Special Products of Polynomials

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 2^nd degree trinomial (polynomial with three terms). In this lesson, we will talk about some special products of binomials.

A special binomial product is the square of a binomial. Consider the following multiplication: \(\begin{align*}(x+4)(x+4)\end{align*}\). We are multiplying the same expression by itself, which means that we are squaring the expression. This means that:

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

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 \(\begin{align*}(a+b)^2=a^2+b^2\end{align*}\). To see why \(\begin{align*}(a+b)^2 \neq a^2+b^2\end{align*}\), 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.

Example 1: Simplify by multiplying: \(\begin{align*}(x+10)^2\end{align*}\).

Solution: Use the square of a binomial formula, substituting \(\begin{align*}a=x\end{align*}\) and \(\begin{align*}b=10\end{align*}\)

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

Source: cK-12, https://www.ck12.org/book/basic-algebra/section/9.3/
This work is licensed under a Creative Commons Attribution-NonCommercial 3.0 License.

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 2: Multiply the following binomial and simplify.

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

Solution: Use the above formula, substituting \(\begin{align*}a=5x\end{align*}\) and \(\begin{align*}b=9\end{align*}\). Multiply.

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

Let's now see how special products of polynomials apply to geometry problems and to mental arithmetic. Look at the following example.

Example: 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 3: 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 \(\begin{align*}43=(50-7)\end{align*}\) 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) \(\begin{align*}45^2 = (40+5)^2 = (40)^2+2(40)(5) +(5)^2 = 1600+400+25=2,025\end{align*}\)

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.

Special Products of Binomials

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

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

14. \(\begin{align*}(2x-1)(2x+1)\end{align*}\)
15. \(\begin{align*}(2x-3)(2x+3)\end{align*}\)
16. \(\begin{align*}(4+6x)(4-6x)\end{align*}\)
17. \(\begin{align*}(6+2r)(6-2r)\end{align*}\)
18. \(\begin{align*}(-2t+7)(2t+7)\end{align*}\)