7.4: Special products
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In the previous chapter, we recognized two special products: Difference of two squares and Perfect square trinomials. In this section, we discuss these special products to factor expressions.
- Difference of two squares: \(a^2 − b^2 = (a + b)(a − b)\)
- Difference of two fourth powers: \(a^4 − b^4 = (a^2 + b^2)(a + b)(a − b)\)
- Perfect square trinomials: \(a^2 − 2ab + b^2 = (a − b)^2\) or \(a^2 + 2ab + b^2 = (a + b)^2\)
Difference of Two Squares
Factor completely: \(x^2-16\)
Solution
Notice we have a difference of two squares. We can use the formula to factor:
\[\begin{array}{rl}x^2-16&\text{Difference of two squares} \\ (x)^2-(4)^2&\text{Factor} \\ (x+4)(x-4)&\text{Factored form}\end{array}\nonumber\]
As long as we determine \(a\) and \(b\) from the formula, we can easily write the expression in factored form. Let’s continue with another example.
Factor completely: \(9a^2-25b^2\)
Solution
Notice we have a difference of two squares. We can use the formula to factor:
\[\begin{array}{rl}9a^2-25b^2 &\text{Difference of two squares} \\ (3a)^2-(5b)^2&\text{Factor} \\ (3a+5b)(3a-5b)&\text{Factored form}\end{array}\nonumber\]
It is important to note that a sum of squares, e.g., \(x^2 + y^2\), is not factorable. Hence, a sum of two squares is always prime unless there is a greatest common factor
Difference of Two Fourth Powers
The difference of two fourth powers is just a difference of two squares with the exception that there is an additional difference of two squares to be factored in order to factor completely
Factor completely: \(a^4-b^4\)
Solution
Notice we have a difference of two squares. We can use the formula to factor:
\[\begin{array}{rl}a^4-b^4&\text{Difference of two squares} \\ (a^2)^2-(b^2)^2&\text{Factor} \\ (a^2+b^2)(a^2-b^2)&\text{Difference of two squares...again} \\ (a^2+b^2)(a+b)(a-b)&\text{Factored form}\end{array}\nonumber\]
Factor completely: \(x^4-16\)
Solution
Notice we have a difference of two squares. We can use the formula to factor:
\[\begin{array}{rl}x^4-16&\text{Difference of two squares} \\ (x^2)^2-(4)^2&\text{Factor} \\ (x^2+4)(x^2-4)&\text{Difference of two squares...again} \\ (x^2+4)(x+2)(x-2)&\text{Factored form}\end{array}\nonumber\]
Perfect Square Trinomials
Students tend to wonder the reason for the word “square” in the formula when there are four sides to the geometric square shape. Well, it happens that these two are related. Let’s take a generic perfect square trinomial: \(a^2 + 2ab + b^2\), and put this in a geometric representation:
Looking at the square to the right, the area of each square and rectangle are labeled within each figure. If we add the areas together, we would obtain \[a^2+2ab+b^2\nonumber\] The reason we call this a perfect square trinomial is because the sum of all the areas is the area of the outer square.
Factor completely: \(x^2-6x+9\)
Solution
Notice we have three terms. We could factor as usual, or recognize that this is a special product, a perfect square trinomial.
\[\begin{array}{rl}x^2-6x+9&\text{Perfect square trinomial} \\ (x)^2-2(x)(3)+(3)^2&\text{Factor} \\ (x-3)^2&\text{Factored form}\end{array}\nonumber\]
Since the middle term of the trinomial was negative, then we have subtraction in the factored form.
Factor completely: \(4x^2+20xy+25y^2\)
Solution
Notice we have three terms. We could factor as usual, or recognize that this is a special product, a perfect square trinomial.
\[\begin{array}{rl}4x^2+20xy+25y^2&\text{Perfect square trinomial} \\ (2x)^2+2(2x)(5y)+(5y)^2&\text{Factor} \\ (2x+5y)^2&\text{Factored form}\end{array}\nonumber\]
Since the middle term of the trinomial was positive, then we have addition in the factored form.
Factoring Special Products with a Greatest Common Factor
Factor completely: \(72x^2-2\)
Solution
We have two terms and subtraction in between. Notice that this isn’t a different of two squares, but that we do have a GCF. Let’s factor the GCF and then see if we can use a special product formula:
\[\begin{array}{rl}72x^2-2&\text{Factor a GCF }2 \\ \color{blue}{2}\color{black}{}(36x^2-1)&\text{Difference of two squares} \\ \color{blue}{2}\color{black}{}((6x)^2-(1)^2)&\text{Factor} \\ \color{blue}{2}\color{black}{}(6x+1)(6x-1)&\text{Factored form}\end{array}\nonumber\]
In Example 7.4.7 , we couldn’t factor using the difference of two squares right away. It wasn’t until after we factored the GCF that we recognized the expression as a difference of two squares. Hence, we always look for a GCF to factor before applying any other factoring methods.
Factor completely: \(48x^2y-24xy+3y\)
Solution
Let’s factor the GCF and then see if we can use a special product formula:
\[\begin{array}{rl}48x^2y-24xy+3y&\text{Factor a GCF }3y \\ \color{blue}{3y}\color{black}{}(16x^2-8x+1)&\text{Perfect square trinomial} \\ \color{blue}{3y}\color{black}{}((4x)^2-2(4x)(1)+(1)^2)&\text{Factor} \\ \color{blue}{3y}\color{black}{}(4x-1)^2&\text{Factored form}\end{array}\nonumber\]
Since the middle term of the trinomial was negative, then we have subtraction in the factored form.
The first known record of work with polynomials comes from the Chinese around 200 BC. Problems would be written as “three sheafs of a good crop, two sheafs of a mediocre crop, and one sheaf of a bad crop sold for 29 dou. This would be the polynomial (trinomial) equation \(3x + 2y + z = 29\).
A Sum or Difference of Two Cubes
There are special formulas for a sum or difference of two cubes.
- Difference of two cubes: \(a^3 − b^3 = (a − b)(a^2 + ab + b^2)\)
- Sum of two cubes: \(a^3 + b^3 = (a + b)(a^2 − ab + b^2)\)
We can also use the acronym SOAP for the formulas for factoring a sum or difference of two cubes.
\[\begin{array}{ll}\textbf{S}\text{ame}&\text{binomial has the same sign as the expression} \\ \textbf{O}\text{pposite}&\text{middle term of the trinomial has the opposite sign than the expression} \\ \textbf{A}\text{lways} \\ \textbf{P}\text{ositive}&\text{last term of the trinomial is always positive}\end{array}\nonumber\]
SOAP is an easier way of remembering the signs in the formula because the formulas for the sum and difference of two cubes are the same except for the signs. Let’s take a look:
\[\begin{array}{c}a^3\color{blue}{\underset{sign}{\underbrace{-}}}\color{black}{}b^3=(a\color{blue}{\underset{same}{\underbrace{-}}}\color{black}{}b)(a^2\color{blue}{\underset{opposite}{\underbrace{+}}}\color{black}{}ab\color{blue}{\underset{positive}{\underbrace{+}}}\color{black}{}b^2) \\ a^3\color{blue}{\underset{sign}{\underbrace{+}}}\color{black}{}b^3=(a\color{blue}{\underset{same}{\underbrace{+}}}\color{black}{}b)(a^2\color{blue}{\underset{opposite}{\underbrace{-}}}\color{black}{}ab\color{blue}{\underset{positive}{\underbrace{+}}}\color{black}{}b^2)\end{array}\nonumber\]
Once we identify \(a\) and \(b\), then we can just plug-n-chug into one of the formulas and use SOAP for the signs.
Factor completely: \(m^3-27\)
Solution
Notice we have a difference of two cubes. We can use the formula to factor:
\[\begin{array}{rl}m^3-27&\text{Difference of two cubes} \\ (m)^3-(3)^3&\text{Factor, where }a=m\text{ and }b=3 \\ (m-3)((m)^2+(3)(m)+(3)^2)&\text{Simplify} \\ (m-3)(m^2+3m+9)&\text{Factored form}\end{array}\nonumber\]
Factor completely: \(125p^3+8r^3\)
Solution
Notice we have a sum of two cubes. We can use the formula to factor:
\[\begin{array}{rl}125p^3+8r^3&\text{Sum of two cubes} \\ (5p)^3+(2r)^3&\text{Factor, where }a=5p\text{ and }b=2r \\ (5p+2r)((5p)^2-(5p)(2r)+(2r)^2)&\text{Simplify} \\ (5p+2r)(25p^2-10pr+4r^2)&\text{Factored form}\end{array}\nonumber\]
Factor completely. \(128a^4b^2+54ab^5\)
Solution
Let’s factor the GCF and then see if we can use a special product formula:
\[\begin{array}{rl}128a^4b^2+54ab^5&\text{Factor a GCF }2ab^2 \\ \color{blue}{2ab^2}\color{black}{}(64a^3+27b^3)&\text{Sum of two cubes} \\ \color{blue}{2ab^2}\color{black}{}((4a)^3+(3b)^3)&\text{Factor, where }a=4a\text{ and }b=3b \\ \color{blue}{2ab^2}\color{black}{}(4a+3b)((4a)^2-(4a)(3b)+(3b)^2)&\text{Simplify} \\ \color{blue}{2ab^2}\color{black}{}(4a+3b)(16a^2-12ab+9b^2)&\text{Factored form}\end{array}\nonumber\]
Special Products Homework
Factor completely by using the special product formulas.
\(r^2-16\)
\(v^2-25\)
\(p^2-4\)
\(9k^2-4\)
\(3x^2-27\)
\(16x^2-36\)
\(18a^2-50b^2\)
\(a^2-2a+1\)
\(x^2+6x+9\)
\(x^2-6x+9\)
\(25p^2-10p+1\)
\(25a^2+30ab+9b^2\)
\(4a^2-20ab+25b^2\)
\(8x^2-24xy+18y^2\)
\(8-m^3\)
\(x^3-64\)
\(216-u^3\)
\(125a^3-64\)
\(64x^3+27y^3\)
\(54x^3+250y^3\)
\(a^4-81\)
\(16-z^4\)
\(x^4-y^4\)
\(m^4-81b^4\)
\(x^2-9\)
\(x^2-1\)
\(4v^2-1\)
\(9a^2-1\)
\(5n^2-20\)
\(125x^2+45y^2\)
\(4m^2+64n^2\)
\(k^2+4k+4\)
\(n^2-8n+16\)
\(k^2-4k+4\)
\(x^2+2x+1\)
\(x^2+8xy+16y^2\)
\(18m^2-24mn+8n^2\)
\(20x^2+20xy+5y^2\)
\(x^3+64\)
\(x^3+8\)
\(125x^3-216\)
\(64x^3-27\)
\(32m^3-108n^3\)
\(375m^3+648n^3\)
\(x^4-256\)
\(n^4-1\)
\(16a^4-b^4\)
\(81c^4-16d^4\)