Section6.4Special Cases of Multiplying Polynomials
¶While we are now able to multiply polynomials together, we need to look a bit further into multiplication to see a few special cases of polynomial multiplication.
Subsection6.4.1Squaring a Binomial
Example6.4.2
Expand the following:
\(\left( x+4 \right)^2\)
\(\left( y-7 \right)^2\)
Solution
To be able to distribute this appropriately, we first need to rewrite the expression explicitly as multiplication.
\begin{equation*} (x+4)^2=(x+4)(x+4) ~~~\text{ and }~~~ (x-7)^2=(x-7)(x-7) \end{equation*}Once this is rewritten, we can use the distribution processes we already know: we can expand this product into four pairs of products, multiply, and then combine the like terms.
- \begin{align*} \left( x+4 \right)^2 \amp= \left( x+4 \right)\left( x+4 \right) \\ \amp= x^2 + 4x + 4x + 16 \\ \amp= x^2 + 8x + 16 \end{align*}
- \begin{align*} \left( y-7 \right)^2 \amp= \left( y-7 \right)\left( y-7 \right)\\ \amp= y^2 -7y -7y + 49 \\ \amp= y^2 -14y + 49 \end{align*}
Both of the previous examples were examples of squaring a binomial. On the one hand, once we wrote each expression without the exponent, we're not doing anything different than what we did in the previous section. On the other hand, something very special did happen.
Look closely at the following two steps of work. (We'll remove the step where we expanded the exponent.)
\begin{align*} \left( x+4 \right)^2 \amp= x^2 \highlight{{}+ 4x + 4x}{} + 16 \\ \left( y-7 \right)^2 \amp= y^2 \highlight{{} - 7y -7y}{} + 49 \end{align*}Notice that the two middle terms are not only the same, they are also exactly the product of the two terms in the binomial!
What we're seeing is a pattern that has two names: "squaring a binomial" and "a perfect square trinomial." The first name is a description of what we're doing, we are literally squaring a binomial. The second name is a description of what you end up with. This second name will become important in a future chapter.
The general way this pattern is presented is by squaring the most general binomial possible, \((a+b)\text{.}\) Once we establish the pattern for \((a+b)^2\text{,}\) we can substitute anything in place of \(a\) and \(b\) and rely upon the general pattern.
We first must expand \((a+b)^2\) as \((a+b)(a+b)\) and then we can mentally use distribution or use the FOIL method:
\begin{align*} (a+b)^2\amp=(a+b)(a+b)\\ \amp=a^2+ab+ab+b^2\\ \amp=a^2+2ab+b^2 \end{align*}Notice the final simplification step was to add \(ab+ab\text{.}\) Since these are like terms, we can combine them into \(2ab\text{.}\) We now have the overall pattern:
\begin{equation*} (a+b)^2 = a^2+2ab+b^2 \end{equation*}Example6.4.3
Simplify \((x+9)^2\) using the squaring a binomial pattern.
To do this, we need to recognize that to apply the pattern \((a+b)^2 = a^2+2ab+b^2\) in our situation, \(a=x\) and \(b=9\text{.}\)
\begin{alignat*}{3} ( a +b)^2 \amp= a^2 \amp \amp+ 2ab \amp \amp+ b^2\\ ( \highlight{x} {}+{} \highlight{9})^2 \amp= \highlight{(x)}^2 \amp \amp+ 2\highlight{(x)}\highlight{(9)} \amp \amp+ \highlight{(9)}^2\\ \amp= x^2 \amp \amp+ 18x \amp \amp+ 81 \end{alignat*}Typically, we would not write the general pattern as our first step of work. That was done to emphasize the pattern in this example.
Suppose we have a binomial with subtraction instead of addition between the terms. How does this change our pattern?
Let's multiply \((a-b)^2\) using generic rectangles. As a first step, we must rewrite the expression \((a-b)^2\) as \((a-b)(a-b)\text{.}\) Then we can set up generic rectangles and find their area:
When we combine the four terms, we find that the product is:
\begin{equation*} (a-b)^2=a^2-2ab+b^2 \end{equation*}Example6.4.5
Multiply \((2x-3)^2\) using the squaring a binomial pattern.
For this example we need to recognize that to apply the pattern \((a-b)^2 = a^2-2ab+b^2\) in our situation, \(a=2x\) and \(b=3\text{.}\)
\begin{alignat*}{3} ( a-b)^2 \amp= a^2 \amp \amp- 2ab \amp \amp+ b^2\\ ( \highlight{2x} {}-{} \highlight{3})^2 \amp= \highlight{(2x)}^2 \amp \amp- 2\highlight{(2x)}\highlight{(3)} \amp \amp+ \highlight{(3)}^2\\ \amp= 4x^2 \amp \amp- 12x \amp \amp+ 9 \end{alignat*}Again, we would not typically write the general pattern as our first step of work.
Squaring a Binomial Formulas:
\begin{align*} (a+b)^2 \amp= a^2+2ab+b^2 \\ (a-b)^2 \amp= a^2-2ab+b^2 \end{align*}Example6.4.6
Multiply the following using the squaring a binomial pattern:
\((2x-3y)^2\)
\((5xy+1)^2\)
Solution
- \begin{align*} (2x-3y)^2 \amp= (2x)^2-2(2x)(3y)+(3y)^2 \\ \amp= 4x^2-12xy+9y^2 \end{align*}
- \begin{align*} (5xy+1)^2 \amp= (5xy)^2+2(5xy)(1)+(1)^2 \\ \amp= 25x^2y^2+10xy+1 \end{align*}
Example6.4.7
A circle's area can be calculated by the formula
\begin{equation*} A=\pi r^2 \end{equation*}where \(A\) stands for area, and \(r\) stands for radius. If a certain circle's radius can be modeled by \(x-5\) feet, use an expanded polynomial to model the circle's area.
Solution
The circle's area would be:
\begin{align*} A\amp=\pi r^2\\ \amp=\pi (x-5)^2\\ \amp=\pi \left[(x)^2-2(x)(5)+(5)^2\right]\\ \amp=\pi \left[x^2-10x+25\right]\\ \amp=\pi x^2-10\pi x+25\pi \end{align*}The circle's area can be modeled by \(\pi x^2-10\pi x+25\pi\) square feet.
Exercise6.4.8
Exercise6.4.9
Subsection6.4.2The Product of the Sum and Difference of Two Terms
Example6.4.10
Multiply the following:
\((x+5)(x-5)\)
\((y-8)(y+8)\)
\((a+b)(a-b)\)
Solution
We could approach these as using distribution, FOIL, or generic rectangles.
- \begin{align*} (x+5)(x-5) \amp= x^2-5x+5x-25 \\ \amp= x^2 -25 \end{align*}
- \begin{align*} (y+8)(y-8) \amp= y^2-8y+8y-49\\ \amp= y^2 - 64 \end{align*}
- \begin{align*} (a+b)(a-b) \amp= a^2-ab+ab-b^2\\ \amp= a^2 - b^2 \end{align*}
Notice in these three examples that the two middle terms were opposites. Unlike squaring binomials, where the middle terms combine to give you \(2ab\text{,}\) the two middle terms in these examples cancel and leave you with no middle term.
As with the previous special case, this one has two names. This can be called "the product of the sum and difference of two terms," because this pattern is built on multiplying two binomials that have the same two terms, except one binomial is a sum and the other binomial is a difference. The second name is "a difference of squares," because the end result of the multiplication is a binomial that is the difference of two perfect squares. As before, the second name will become useful in a future chapter when using exactly the technique described in this section will be absolutely necessary.
Product of the Sum and Difference of Two Terms Formula:
\begin{equation*} (a+b)(a-b)=a^2-b^2 \end{equation*}Example6.4.11
Multiply the following using the product of the sum and difference of two terms pattern:
\((4x-7y)(4x+7y)\)
\((3+7xy)(3-7xy)\)
\((x^2+2y^4)(x^2-2y^4)\)
Solution
The first step to understanding this method is to identify the values of \(a\) and \(b\) in \((a+b)(a-b)=a^2-b^2\text{.}\)
-
In this instance, \(a=4x\) and \(b=7y\text{.}\) Filling in the pattern,
\begin{align*} (4x-7y)(4x+7y) \amp= (4x)^2-(7y)^2 \\ \amp= 16x^2-49y^2 \end{align*} -
Here, \(a=3\) and \(b=8xy\text{.}\) So,
\begin{align*} (3+8xy)(3-8xy) \amp= (3)^2-(8xy)^2 \\ \amp= 9 -64x^2y^2 \end{align*} -
Last, \(a=x^2\) and \(b=2y^4\) and the expansion is,
\begin{align*} (x^2+2y^4)(x^2-2y^4) \amp= (x^2)^2-(2y^4)^2 \\ \amp= x^4 -4y^8 \end{align*}
Exercise6.4.12
Exercise6.4.13
Exercise6.4.14
Subsection6.4.3Binomials Raised to Other Powers
Example6.4.15
Multiply \((x+5)^3\text{.}\)
Before we start multiplying, it is important to recognize that \((x+5)^3\neq x^3+ 5^3\text{.}\) We can show this doesn't work by picking a value of \(1\) for \(x\text{:}\)
\begin{align*} (1+5)^3\amp=6^3\\ \amp=216\\ \\ 1^3+5^3\amp=1+125\\ \amp=126 \end{align*}The reason \((x+5)^3\neq x^3+ 5^3\) is that in trying to do this, we're applying an exponent rule, \((a\cdot b)^n=a^n\cdot b^n\text{,}\) in a situation that requires distribution. The fact that there is addition and subtraction in the parentheses indicates the exponent rule doesn't apply.
The first step in multiplying \((x+5)^3\) is to remember that the exponent of \(3\) indicates \((x+5)^3=(x+2)\cdot (x+5)\cdot (x+2)\text{.}\) Once we rewrite this in an expanded form, we next multiply the two binomials on the left and then finish by multiplying that result by the remaining binomial:
\begin{align*} (x+5)^3\amp=\highlight{\left[(x+5)(x+5)\right]}(x+5)\\ \amp=\highlight{\left[x^2+10x+25\right]}(x+5)\\ \amp=x^3 + 5x^2 + 10x^2 +50x+25x+125\\ \amp=x^3 + 15x^2 + 75x + 125 \end{align*}Exercise6.4.16
If we wanted to multiply a binomial to any power, we always will start by rewriting the expression without an exponent.
To multiply \((x-3)^4\text{,}\) we'd start by rewriting \((x-3)^4\) as \((x-3)(x-3)(x-3)(x-3)\) and then multiplying pairs of polynomials from the left to the right.
\begin{align*} (x-3)^4\amp=\highlight{\left[(x-3)(x-3)\right]}(x-3)(x-3)\\ \amp=\highlight{\left[(x^2-6x+9)(x-3)\right]}(x-3)\\ \amp= \left[x^3-9x^2+27x-27\right](x-3)\\ \amp=x^4 - 12 x^3 + 54 x^2- 108 x + 81 \end{align*}Subsection6.4.4Exercises
Perfect Square Trinomial Formula
Difference of Squares Formula
Application Problems
Binomials Raised to Other Powers