# 9.3: Special Products of Polynomials

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Find the square of a binomial
• Find the product of binomials using sum and difference formula
• Solve problems using special products of polynomials

## Introduction

We saw that when we multiply two binomials we need to make sure to multiply each term in the first binomial with each term in the second binomial. Let’s look at another example.

Multiply two linear binomials (binomials whose degree is 1):

\begin{align*}(2x+3)(x+4)\end{align*}

When we multiply, we obtain a quadratic polynomial (one with degree 2) with four terms:

\begin{align*}2x^2+8x+3x+12\end{align*}

The middle terms are like terms and we can combine them. We simplify and get \begin{align*}2x^2+11x+12\end{align*}. This is a quadratic, or second-degree, trinomial (polynomial with three terms).

You can see that every time we multiply two linear binomials with one variable, we will obtain a quadratic polynomial. In this section we’ll talk about some special products of binomials.

## Find the Square of a Binomial

One special binomial product is the square of a binomial. Consider the product \begin{align*}(x+4)(x+4)\end{align*}.

Since we are multiplying the same expression by itself, that means we are squaring the expression. \begin{align*}(x+4)(x+4)\end{align*} is the same as \begin{align*}(x+4)^2\end{align*}.

When we multiply it out, we get \begin{align*}x^2+4x+4x+16\end{align*}, which simplifies to \begin{align*}x^2+8x+16\end{align*}.

Notice that the two middle terms—the ones we added together to get \begin{align*}8x\end{align*}—were the same. Is this a coincidence? In order to find that out, let’s square a general linear binomial.

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

Sure enough, the middle terms are the same. How about if the expression we square is a difference instead of a sum?

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

It looks like the middle two terms are the same in general whenever we square a binomial. The general pattern is: to square a binomial, take the square of the first term, add or subtract twice the product of the terms, and add the square of the second term. You should remember these formulas:

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

Remember! Raising a polynomial to a power means that we multiply the polynomial by itself however many times the exponent indicates. For instance, \begin{align*}(a+b)^2=(a+b)(a+b)\end{align*}. Don’t make the common mistake of thinking that \begin{align*}(a+b)^2=a^2+b^2\end{align*}! To see why that’s not true, try substituting numbers for \begin{align*}a\end{align*} and \begin{align*}b\end{align*} into the equation (for example, \begin{align*}a = 4\end{align*} and \begin{align*}b = 3\end{align*}), and you will see that it is not a true statement. The middle term, \begin{align*}2ab\end{align*}, is needed to make the equation work.

We can apply the formulas for squaring binomials to any number of problems.

Example 1

Square each binomial and simplify.

a) \begin{align*}(x+10)^2\end{align*}

b) \begin{align*}(2x-3)^2\end{align*}

c) \begin{align*}(x^2+4)^2\end{align*}

d) \begin{align*}(5x-2y)^2\end{align*}

Solution

Let’s use the square of a binomial formula to multiply each expression.

a) \begin{align*}(x+10)^2\end{align*}

If we let \begin{align*}a = x\end{align*} and \begin{align*}b = 10\end{align*}, then our formula \begin{align*}(a+b)^2=a^2+2ab+b^2\end{align*} becomes \begin{align*}(x+10)^2=x^2+2(x)(10)+10^2\end{align*}, which simplifies to \begin{align*}x^2+20x+100\end{align*}.

b) \begin{align*}(2x-3)^2\end{align*}

If we let \begin{align*}a = 2x\end{align*} and \begin{align*}b = 3\end{align*}, then our formula \begin{align*}(a-b)^2=a^2-2ab+b^2\end{align*} becomes \begin{align*}(2x-3)^2=(2x^2)-2(2x)(3)+(3)^2\end{align*}, which simplifies to \begin{align*}4x^2-12x+9\end{align*}.

c) \begin{align*}(x^2+4)^2\end{align*}

If we let \begin{align*}a = x^2\end{align*} and \begin{align*}b = 4\end{align*}, then

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

d) \begin{align*}(5x-2y)^2\end{align*}

If we let \begin{align*}a = 5x\end{align*} and \begin{align*}b = 2y\end{align*}, then

\begin{align*}(5x-2y)^2 & = (5x)^2-2(5x)(2y)+(2y)^2\\ & = 25x^2-20xy+4y^2\end{align*}

## Find the Product of Binomials Using Sum and Difference Patterns

Another special binomial product is the product of a sum and a difference of terms. For example, let’s multiply the following binomials.

\begin{align*}(x+4)(x-4)&=x^2-4x+4x-16\\ & = x^2-16\end{align*}

Notice that the middle terms are opposites of each other, so they cancel out when we collect like terms. This is not a coincidence. This always happens when we multiply a sum and difference of the same terms. In general,

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

When multiplying a sum and difference of the same two terms, the middle terms cancel out. We get the square of the first term minus the square of the second term. You should remember this formula.

Sum and Difference Formula: \begin{align*}(a+b)(a-b)=a^2-b^2\end{align*}

Let’s apply this formula to a few examples.

Example 2

Multiply the following binomials and simplify.

a) \begin{align*}(x+3)(x-3)\end{align*}

b) \begin{align*}(5x+9)(5x-9)\end{align*}

c) \begin{align*}(2x^3+7)(2x^3-7)\end{align*}

d) \begin{align*}(4x+5y)(4x-5y)\end{align*}

Solution

a) Let \begin{align*}a = x\end{align*} and \begin{align*}b = 3\end{align*}, then:

\begin{align*}(a+b)(a-b) & = a^2-b^2\\ (x+3)(x-3) & = x^2-3^2\\ & = x^2-9\end{align*}

b) Let \begin{align*}a = 5x\end{align*} and \begin{align*}b = 9\end{align*}, then:

\begin{align*}(a+b)(a-b)&=a^2-b^2\\ (5x+9)(5x-9)&=(5x)^2-9^2\\ & = 25x^2-81\end{align*}

c) Let \begin{align*}a = 2x^3\end{align*} and \begin{align*}b = 7\end{align*}, then:

\begin{align*}(2x^3+7)(2x^3-7) & = (2x^3)^2-(7)^2\\ & = 4x^6-49\end{align*}

d) Let \begin{align*}a = 4x\end{align*} and \begin{align*}b = 5y\end{align*}, then:

\begin{align*}(4x+5y)(4x-5y)&=(4x)^2-(5y)^2\\ & = 16x^2-25y^2\end{align*}

## Solve Real-World Problems Using Special Products of Polynomials

Now let’s see how special products of polynomials apply to geometry problems and to mental arithmetic.

Example 3

Find the area of the following square:

Solution

The length of each side is \begin{align*}(a+b)\end{align*}, so the area is \begin{align*}(a+b)(a+b)\end{align*}.

Notice that this gives a visual explanation of the square of a binomial. The blue square has area \begin{align*}a^2\end{align*}, the red square has area \begin{align*}b^2\end{align*}, and each rectangle has area \begin{align*}ab\end{align*}, so added all together, the area \begin{align*}(a+b)(a+b)\end{align*} is equal to \begin{align*}a^2+2ab+b^2\end{align*}.

The next example shows how you can use the special products to do fast mental calculations.

Example 4

Use the difference of squares and the binomial square formulas to find the products of the following numbers without using a calculator.

a) \begin{align*}43 \times 57\end{align*}

b) \begin{align*}112 \times 88\end{align*}

c) \begin{align*}45^2\end{align*}

d) \begin{align*}481 \times 319\end{align*}

Solution

The key to these mental “tricks” is to rewrite each number as a sum or difference of numbers you know how to square easily.

a) Rewrite 43 as \begin{align*}(50-7)\end{align*} and 57 as \begin{align*}(50 + 7)\end{align*}.

Then \begin{align*}43 \times 57 = (50-7)(50 + 7) = (50)^2 - (7)^2 = 2500-49 = 2451\end{align*}

b) Rewrite 112 as \begin{align*}(100 + 12)\end{align*} and 88 as \begin{align*}(100-12)\end{align*}.

Then \begin{align*}112 \times 88 = (100 + 12)(100-12) = (100)^2 - (12)^2 = 10000 - 144 = 9856\end{align*}

c) \begin{align*}45^2 = (40 + 5)^2 = (40)^2 + 2(40)(5) + (5)^2 = 1600 + 400 + 25 = 2025\end{align*}

d) Rewrite 481 as \begin{align*}(400 + 81)\end{align*} and 319 as \begin{align*}(400-81)\end{align*}.

Then \begin{align*}481 \times 319 = (400 + 81)(400-81) = (400)^2-(81)^2\end{align*}

\begin{align*}(400)^2\end{align*} is easy - it equals 160000.

\begin{align*}(81)^2\end{align*} is not easy to do mentally, so let’s rewrite 81 as \begin{align*}80 + 1\end{align*}.

\begin{align*}(81)^2 = (80 + 1)^2 = (80)^2 + 2(80)(1) + (1)^2 = 6400 + 160 + 1 = 6561\end{align*}

Then \begin{align*}481 \times 319 = (400)^2 - (81)^2 = 160000 - 6561 = 153439\end{align*}

## Review Questions

Use the special product rule for squaring binomials to multiply these expressions.

1. \begin{align*}(x+9)^2\end{align*}
2. \begin{align*}(3x-7)^2\end{align*}
3. \begin{align*}(5x-y)^2\end{align*}
4. \begin{align*}(2x^3-3)^2\end{align*}
5. \begin{align*}(4x^2+y^2)^2\end{align*}
6. \begin{align*}(8x-3)^2\end{align*}
7. \begin{align*}(2x+5)(5+2x)\end{align*}
8. \begin{align*}(xy-y)^2\end{align*}

Use the special product of a sum and difference to multiply these expressions.

1. \begin{align*}(2x-1)(2x+1)\end{align*}
2. \begin{align*}(x-12)(x+12)\end{align*}
3. \begin{align*}(5a-2b)(5a+2b)\end{align*}
4. \begin{align*}(ab-1)(ab+1)\end{align*}
5. \begin{align*}(z^2+y)(z^2-y)\end{align*}
6. \begin{align*}(2q^3+r^2)(2q^3-r^2)\end{align*}
7. \begin{align*}(7s-t)(t+7s)\end{align*}
8. \begin{align*}(x^2y+xy^2)(x^2y-xy^2)\end{align*}

Find the area of the lower right square in the following figure.

Multiply the following numbers using special products.

1. \begin{align*}45 \times 55\end{align*}
2. \begin{align*}56^2\end{align*}
3. \begin{align*}1002 \times 998\end{align*}
4. \begin{align*}36 \times 44\end{align*}
5. \begin{align*}10.5 \times 9.5\end{align*}
6. \begin{align*}100.2 \times 9.8\end{align*}
7. \begin{align*}-95 \times -105\end{align*}
8. \begin{align*}2 \times -2\end{align*}

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CK.MAT.ENG.SE.2.Algebra-I.9.3