6.5: Special products
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There are a few shortcuts that we can take when multiplying polynomials. If we can recognize when to use them, we should so that we can obtain the results even quicker. In future chapters, we will need to be efficient in these techniques since multiplying polynomials will only be one of the steps in the problem.
Difference of Two Squares
The first shortcut is called a difference of two squares. A sum and a difference is easily recognized as the numbers and variables are exactly the same, but the signs in the middle are opposites.
Multiply: \((a+b)(a-b)\)
Solution
We can multiply these binomials by distribution.
\[\begin{array}{rl}(a+b)(a-b)&\text{Distribute }a\text{ and }b\text{ to }(a-b) \\ a(a-b)+b(a-b)&\text{Distribute} \\ a^2-ab+ba-b^2&\text{Combine like terms} \\ a^2\cancel{-ab}\cancel{+ba}-b^2&\text{Simplify} \\ a^2-b^2&\text{Product}\end{array}\nonumber\]
Notice the middle terms cancelled and the product is a difference of two squares: \(a^2 − b^2\).
Given a product of two binomials, where the terms are the same but opposite middle signs, the product results in a difference of two squares, the squares of the terms: \[(a+b)(a-b)=a^2-b^2\nonumber\]
So awesome, right? This means if we are given any product of two binomials of this form, we can just square the terms and put a subtraction sign in between. Let’s look at a couple of examples.
Multiply: \((3x+7)(3x-7)\)
Solution
Notice the terms are \(3x\) and \(7\) and have opposite middle signs. Hence, we can use the difference of two squares formula to arrive at the product quickly:
\[\begin{array}{rl}(3x+7)(3x-7)&\text{Terms are }3x\text{ and }7\\ (3x)^2-(7)^2&\text{Square the terms and put a subtraction sign in between} \\ 9x^2-49&\text{Product}\end{array}\nonumber\]
Multiply: \((2x-6y)(2x+6y)\)
Solution
Notice the terms are \(2x\) and \(6y\) and have opposite middle signs. Hence, we can use the difference of two squares formula to arrive at the product quickly:
\[\begin{array}{rl}(2x-6y)(2x+6y)&\text{Terms are }2x\text{ and }6y \\ (2x)^2-(6y)^2&\text{Square the terms and put a subtraction sign in between} \\ 4x^2-36y^2&\text{Product}\end{array}\nonumber\]
It is interesting to note that while we can obtain a product like \(a^2 − b^2\), it is impossible to obtain a product like \(a^2 + b^2\). There are no two binomials in the real number system in which are multiplied to obtain a sum of two squares. Do not be fooled, though. There are products of two binomials out in the world that will result in a sum of two squares, but just not in this course.
Perfect Square Trinomials
Another shortcut used to multiply binomials is called perfect square trinomials. These are easy to recognize because this product is the square of a binomial. Let’s take a look at an example.
Multiply: \((a+b)^2\)
Solution
We can multiply these binomials by distribution.
\[\begin{array}{rl}(a+b)^2&\text{Rewrite as a product of two binomials} \\ (a+b)(a+b)&\text{Distribute }a\text{ and }b\text{ to }(a+b) \\ a(a+b)+b(a+b)&\text{Distribute} \\ a^2+ab+ba+b^2&\text{Combine like terms} \\ a^2+2ab+b^2&\text{Product}\end{array}\nonumber\]
Notice the first term is the square of \(a\), the middle term is \(2\) times the product of \(a\) and \(b\), and the last term is the square of \(b\). I.e., square the first, twice the product, square the last. Hence, the square of a binomial is a perfect square trinomial.
Given a square of a binomial, where the terms are the same but can have addition or subtraction middle signs, the product results in a perfect square trinomial:
\[\begin{array}{c}(a+b)^2=a^2+2ab+b^2 \\ (a-b)^2=a^2-2ab+b^2\end{array}\nonumber\]
Simplify: \((x-5)^2\)
Solution
Notice this is the square of binomial \((x − 5)\). We can use the perfect square trinomial formula to simplify.
\[\begin{array}{rl}(x-5)^2&\text{Terms are }x\text{ and }5 \\ (x)^2-2(x)(5)+(5)^2&\text{Follow the formula for }(a-b)^2 \\ x^2-10x+25&\text{Product}\end{array}\nonumber\]
Simplify: \((2x+9)^2\)
Solution
Notice this is the square of binomial \((2x + 9)\). We can use the perfect square trinomial formula to simplify.
\[\begin{array}{rl}(2x+9)^2&\text{Terms are }2x\text{ and }9 \\ (2x)^2+2(2x)(9)+(9)^2&\text{Follow the formula for }(a+b)^2 \\ 4x^2+36x+81&\text{Product}\end{array}\nonumber\]
Simplify: \((3x-7y)^2\)
Solution
Notice this is the square of binomial \((3x − 7y)\). We can use the perfect square trinomial formula to simplify.
\[\begin{array}{rl}(3x-7y)^2&\text{Terms are }3x\text{ and }7y \\ (3x)^2-2(3x)(7y)+(7y)^2&\text{Follow the formula for }(a-b)^2 \\ 9x^2-42x+49y^2&\text{Product}\end{array}\nonumber\]
Be very careful when we are squaring a binomial. Be sure to avoid the common error of only squaring the first and last terms. A common error is to do the following: \((x − 5)^2 = x^2 − 25\) (or \(x^2 + 25\)). Notice both of these are missing the middle term, \(−10x\).
Another important observation is that the middle term in the answer always has the same sign as the middle term in the given problem.
These two formulas are important to commit to memory. The more familiar we are with them, the next two chapters will be so much easier. The final example covers both types of problems. Be sure to notice the difference between the examples.
Let’s take a look at three examples side-by-side to see the difference between all the formulas. Let’s multiply \[(4x-7)(4x+7),\: (4x+7)^2,\: (4x-7)^2\nonumber\]
Solution
We apply the formulas to simplify each product.
| \((4x-7)(4x+7)\) | \((4x+7)^2\) | \((4x-7)^2\) |
|---|---|---|
| \((4x)^2-(7)^2\) | \((4x)^2+2(4x)(7)+(7)^2\) | \((4x)^2-2(4x)(7)+(7)^2\) |
| \(16x^2-49\) | \(16x^2+56x+49\) | \(16x^2-56x+49\) |
We see that the first product is a difference of two squares and the product is two terms. The second and third products are squares of binomials that results in perfect square trinomials and are three terms each.
There are also formulas for higher powers of binomials as well, such as \((a+b)^3 = a^3 + 3a^2 b+ 3ab^2 +b^3\). While French mathematician, Blaise Pascal, often gets credit for working with these expansions of binomials in the \(17^{\text{th}}\) century, Chinese mathematicians had been working with them almost 400 years earlier.
Special Products Homework
Find each product by applying the special product formulas.
\((x + 8)(x − 8)\)
\((1 + 3p)(1 − 3p)\)
\((1 − 7n)(1 + 7n)\)
\((5n − 8)(5n + 8)\)
\((4x + 8)(4x − 8)\)
\((4y − x)(4y + x)\)
\((4m − 8n)(4m + 8n)\)
\((6x − 2y)(6x + 2y)\)
\((a + 5)^2\)
\((x − 8)^2\)
\((p + 7)^2\)
\((7 − 5n)^2\)
\((5m − 8)^2\)
\((5x + 7y)^2\)
\((2x + 2y)^2\)
\((5 + 2r)^2\)
\((2 + 5x)^2\)
\((4v − 7)(4v + 7)\)
\((n − 5)(n + 5)\)
\((4k + 2)^2\)
\((a − 4)(a + 4)\)
\((x − 3)(x + 3)\)
\((8m + 5)(8m − 5)\)
\((2r + 3)(2r − 3)\)
\((b − 7)(b + 7)\)
\((7a + 7b)(7a − 7b)\)
\((3y − 3x)(3y + 3x)\)
\((1 + 5n)^2\)
\((v + 4)^2\)
\((1 − 6n)^2\)
\((7k − 7)^2\)
\((4x − 5)^2\)
\((3a + 3b)^2\)
\((4m − n)^2\)
\((8x + 5y)^2\)
\((m − 7)^2\)
\((8n + 7)(8n − 7)\)
\((b + 4)(b − 4)\)
\((7x + 7)^2\)
\((3a − 8)(3a + 8)\)