# 9.5: Special Products of Polynomials

**Basic**Created by: CK-12

**Practice**Special Products of Polynomials

Suppose that the entrance to a dog house is a square, with a height of \begin{align*}x\end{align*}

### Guidance

When we multiply two linear (degree of 1) binomials, we create a quadratic (degree of 2) polynomial with four terms. The middle terms are like terms so we can combine them and simplify to get a quadratic or \begin{align*}2^{nd}\end{align*}**trinomial** (polynomial with three terms). In this Concept, we will talk about some special products of binomials.

**Finding 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*}

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

This follows the general pattern of the following rule.

**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*}

Stay aware of the common mistake \begin{align*}(a+b)^2=a^2+b^2\end{align*}*not* a true statement. The middle term, \begin{align*}2ab\end{align*}

#### Example A

*Simplify by multiplying:* \begin{align*}(x+10)^2\end{align*}

**Solution:** Use the square of a binomial formula, substituting \begin{align*}a=x\end{align*}

\begin{align*}(a+b)^2&=a^2+2ab+b^2\\
(x+10)^2 & =(x)^2+2(x)(10)+(10)^2=x^2+20x+100\end{align*}

**Finding 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 always happens when we multiply a sum and difference of the same terms.

\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*}

#### Example B

*Multiply the following binomias and simplify.*

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

**Solution:** Use the above formula, with \begin{align*}a=5x\end{align*}

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

**Solving 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. Look at the following example.

#### Example C

*Find the area of the square.*

Solution: \begin{align*}The \ area \ of \ the \ square = side \times side\end{align*}

\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*}Area \ of \ big \ square: (a+b)^2 = Area \ of \ blue \ square = a^2+2 \ (area \ of \ yellow) = 2ab + area \ of \ red \ square = b^2\end{align*}

The next example shows how to use the special products in doing fast mental calculations.

#### Example D

*Find the products of the following numbers without using a calculator.*

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

(b) \begin{align*}45^2\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*}

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

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

### Guided Practice

*Multiply \begin{align*}(2x+3y)(2x-3y)\end{align*} (2x+3y)(2x−3y).*

**Solution:**

In order to get the hang of the patterns involved in ** special products,** apply the distributive property to see what will happen:

\begin{align*}(2x+3y)(2x-3y)&=\\
2x(2x-3y)+3y(2x-3y)&=\\
2x(2x)+2x(-3y)+3y(2x)+3y(-3y)&=\\
4x^2-6xy+6xy-9y^2&=\\
4x^2-9y^2\end{align*}

Notice how the two **middle terms** canceled each other out. This always happens, which is where we get the sum and difference product. Compare the answer above to that from using the sum and difference product:

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

The two answers are the same. You can use the sum and difference product as a shortcut, so you don't always have to go through the whole process of multiplying out using the distributive property.

### Practice

Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Special Products of Binomials (10:36)

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

- \begin{align*}(x+9)^2\end{align*}
- \begin{align*}(x-1)^2\end{align*}
- \begin{align*}(2y+6)^2\end{align*}
- \begin{align*}(3x-7)^2\end{align*}
- \begin{align*}(7c+8)^2\end{align*}
- \begin{align*}(9a^2+6)^2\end{align*}
- \begin{align*}(b^2-1)^2\end{align*}
- \begin{align*}(m^3+4)^2\end{align*}
- \begin{align*}\left ( \frac{1}{4} t+2 \right )^2\end{align*}
- \begin{align*}(6k-3)^2\end{align*}
- \begin{align*}(a^3-7)^2\end{align*}
- \begin{align*}(4x^2+y^2)^2\end{align*}
- \begin{align*}(8x-3)^2\end{align*}

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

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

- \begin{align*}45\times 55\end{align*}
- \begin{align*}97 \times 83\end{align*}
- \begin{align*}19^2\end{align*}
- \begin{align*}56^2\end{align*}
- \begin{align*}876 \times 824\end{align*}
- \begin{align*}1002 \times 998\end{align*}
- \begin{align*}36 \times 44\end{align*}

### Image Attributions

Here you'll learn how to find special products of two binomials.