ORCCA Special Cases of Multiplying Polynomials

Section 5.5 Special Cases of Multiplying Polynomials

Objectives: PCC Course Content and Outcome Guide

MTH 65 CCOG 1.b

permalinkSince we are now able to multiply polynomials together in general, we will look at a few special patterns with polynomial multiplication where there are some shortcuts worth knowing about.

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Figure 5.5.1. Alternative Video Lesson

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Subsection 5.5.1 Squaring a Binomial

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Example 5.5.2.

To “square a binomial” is to take a binomial and multiply it by itself. In the same way that $4^2=4\cdot 4\text{,}$ it's also true that $(x+4)^2=(x+4)(x+4)\text{.}$ To expand this expression, we'll simply distribute $(x+4)$ across $(x+4)\text{:}$

\begin{aligned} (x + 4)^{2} & = (x + 4) (x + 4) \\ = x (x + 4) + 4 (x + 4) \\ = x^{2} + 4 x + 4 x + 16 \\ = x^{2} + 8 x + 16 \end{aligned}

$\begin{align*} (x+4)^2 \amp= (x+4)(x+4)\\ \amp=x(x+4) + 4(x+4)\\ \amp= x^2 + 4x + 4x + 16\\ \amp= x^2 + 8x + 16 \end{align*}$

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

\begin{aligned} (y - 7)^{2} & = (y - 7) (y - 7) \\ = y (y - 7) - 7 (y - 7) \\ = y^{2} - 7 y - 7 y + 49 \\ = y^{2} - 14 y + 49 \end{aligned}

$\begin{align*} (y-7)^2 \amp= (y-7)(y-7)\\ \amp=y(y-7) - 7(y-7)\\ \amp= y^2 - 7y - 7y + 49\\ \amp= y^2 - 14y + 49 \end{align*}$

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

\begin{aligned} (x + 4)^{2} & = x^{2} + 2 (4 x) + 4^{2} \end{aligned}

$\begin{align*} (\firsthighlight{x}+\secondhighlight{4})^2 \amp= \firsthighlight{x}^2 +2(\firsthighlight{4}\secondhighlight{x}) + \secondhighlight{4}^2\\ \end{align*}$

And:

\begin{aligned} (y - 7)^{2} & = y^{2} - 2 (7 y) + 7^{2} \end{aligned}

$\begin{align*} (\firsthighlight{y}-\secondhighlight{7})^2 \amp= \firsthighlight{y}^2 -2(\firsthighlight{7}\secondhighlight{y}) + \secondhighlight{7}^2 \end{align*}$

Either way, we have:

(first)^{2} \pm 2 (first) (second) + (second)^{2}

$\begin{equation*} (\firsthighlight{\text{first}})^2\pm2(\firsthighlight{\text{first}})(\secondhighlight{\text{second}})+(\secondhighlight{\text{second}})^2 \end{equation*}$

and the choice of $+$ or $-$ matches the original binomial.

permalinkWhat we're seeing is a pattern relating two things. The left side is the square of a binomial, and the result on the right is called a perfect square trinomial, a trinomial that was born from something getting squared.

The general way this pattern is established is by squaring each of the two most general binomials, $(a+b)$ and $(a-b)\text{.}$ Once we have done so, we can substitute anything in place of $a$ and $b$ and rely upon the general pattern to simplify squared binomials.

We can write $(a+b)^2$ as $(a+b)(a+b)$ and then multiply those binomials:

\begin{aligned} (a + b)^{2} & = (a + b) (a + b) \\ = a^{2} + a b + b a + b^{2} \\ = a^{2} + 2 a b + b^{2} \end{aligned}

$\begin{align*} (\firsthighlight{a}+\secondhighlight{b})^2\amp=(\firsthighlight{a}+\secondhighlight{b})(\firsthighlight{a}+\secondhighlight{b})\\ \amp=\firsthighlight{a}^2+\firsthighlight{a}\secondhighlight{b}+\secondhighlight{b}\firsthighlight{a}+\secondhighlight{b}^2\\ \amp=\firsthighlight{a}^2+2\firsthighlight{a}\secondhighlight{b}+\secondhighlight{b}^2 \end{align*}$

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{aligned} (a - b)^{2} & = (a - b) (a - b) \\ = a^{2} - a b - b a + b^{2} \\ = a^{2} - 2 a b + b^{2} \end{aligned}

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

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Fact 5.5.3. Binomial Squared Formulas.

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

(a + b)^{2} = a^{2} + 2 a b + b^{2}

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

(a - b)^{2} = a^{2} - 2 a b + b^{2}

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

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

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Remark 5.5.4.

Notice that when both $(a+b)^2$ and $(a-b)^2$ are expanded, the last term is adding $b^2$ either way. This is because any number or expression, regardless of its sign, is positive after it is squared.

Some students will prefer to memorize the Binomial Squared Formulas and apply them by substuting expressions in for $a$ and $b\text{.}$ An alternative visualization is presented in Figure 5.5.5.

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Figure 5.5.5. Visualizing the Squaring of a Binomial

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Example 5.5.6.

Expand $(2x-3)^2$ using the Binomial Squared Formulas.

To apply the formula for squaring a binomial, we take $\firsthighlight{a=2x}$ and $\secondhighlight{b=3}\text{.}$ Expanding this, we have:

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

$\begin{alignat*}{3} (\firsthighlight{2x} {}-{} \secondhighlight{3})^2 \amp= \firsthighlight{(2x)}^2 \amp \amp- 2\firsthighlight{(2x)}\secondhighlight{(3)} \amp \amp+ \secondhighlight{(3)}^2\\ \amp= 4x^2 \amp \amp- 12x \amp \amp+ 9 \end{alignat*}$

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Checkpoint 5.5.7.

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Example 5.5.8.

Use the visualization in Figure 5.5.5 to expand these binomials squared.

$(x+8)^2$
$(2t-7)^2$

Explanation

Diagramming the process:

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Diagramming the process:

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Example 5.5.9.

A circle's area can be calculated using the formula

A = π r^{2}

$\begin{equation*} A=\pi r^2 \end{equation*}$

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.

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Explanation

The circle's area would be:

\begin{align*} A\amp=\pi r^2\\ \amp=\pi (x-5)^2\amp\text{Now use a method for squaring this binomial}\ldots\\ \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.

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Checkpoint 5.5.10.

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Warning 5.5.11. Common Mistakes.

Now we know how to expand $(a+b)^2$ and $(a-b)^2\text{.}$ It is a common mistake to think that these are equal to $a^2+b^2$ and $a^2-b^2\text{,}$ respectively, as if you could just “distribute” the exponent. Now we know that actually you get $a^2+2ab+b^2$ and $a^2-2ab+b^2\text{.}$

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Subsection 5.5.2 The Product of the Sum and Difference of Two Terms

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

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Example 5.5.12.

Multiply the following binomials:

$(x+5)(x-5)$
$(y-8)(y+8)$

Explanation

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

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

permalinkNotice 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{aligned} (a + b) (a - b) & = a^{2} - a b + a b - b^{2} \\ = a^{2} - b^{2} \end{aligned}

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

permalinkAs with the square of a binomial producing a perfect suqare trinomial, this pattern also has two things we can give a name to. The left side is the product of a sum and its conjugate, and the result on the right is a difference of squares.

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Fact 5.5.13. The Product of a Sum and Its Conjugate Formula.

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

(a + b) (a - b) = a^{2} - b^{2}

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

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Example 5.5.14.

Multiply the following using Fact 5.5.13.

$(4x-7y)(4x+7y)$
$-2(3x+1)(3x-1)$

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*}

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Example 5.5.17.

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

Before we start expanding this expression, it is important to recognize that $(x+5)^3\neq x^3+ 5^3\text{,}$ similar to the message in Warning 5.5.11. To be sure, we can see that if we evaluate at $x=1\text{,}$ we get different results.

\begin{aligned} (1 + 5)^{3} & = 6^{3} & 1^{3} + 5^{3} & = 1 + 125 \\ = 216 & = 126 \end{aligned}

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

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

(x + 5)^{3} = \overset{3 times}{\overset{⏞}{(x + 5) (x + 5) (x + 5)}}

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

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{aligned} (x + 5)^{3} & = \overset{a binomial squared}{\overset{⏞}{(x + 5) (x + 5)}} (x + 5) \\ = (x^{2} + 10 x + 25) (x + 5) \\ = x^{3} + 5 x^{2} + 10 x^{2} + 50 x + 25 x + 125 \\ = x^{3} + 15 x^{2} + 75 x + 125 \end{aligned}

$\begin{align*} (x+5)^3\amp=\overbrace{(x+5)(x+5)}^{\text{a binomial squared}}(x+5)\\ \amp=\highlight{\left(x^2+10x+25\right)}(x+5)\\ \amp=x^3 + 5x^2 + 10x^2 +50x+25x+125\\ \amp=x^3 + 15x^2 + 75x + 125 \end{align*}$

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Checkpoint 5.5.18.

permalinkGeneralizing, if we want to expand a binomial raised to a high whole number power, we can start by rewriting the expression without an exponent. Then it will help some to use the formula for the square of a binomial.

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Example 5.5.19.

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

(x - 3)^{4} = \overset{4 times}{\overset{⏞}{(x - 3) (x - 3) (x - 3) (x - 3)}}

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

We will then multiply pairs of polynomials from the left to the right.

\begin{aligned} (x - 3)^{4} & = \overset{a perfect square}{\overset{⏞}{(x - 3) (x - 3)}} \overset{a perfect square}{\overset{⏞}{(x - 3) (x - 3)}} \\ = (x^{2} - 6 x + 9) (x^{2} - 6 x + 9) \\ = x^{4} - 6 x^{3} + 9 x^{2} - 6 x^{3} + 36 x^{2} - 54 x + 9 x^{2} - 54 x + 81 \\ = x^{4} - 12 x^{3} + 54 x^{2} - 108 x + 81 \end{aligned}

$\begin{align*} (x-3)^4\amp=\overbrace{(x-3)(x-3)}^{\text{a perfect square}}\overbrace{(x-3)(x-3)}^{\text{a perfect square}}\\ \amp=\highlight{\left(x^2-6x+9\right)}\highlight{\left(x^2-6x+9\right)}\\ \amp=x^4 - 6x^3 + 9x^2 - 6x^3 + 36x^2 - 54x + 9x^2 - 54x + 81 \\ \amp=x^4 - 12x^3 + 54x^2 - 108x + 81 \end{align*}$

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Reading Questions 5.5.4 Reading Questions

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1.

How many special patterns should you be on the lookout for when multiplying and/or squaring binomials?

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2.

Do you prefer to memorize the formula for the square of a binomial or to visualize the process?

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Exercises 5.5.5 Exercises

Review and Warmup

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1.

Use the properties of exponents to simplify the expression.

$\left(3y^{11}\right)^4$

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2.

Use the properties of exponents to simplify the expression.

$\left(5x^{12}\right)^3$

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3.

Use the properties of exponents to simplify the expression.

$\displaystyle{\left(2x\right)^2}$

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4.

Use the properties of exponents to simplify the expression.

$\displaystyle{\left(4r\right)^2}$

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5.

Use the properties of exponents to simplify the expression.

$\left(-10y^{5}\right)^2$

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6.

Use the properties of exponents to simplify the expression.

$\left(-7x^{6}\right)^3$

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7.

Use the properties of exponents to simplify the expression.

$-3\left(-3t^{8}\right)^3$

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8.

Use the properties of exponents to simplify the expression.

$-4\left(-8t^{9}\right)^2$

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9.

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

${-9s+2}$
${7x+2}$
${9y^{2}+3y}$
${-x+8s}$

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10.

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

${-2t+7x}$
${-y-2s}$
${-9x^{2}+5y}$
${t-8y}$

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11.

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

${7t^{2}+7-5t^{2}-4}$
${-t^{2}-2t^{2}}$
${3y+6y-2x}$
${-4s-3z+7z}$

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12.

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

${t+6t}$
${9z-6t+2+7z}$
${6x-5x-4z+2s^{2}}$
${3z^{2}+2}$