ORCCA Special Cases of Multiplying Polynomials

Section 6.6 Special Cases of Multiplying Polynomials

Since we are now able to multiply polynomials together, we will look at a few special cases of polynomial multiplication.

Subsection 6.6.1 Squaring a Binomial

Example 6.6.1.

To “square a binomial” is to take a binomial and multiply it by itself. We know that exponent notation means that \(4^2=4\cdot 4\text{.}\) Applying this to a binomial, we'll see that \((x+4)^2=(x+4)(x+4)\text{.}\) To expand this expression, we'll simply distribute \((x+4)\) across \((x+4)\text{:}\)

\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*}

Similarly, to expand \((y-7)^2\text{,}\) we'll have:

\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*}

These two examples might look like any other example of multiplying binomials, but looking closely we can see that something very specific (or special) happened. Focusing on the original expression and the simplified one, we can see that a specific pattern occurred in each:

\begin{align*} \left( x+4 \right)^2 \amp= x^2 + \highlight{4}x + \highlight{4}x + \highlight{4\cdot 4}\\ \left( x+\highlight{4} \right)^2 \amp= x^2 +2(\highlight{4}x) + \highlight{4}^2\\ \end{align*}

And:

\begin{align*} \left( y-7 \right)^2 \amp= y^2 -\highlight{7}y - \highlight{7}y + \highlight{7\cdot 7}\\ \left( y-\highlight{7} \right)^2 \amp= y^2 -2(\highlight{7}y) + \highlight{7}^2 \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. Furthermore, the last term is the square of the second term in each original binomial.

What we're seeing is a pattern that relates to two important phrases: The process is called squaring a binomial, and the result is called a perfect square trinomial. The first phrase is a description of what we're doing, we are literally squaring a binomial. The second phrase is a description of what you end up with. This second name will become important in a future chapter.

Example 6.6.2.

The general way this pattern is presented is by squaring the two most general binomials possible, \((a+b)\) and \((a-b)\text{.}\) We will establish the pattern for \((a+b)^2\) and \((a-b)^2\text{.}\) Once we have done so, we will be able to substitute anything in place of \(a\) and \(b\) and rely upon the general pattern to simplify squared binomials.

We first must expand \((a+b)^2\) as \((a+b)(a+b)\) and then we can multiply those binomials:

\begin{align*} (a+b)^2\amp=(a+b)(a+b)\\ \amp=a^2+ab+ba+b^2\\ \amp=a^2+2ab+b^2 \end{align*}

Notice the final simplification step was to add \(ab+ba\text{.}\) Since these are like terms, we can combine them into \(2ab\text{.}\)

Similarly, we can find a general formula for \((a-b)^2\text{:}\)

\begin{align*} (a-b)^2\amp=(a-b)(a-b)\\ \amp=a^2-ab-ba+b^2\\ \amp=a^2-2ab+b^2 \end{align*}

Fact 6.6.3. Squaring a Binomial Formulas.

If \(a\) and \(b\) are real numbers or variable expressions, then we have the following formulas:

\begin{equation*} (a+b)^2 = a^2+2ab+b^2 \end{equation*}

\begin{equation*} (a-b)^2 = a^2-2ab+b^2 \end{equation*}

These formulas will allow us to multiply this type of special product more quickly.

Remark 6.6.4.

Notice that when both \((a+b)^2\) and \((a-b)^2\) are expanded in Example 6.6.2, the last term was a positive \(b^2\) in both. This is because any number or expression, regardless of its sign, is positive after it is squared.

Subsection 6.6.2 Further Examples of Squaring Binomials

Example 6.6.5.

Expand \((2x-3)^2\) using the squaring a binomial formula.

For this example we need to recognize that to apply the formula \((a-b)^2 = a^2-2ab+b^2\) in this situation, \(a=2x\) and \(b=3\text{.}\) Expanding this, we have:

\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*}

Remark 6.6.6.

While we rely on the formula for squaring a binomial in Example 6.6.5, we will often omit the step of formally writing the formula and jump to the simplification, in this way:

\begin{equation*} (2x-3)^2=4x^2-12x+9 \end{equation*}

Example 6.6.7.

Multiply the following using the squaring a binomial formula:

\(\displaystyle (5xy+1)^2\)
\(\displaystyle 4(3x-7)^2\)

Explanation

\(\displaystyle \begin{aligned}[t] (5xy+1)^2 \amp= (5xy)^2+2(5xy)(1)+1^2\\ \amp= 25x^2y^2+10xy+1\end{aligned} \)
With this expression, we will first note that the factor of \(4\) is outside the portion of the expression that is squared. Using the order of operations, we will first expand \((3x-7)^2\) and then multiply that expression by \(4\text{:}\)

\begin{align*} 4(3x-7)^2 \amp= 4\left((3x)^2-2(3x)(7)+7^2\right)\\ \amp= 4\left(9x^2-42x+49\right)\\ \amp= 36x^2-168x+196 \end{align*}

Example 6.6.8.

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.

Explanation

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.

Checkpoint 6.6.9.

Subsection 6.6.3 The Product of the Sum and Difference of Two Terms

To identify the next “special case” for multiplying polynomials, we'll look at a couple of examples.

Example 6.6.10.

Multiply the following binomials:

\(\displaystyle (x+5)(x-5)\)
\(\displaystyle (y-8)(y+8)\)

Explanation

We can approach these as using distribution, FOIL, or generic rectangles, and obtain the following:

\(\displaystyle \begin{aligned}[t] (x+5)(x-5) \amp= x^2-5x+5x-25\\ \amp= x^2 -25\end{aligned} \)
\(\displaystyle \begin{aligned}[t] (y+8)(y-8) \amp= y^2-8y+8y-4\\ \amp= y^2 - 64\end{aligned} \)

Notice that for each of these products, we multiplied the sum of two terms by the difference of the same two terms. Notice also in these three examples that once these expressions were multiplied, the two middle terms were opposites and thus canceled to zero.

These pairs, generally written as \((a+b)\) and \((a-b)\text{,}\) are known as conjugates. If we multiply \((a+b)(a-b)\text{,}\) we can see this general pattern more clearly:

\begin{align*} (a+b)(a-b) \amp= a^2-ab+ab-b^2\\ \amp= a^2 - b^2 \end{align*}

As with the previous special case, this one also 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 pertinent.

Fact 6.6.11. The Product of the Sum and Difference of Two Terms Formula.

If \(a\) and \(b\) are real numbers or variable expressions, then we have the following formula:

\begin{equation*} (a+b)(a-b)=a^2-b^2 \end{equation*}

Checkpoint 6.6.12.

Example 6.6.13.

Multiply the following using Fact 6.6.11.

\(\displaystyle (4x-7y)(4x+7y)\)
\(\displaystyle -2(3x+1)(3x-1)\)

Explanation

The first step to using this method is to identify the values of \(a\) and \(b\text{.}\)

In this instance, \(a=4x\) and \(b=7y\text{.}\) Using the formula,

\begin{align*} (4x-7y)(4x+7y) \amp= (4x)^2-(7y)^2\\ \amp= 16x^2-49y^2 \end{align*}
In this instance, we have a constant factor as well as a product in the form \((a+b)(a-b)\text{.}\) We will first expand \((3x+1)(3x-1)\) by identifying \(a=3x\) and \(b=1\) and using the formula. Then we will multiply the factor of \(-2\) through this expression. So,

\begin{align*} -2(3x+1)(3x-1)\amp= -2\left((3x)^2-1^2\right)\\ \amp= -2\left(9x^2-1\right)\\ \amp= -18x^2+2 \end{align*}

Checkpoint 6.6.14.

List 6.6.15. Special Cases of Multiplication Formulas

If \(a\) and \(b\) are real numbers or variable expressions, then we have the following formulas:

Squaring a Binomial (Sum): \(\displaystyle (a+b)^2 = a^2+2ab+b^2 \)
Squaring a Binomial (Difference): \(\displaystyle (a-b)^2 = a^2-2ab+b^2\)
Product of the Sum and Difference of Two Terms: \(\displaystyle (a+b)(a-b)=a^2-b^2\)

Warning 6.6.16. Common Mistakes.

We've found that

\begin{equation*} (a+b)(a-b)=a^2-b^2 \end{equation*}

However,

\begin{equation*} (a-b)^2\neq a^2-b^2\text{ because }(a-b)^2=a^2\highlight{{}-{}2ab{}+{}}b^2 \end{equation*}

Similarly,

\begin{equation*} (a+b)^2\neq a^2+b^2\text{ because }(a+b)^2=a^2\highlight{{}+{}2ab}+b^2 \end{equation*}

Subsection 6.6.4 Binomials Raised to Other Powers

Example 6.6.17.

Simplify the expression \((x+5)^3\) into an expanded polynomial.

Before we start expanding this expression, it is important to recognize that \((x+5)^3\neq x^3+ 5^3\text{.}\) We can see that this doesn't work by inputting \(1\) for \(x\) and applying the order of operations:

\begin{align*} (\substitute{1}+5)^3\amp=6^3\amp \substitute{1}^3+5^3\amp=1+125\\ \amp=216\amp\amp=126 \end{align*}

With this in mind, we will need to rely on distribution to expand this expression. The first step in expanding \((x+5)^3\) is to remember that the exponent of \(3\) indicates that

\begin{equation*} (x+5)^3=\overbrace{(x+5)(x+5)(x+5)}^{3\text{ times }} \end{equation*}

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*}

Checkpoint 6.6.18.

If we wanted to expand a binomial raised to any power, we always start by rewriting the expression without an exponent.

To multiply \((x-3)^4\text{,}\) we'd start by rewriting \((x-3)^4\) in expanded form as:

\begin{equation*} (x-3)^4=\overbrace{(x-3)(x-3)(x-3)(x-3)}^{4\text{ times}} \end{equation*}

We will then multiply 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^2-9x^3+27x^2-27x-3x^3+27x^2-81x+81\\ \amp=x^4 - 12 x^3 + 54 x^2- 108 x + 81 \end{align*}

Exercises 6.6.5 Exercises

Review and Warmup

1.

Use the properties of exponents to simplify the expression.

\(\left(4y^{2}\right)^4\)

2.

Use the properties of exponents to simplify the expression.

\(\left(5x^{3}\right)^3\)

3.

Use the properties of exponents to simplify the expression.

\(\displaystyle{\left(2y\right)^2}\)

4.

Use the properties of exponents to simplify the expression.

\(\displaystyle{\left(5t\right)^3}\)

5.

Use the properties of exponents to simplify the expression.

\(\left(-8x^{7}\right)^3\)

6.

Use the properties of exponents to simplify the expression.

\(\left(-5t^{8}\right)^2\)

7.

Use the properties of exponents to simplify the expression.

\(-3\left(-10r^{9}\right)^3\)

8.

Use the properties of exponents to simplify the expression.

\(-4\left(-6r^{10}\right)^2\)

9.

Simplify each expression, if possible, by combining like terms.

\({8z^{2}-2s^{2}}\)
\({-4s^{2}-8}\)
\({4y-3y}\)
\({3x+2z^{2}}\)

10.

Simplify each expression, if possible, by combining like terms.

\({2z+z}\)
\({6x-8s^{2}}\)
\({-8s+5s}\)
\({-9s-3s}\)

11.

Simplify each expression, if possible, by combining like terms.

\({3z^{2}-9z}\)
\({4t-9s+7s}\)
\({-t+8s+8y+3s}\)
\({-2t^{2}+7t^{2}-5t}\)

12.

Simplify each expression, if possible, by combining like terms.

\({5z-z^{2}}\)
\({2y+9s}\)
\({-4x-8x}\)
\({-4s^{2}-6x-s}\)

13.

Determine if the following statements are true or false.

\((a-b)^2=a^2-b^2\)
- True
- False
\((a+b)^2=a^2+b^2\)
- True
- False
\((a+b)(a-b)=a^2-b^2\)
- True
- False

14.

Determine if the following statements are true or false.

\((2(a-b))^2=4(a-b)^2\)
- True
- False
\(2(a+b)^2=2a^2+2b^2\)
- True
- False
\(2(a+b)(a-b)=2a^2-2b^2\)
- True
- False