# 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 that each term in the first binomial multiplies with each term in the second binomial. Let’s look at another example.

Multiply two linear (i.e. with \begin{align*}\text{degree} = 1\end{align*}) binomials:

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

When we multiply, we obtain a quadratic (i.e. with \begin{align*}\text{degree} = 2\end{align*}) polynomial 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 \begin{align*}2^{nd}\end{align*} 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 will talk about some special products of binomials.

## Find the Square of a Binomial

A special binomial product is the square of a binomial. Consider the following multiplication.

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

Since we are multiplying the same expression by itself that means that we are squaring the expression. This means that:

\begin{align*}&&(x + 4)(x + 4) &= (x + 4)^2\\ \text{Let’s multiply:} && (x + 4)(x + 4) &= x^2 + 4x + 4x + 16\\ \text{And combine like terms:} && & = x^2 + 8x + 16\end{align*}

Notice that the middle terms are 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*}

It looks like the middle terms are the same again. So far we have squared the sum of binomials. Let’s now square a difference of binomials.

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

We notice a pattern when squaring binomials. To square a binomial, add the square of the first term, add or subtract twice the product of the terms, and the square of the second term. You should remember these formulas:

Square of a Binomial

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

Remember! A polynomial that is raised to an exponent 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 \begin{align*}(a+b)^{2}=a^{2}+b^{2}\end{align*}. To see why \begin{align*}(a+b)^2 \neq a^2+b^2\end{align*} try substituting numbers for a and b 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

\begin{align*}&({\color{red}a}^2 \ + \ {\color{blue}b}) \ = \ {\color{red}a}^2 \ \ + \ 2{\color{red}a} \quad \ {\color{blue}b} \ + \ \ {\color{blue}b}^2\\ & \ \downarrow \quad \quad \downarrow \quad \quad \ \downarrow \quad \quad \quad \downarrow \ \quad \downarrow \quad \quad \ \downarrow \\ & ({\color{red}x} \ + \ {\color{blue}10})^2 = ({\color{red}x})^2 + 2({\color{red}x})({\color{blue}10}) + ({\color{blue}10})^2\\ & \ \ \qquad \qquad =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

\begin{align*}(a-b)^2 & =a^2 -2ab + b^2 \\ (2x-3)^2 & = (2x)^2 - 2(2x)(3) + (3)^2 \\ & = 4x^2 - 12x + 9\end{align*}

c) \begin{align*}(x + 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.

\begin{align*}(a + b) (a - b) & = a^2 - 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*}&({\color{red}a} + {\color{blue}b}) ({\color{red}a} - {\color{blue}b}) = \ {\color{red}a}^2 \ - \ \ {\color{blue}b}^2\\ & \ \downarrow \quad \downarrow \quad \downarrow \quad \downarrow \ \ \quad \ \downarrow \quad \quad \ \ \downarrow \\ & ({\color{red}x} + {\color{blue}3}) ({\color{red}x} - {\color{blue}3}) = ({\color{red}x})^2 - ({\color{blue}3})^2\\ & \ \qquad \qquad \qquad =x^2 - 9\end{align*}

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

\begin{align*}&({\color{red}a} \ + \ {\color{blue}b}) ({\color{red}a} \ - \ {\color{blue}b}) = \ {\color{red}a}^2 \ - \ \ {\color{blue}b}^2\\ & \ \downarrow \quad \ \ \downarrow \quad \downarrow \quad \ \ \downarrow \ \ \quad \ \downarrow \quad \quad \ \ \downarrow \\ & ({\color{red}5x} + {\color{blue}9}) ({\color{red}5x} - {\color{blue}9}) = ({\color{red}5x})^2 - ({\color{blue}9})^2\\ & \ \qquad \qquad \qquad \quad =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

Let’s now 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 area of the square = side \begin{align*}\times\end{align*} side

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

Notice that this gives a visual explanation of the square of binomials product.

\begin{align*}& \text{Area of the big square} && = \text{area of the blue square} && + 2 \ \text{(area of yellow rectangle)} && + \text{area of red square}\\ & (a + b)^2 && = a^2 && + 2ab && + b^2\end{align*}

The next example shows how to use the special products in doing 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 309\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 \begin{align*}43 = (50 - 7)\end{align*} and \begin{align*}57 = (50 + 7).\end{align*}

Then \begin{align*}43 \times 57 = (50 - 7)(50 + 7) = (50)^2 - (7)^2 = 2500 - 49 = 2,451\end{align*}

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

Then \begin{align*}112 \times 88 = (100 + 12)(100 - 12) = (100)^2 - (12)^2 = 10,000 - 144 = 9,856\end{align*}

(c) \begin{align*}45^2 = (40 + 5)^2 = (40)^2 + 2(40)(5) + (5)^2 = 1600 + 400 + 25 = 2,025\end{align*}

(d) Rewrite \begin{align*}481 = (400 + 81)\end{align*} and \begin{align*}319 = (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 160,000

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

\begin{align*}(81)^2 = (80 + 1)^2 = (80)^2 + 2(80)(1) + (1)^2 = 6400 + 160 + 1 = 6,561\end{align*}

Then, \begin{align*}481 \times 309 = (400)^2 - (81)^2 = 160,000 - 6,561 = 153, 439\end{align*}

## Review Questions

Use the special product 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*}(4x^2 + y^2)^2\end{align*}
4. \begin{align*}(8x - 3)^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*}

Find the area of the orange square in the following figure. It is the lower right shaded box.

Multiply the following numbers using the 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*}

1. \begin{align*}x^2 + 18x + 81\end{align*}
2. \begin{align*}9x^2 - 42x + 49\end{align*}
3. \begin{align*}16x^4 + 8x^2y^2 + y^4\end{align*}
4. \begin{align*}64x^2 - 48x + 9\end{align*}
5. \begin{align*}4x^2 - 1\end{align*}
6. \begin{align*}x^2 - 144\end{align*}
7. \begin{align*}25a^2 - 4b^2\end{align*}
8. \begin{align*}a^2b^2 - 1\end{align*}
9. \begin{align*}\text{Area} = (a - b)^2 = a^2 - 2ab + b^2\end{align*}
10. \begin{align*}(50 - 5) (50 + 5) = 2475\end{align*}
11. \begin{align*}(50 + 6)^2 = 3136\end{align*}
12. \begin{align*}(1000 + 2)(1000 - 2) = 999,996\end{align*}
13. \begin{align*}(40 - 4)(40 + 4) = 1584\end{align*}

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