, look for the following features:
- The first and last terms are perfect squares.
- The middle term is divisible by 2, and if we divide the middle term in half (the opposite of doubling it), we will get the product of the terms that when squared produce the first and last terms.
In other words, factoring a perfect square trinomial amounts to finding the terms that, when squared, produce the first and last terms of the trinomial, and substituting into one of the formula
\(a^2+2ab+b^2=(a+b)^2a^2−2ab+b^2=(a−b)^2\)
Sample Set C
Factor each perfect square trinomial.
\(x^2 + 6x + 9\). This expression is a perfect square trinomial. The \(x^2\) and \(9\) are perfect squares.
The terms that when squared produce \(x^2\) and \(9\) are \(x\) and \(3\), respectively.
The middle term is divisible by \(2\), and \(\dfrac{6x}{2} = 3x\). The \(3x\) is the produce of \(x\) and \(3\), which are the terms that produce the perfect squares.
\(x^2 + 6x + 9 = (x+3)^2\)
\(x^4 - 10x^2y^3 + 25y^6\). This expression is a perfect square trinomial. The \(x^4\) and \(25y^6\) are both perfect squares. The terms that when squared produce \(x^4\) and \(25y^6\) are \(x^2\) and \(5y^3\), respectively.
The middle term \(-10x^2y^3\) is divisble by \(2\). In fact, \(\dfrac{-10x^2y^3}{2} = -5x^2y^3\). Thus,
\(x^4-10x^2y^3 + 25y^6 = (x^2-5y^3)^2\)
\(x^2 + 10x + 16\). This expression is not a perfect square trinomial. Although the middle term is divisible by \(2\), \(\dfrac{10x}{2} = 5x\), the \(5\) and \(x\) are not the terms that when squared produce the first and last terms. (This expression would be a perfect square trinomial if the middle term were \(8x\).)
\(4a^4 + 32a^2b - 64b^2\). This expression is not a perfect square trinomial since the last term \(-64b^2\) is not a perfect square (since any quantity squared is always positive or zero and never negative).
Thus, \(4a^4 + 32a^2b - 64b^2\) cannot be factored using this method.
Practice Set C
Factor, if possible, the following trinomials.
\(m^2−8m+16\)
- Answer
-
\((m−4)^2\)
\(k^2+10k+25\)
- Answer
-
\((k+5)^2\)
\(4a^2+12a+9\)
- Answer
-
\((2a+3)^2\)
\(9x^2−24xy+16y^2\)
- Answer
-
\((3x−4y)^2\)
\(2w^3z+16w^2z^2+32wz^3\)
- Answer
-
\(2wz(w+4z)^2\)
\(x^2+12x+49\)
- Answer
-
not possible
Exercises
For the following problems, factor the binomials.
\(a^2−9\)
- Answer
-
\((a+3)(a−3)\)
\(a^2−25\)
\(x^2−16\)
- Answer
-
\((x+4)(x−4)\)
\(y^2−49\)
\(a^2−100\)
- Answer
-
\((a+10)(a−10)\)
\(b^2−36\)
\(4a^2−64\)
- Answer
-
\(4(a+4)(a−4)\)
\(2b^2−32\)
\(3x^2−27\)
- Answer
-
\(3(x+3)(x−3)\)
\(5x^2−125\)
\(4a^2−25\)
- Answer
-
\((2a+5)(2a−5)\)
\(9x^2−100\)
\(36y^2−25\)
- Answer
-
\((6y+5)(6y−5)\)
\(121a^2−9\)
\(12a^2−75\)
- Answer
-
\(3(2a+5)(2a−5)\)
\(10y^2−320\)
\(8y^2−50\)
- Answer
-
\(2(2y+5)(2y−5)\)
\(a^2b^2−9\)
\(x^2y^2−25\)
- Answer
-
\((xy+5)(xy−5)\)
\(x^4y^4−36\)
\(x^4y^4−9a^2\)
- Answer
-
\((x^2y^2+3a)(x^2y^2−3a)\)
\(a^2b^4−16y^4\)
\(4a^2b^2−9b^2\)
- Answer
-
\(b^2(2a+3)(2a−3)\)
\(16x^2−25y^2\)
\(a^2−b^2\)
- Answer
-
\((a+b)(a−b)\)
\(a^4−b^4\)
\(x^4−y^4\)
- Answer
-
\((x^2+y^2)(x+y)(x−y)\)
\(x^8−y^2\)
\(a^8−y^2\)
- Answer
-
\((a^4+y)(a^4−y)\)
\(b^6−y^2\)
\(b^6−x^4\)
- Answer
-
\((b^3+x^2)(b^3−x^2)\)
\(9−x^2\)
\(25−a^2\)
- Answer
-
\((5+a)(5−a)\)
\(49−16a^2\)
\(100−36b^4\)
- Answer
-
\(4(5+3b^2)(5−3b^2)\)
\(128−32x^2\)
\(x^4−16\)
- Answer
-
\((x^2+4)(x+2)(x−2)\)
\(2ab^3−a^3b\)
\(a^4−b^4\)
- Answer
-
\((a^2+b^2)(a+b)(a−b)\)
\(a^{16}−b^4\)
\(x^{12} - y^{12}\)
- Answer
-
\((x^6+x^6)(x^3+y^3)(x^3−y^3)\)
\(a^2c−9c\)
\(a^3c^2−25ac^2\)
- Answer
-
\(ac^2(a+5)(a−5)\)
\(a^4b^4c^2d^2−36x^2y^2\)
\(49x^2y^4z^6−64a^4b^2c^8d^{10}\)
- Answer
-
\((7xy^2z^3+8a^2bc^4d^5)(7xy^2z^3−8a^2bc^4d^5)\)
For the following problems, factor, if possible, the trinomials.
\(x^2+8x+16\)
\(x2+10x+25\)
- Answer
-
\((x+5)^2\)
\(a^2+4a+4\)
\(a^2+12a+36\)
- Answer
-
\((a+6)^2\)
\(b^2+18b+81\)
\(y^2+20y+100\)
- Answer
-
\((y+10)^2\)
\(c^2+6c+9\)
\(a^2−4a+4\)
- Answer
-
\((a−2)^2\)
\(b^2−6b+9\)
\(x^2−10x+25\)
- Answer
-
\((x−5)^2\)
\(b^2−22b+121\)
\(a^2−24a+144\)
- Answer
-
\((a−12)^2\)
\(a^2+2a+1\)
\(x^2+2x+1\)
- Answer
-
\((x+1)^2\)
\(x^2−2x+1\)
\(b^2−2b+1\)
- Answer
-
\((b−1)^2\)
\(4a^2+12a+9\)
\(9x^2+6x+1\)
- Answer
-
\((3x+1)^2\)
\(4x^2+28x+49\)
\(16a^2−24a+9\)
- Answer
-
\((4a−3)^2\)
\(25a^2−20a+4\)
\(9x^2+6xy+y^2\)
- Answer
-
\((3x+y)^2\)
\(16x^2+24xy+9y^2\)
\(36a^2+60ab+25b^2\)
- Answer
-
\((6a+5b)^2\)
\(4x^2−12xy+9y^2\)
\(12a^2−60a+75\)
- Answer
-
\(3(2a−5)^2\)
\(16x^2+8x+1\)
\(32x^2+16x+2\)
- Answer
-
\(2(4x+1)^2\)
\(x^2+x+1\)
\(4a^2+a+9\)
- Answer
-
not factorable
\(9a^2−21a+49\)
\(x^5+8x^4+16x^3\)
- Answer
-
\(x^3(x+4)^2\)
\(12a^3b−48a^2b^2+48ab^3\)
EXERCISES FOR REVIEW
Factor \((m−3)x−(m−3)y\).
- Answer
-
\((m−3)(x−y)\)
Factor \(8xm+16xn+3ym+6yn\) by grouping.