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12.4: Multiplying Binomials

Difficulty Level: At Grade Created by: CK-12
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The Community Garden

The students loved walking through the community garden which was adjacent to the town hall. All of the flowers were in full bloom. One of the community workers was there and he presented the students with the following problem. The students suspected that this was the work of their teacher Mr. Travis, but they began to work on solving the problem anyway. Here it is.

A farmer has two rectangular fields. One measures \begin{align*}3x + 7\end{align*}3x+7 by \begin{align*}2x - 4\end{align*}2x4. The other measures \begin{align*}x^2 + 1\end{align*}x2+1 by \begin{align*}6x + 5\end{align*}6x+5. Find the combined area of the two fields.

To figure this out, you will need to understand multiplying binomials. Because we find the area of a rectangle through multiplication and both of these fields have measurements written in binomials, understanding binomials will be important in solving this problem.

What You Will Learn

In this lesson, you will learn how to complete the following skills.

  • Multiply binomials using a table.
  • Multiply binomials vertically.
  • Multiply binomials horizontally (FOIL)
  • Solve real – world problems by writing, simplifying and evaluating polynomial expressions involving linear dimensions, surface areas and volumes of rectangular prisms.

Teaching Time

I. Multiply Binomials Using a Table

We defined binomials as two-term polynomials. When we added and subtracted polynomials, we were careful to combine like terms. When we multiply polynomials we will carefully apply the rules of exponents, as well.

When we multiply binomials, we can use a table to help us to organize and keep track of the information. This table will help you to organize and keep track of your work. Let’s take a look.


Multiply the binomials \begin{align*}(x+5)(x+3)\end{align*}(x+5)(x+3).

We can use a table like a rectangle, as if each of the binomials were a dimension of the rectangle.

We will insert the two binomials along the sides of the table like a rectangle.

Now, we will find the area of the four separate rectangles.

The dimensions of the first rectangle is \begin{align*}x \cdot x\end{align*}xx, the second is \begin{align*}5 \cdot x\end{align*}5x, the third is \begin{align*}3 \cdot x\end{align*}3x, and the fourth is \begin{align*}3 \cdot 5\end{align*}35.

In order to find the total, we will add the four areas: \begin{align*}x^2 + 5x +3x +15\end{align*}x2+5x+3x+15

Now, combine like terms carefully: \begin{align*}x^2 + 8x + 15\end{align*}x2+8x+15.

This is our answer.


Multiply \begin{align*}(x-4)(x+6)\end{align*}(x4)(x+6)

\begin{align*}x\end{align*}x \begin{align*}-4\end{align*}4
\begin{align*}x\end{align*}x \begin{align*}x^2\end{align*}x2 \begin{align*}-4x\end{align*}4x
\begin{align*}+6\end{align*}+6 \begin{align*}6x\end{align*}6x \begin{align*}-24\end{align*}24

\begin{align*}& x^2-4x+6x-24 \\ &= x^2+2x-24\end{align*}x24x+6x24=x2+2x24

Notice the careful work with the negative and positive signs.

Copy one of these examples down in your notebook.

Here is one that is a little different. Notice the position of the exponent in this example.


Multiply \begin{align*}(5x - 8)^2\end{align*}(5x8)2.

Remember that the exponent applies to the entire binomial such that \begin{align*}(5x -8)^2 = (5x - 8)(5x - 8)\end{align*}(5x8)2=(5x8)(5x8).

\begin{align*}5x\end{align*}5x \begin{align*}-8\end{align*}8
\begin{align*}5x\end{align*}5x \begin{align*}25x^2\end{align*}25x2 \begin{align*}-40x\end{align*}40x
\begin{align*}-8\end{align*}8 \begin{align*}-40x\end{align*}40x \begin{align*}64\end{align*}64

\begin{align*}& 25x^2-40x-40x+64 \\ &=25x^2-80x +64\end{align*}25x240x40x+64=25x280x+64

II. Multiply Binomials Vertically

In the last section, you learned how to multiply binomials by using a table. In this lesson you will learn how to multiply binomials vertically.

A second method for multiplying binomials is similar to the algorithm that we commonly use for multiplying two-digit numbers.

\begin{align*}& \qquad \qquad \qquad \quad \text{thousands} \qquad \quad \text{hundreds} \qquad \quad \text{tens} \qquad \quad \text{ones} \\ & \quad \quad 73 \ \quad \rightarrow \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 70 \ \quad + \qquad 3 \\ & \ \underline{\quad \times 81} \ \quad \rightarrow \ \ \underline{\qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \quad 81 \ \quad + \qquad 1} \\ & \quad \quad 73\ \quad \rightarrow \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 70 \ \quad + \qquad 3 \\ & \ \underline{+4580} \ \quad \rightarrow \ \ \underline{\ \ \qquad 4000 \qquad + \qquad \ 500 \quad + \ \quad 80 \ \quad + \qquad 0} \\ & \quad 4653 \ \quad \leftarrow \ \quad \qquad 4000 \qquad + \ \qquad 500 \quad + \quad 150 \ \quad + \qquad 3 \end{align*}thousandshundredstensones73 70 +3 ×81    81 +173 70 +3 +4580     4000+ 500+ 80 +04653  4000+ 500+150 +3

When you expand the multiplication like is done on the right, you can see that in our multiplication algorithm for two-digit numbers, we line up numbers by similar places.

We will use this same idea to multiply binomials but instead of using decimal places, we will line up the products by like terms.


Multiply \begin{align*}(3x + 2)(5x + 4)\end{align*}(3x+2)(5x+4)

\begin{align*}& 2^{nd} \text{power} \qquad \ \quad 1^{st} \text{power} \qquad \quad 0 \ \text{power} \\ & \qquad \qquad \qquad \qquad \ \ \ \quad 3x \quad + \qquad \ \quad 2 \\ & \underline{\qquad \qquad \qquad \qquad \qquad 5x \quad + \qquad \ \quad 4 \;\;} \\ & \qquad \qquad \qquad \qquad \ \ \quad 12x \quad + \qquad \ \quad 8 \\ & \underline{\qquad 15x^2 \quad + \qquad \ \quad 10x \ \quad + \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\ & \qquad 15x^2 \quad + \qquad \ \quad 22x \quad + \qquad \ \quad 8 \quad \rightarrow \quad 15x^2 + 22x + 8\end{align*}2ndpower 1stpower0 power   3x+ 25x+ 4  12x+ 815x2+ 10x +15x2+ 22x+ 815x2+22x+8

Just as in multiplication with numerals, we first multiply 4 by 2 and get 8. We then multiply 4 by \begin{align*}3x\end{align*}3x and get \begin{align*}12x\end{align*}12x. When we multiply \begin{align*}5x\end{align*}5x by 2, we get \begin{align*}10x\end{align*}10x. This is not a like term of 8 but of \begin{align*}12x\end{align*}12x so the \begin{align*}10x\end{align*}10x was aligned beneath the \begin{align*}12x\end{align*}12x. Finally, the product of \begin{align*}5x\end{align*}5x and \begin{align*}3x\end{align*}3x is \begin{align*}15x^2\end{align*}15x2. This is to the \begin{align*}2^{nd}\end{align*}2nd power so how no like terms. Our sum on the bottom-right shows the final product.


Multiply \begin{align*}(7x - 3)^2\end{align*}(7x3)2

\begin{align*}& 2^{nd} \text{power} \qquad \quad 1^{st} \text{power} \qquad \quad 0 \ \ \text{power} \\ & \qquad \qquad \qquad \qquad \ \ \quad 7x \quad + \qquad \quad -3 \\ & \underline{\qquad \qquad \qquad \qquad \ \ \quad 7x \quad + \qquad \quad -3} \\ & \qquad \qquad \qquad \qquad -21x \quad + \qquad \ \ \quad 9 \\ & \underline{\qquad 49x^2 \quad + \qquad -21x \ \quad + \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\ & \qquad 49x^2 \quad + \qquad -42x \quad + \qquad \qquad 9 \quad \rightarrow \quad 49x^2 -42x + 9\end{align*}2ndpower1stpower0  power  7x+3  7x+321x+  949x2+21x +49x2+42x+949x242x+9

This is another special case called a perfect square trinomial. Because the binomial was taken to the second power, we had \begin{align*}(7x - 3)(7x - 3)\end{align*}. This resulted in our product’s first term, \begin{align*}49x^2\end{align*}, and its last term, 9, being perfect squares. Also, the middle term, \begin{align*}-42x\end{align*}, was equal to \begin{align*}-21x + -21x\end{align*}, or \begin{align*}2 \cdot(-21x)\end{align*}.

Yes. You are thinking of what we often call “FOIL”. We will multiply binomials in that way right now.

III. Multiply Binomials Horizontally (FOIL)

In the previous sections, we noticed that when you multiply two binomials, the product is actually the sum of four multiplications, like in the table method or the vertical method of multiplication. In other words, the product \begin{align*}(3x+2)(5x+4)\end{align*} was found by multiplying \begin{align*}3x\end{align*} by \begin{align*}5x,3x\end{align*} by 4, 2 by \begin{align*}5x\end{align*} and 2 by 4. The four resulting products were combined if they were like terms and that was our final product.

If we notice this pattern, then a third option for multiplying will be to simply show those four multiplications without drawing a box or setting up the multiplication vertically.

This third method is called the FOIL method. “FOIL” is an acronym which tells us which terms to multiply in order to get our product—

F—First terms in the binomials

O—Outside terms in the binomials

I—Inside terms in the binomials

L—Last terms in the binomials

Let’s multiply \begin{align*}(2x+8)(5x-13)\end{align*} using the FOIL method.

F First terms are \begin{align*}2x\end{align*} and \begin{align*}5x\end{align*} \begin{align*}(\mathbf{\underline{2x}}+8)(\mathbf{\underline{5x}}-13)\end{align*} \begin{align*}10x^2\end{align*}
O Outside terms are \begin{align*}2x\end{align*} and -13 \begin{align*}(\mathbf{\underline{2x}}+8)(5x\underline{\mathbf{-13}})\end{align*} \begin{align*}-26x\end{align*}
I Inside terms are 8 and \begin{align*}5x\end{align*} \begin{align*}(2x\underline{\mathbf{+8}})(\mathbf{\underline{5x}}-13)\end{align*} \begin{align*}40x\end{align*}
L Last terms are 8 and -13 \begin{align*}(2x\underline{\mathbf{+8}})(5x\underline{\mathbf{-13}})\end{align*} -104

The table above helps to illustrate which terms must be multiplied. However, we don’t need to make a table like that for each multiplication. We can show it like this:

\begin{align*}(2x+8)(5x-13)=10x^2 - 26x + 40x - 104 = 10x^2 + 14x - 104\end{align*}

Of the three methods we’ve seen for multiplication, you might agree that this is the quickest method. Of course, all three methods should give us the same product.

Write the acronym FOIL and what each letter stands for in your notebook.


Multiply \begin{align*}(6x+5)(3x-1)\end{align*}

\begin{align*}&(6x+5)(3x-1) \\ &= 6x \cdot 3x + 6x \cdot -1 + 5\cdot 3x + 5 \cdot -1 \\ &= 18x^2 -6x +15x -5 \\ &= 18x^2 +9x -5 \end{align*}


Multiply \begin{align*}(5x^3+2x)(7x^2+8)\end{align*}

\begin{align*}&(5x^3+2x)(7x^2+8) \\ &= 5x^3 \cdot 7x^2 + 5x^3 \cdot 8 + 2x \cdot 7x^2 + 2x \cdot 8 \\ &= 35x^5 +40x^3 +14x^3 +16x \\ &= 35x^5 +54x^3 +16x \end{align*}

Now let’s look at how we can work with these concepts in real – life.

IV. Solve Real – World Problems by Writing, Simplifying and Evaluating Polynomial Expressions Involving Linear Dimensions, Surface Areas and Volume of Rectangular Prisms

Now let’s apply what we have learned to three dimensional figures.

To find the volume of a rectangular prism, use the formula \begin{align*}V=lwh\end{align*}.


A rectangular prism has the following dimensions: \begin{align*}l = x, w = 2x + 1,\end{align*} and \begin{align*}h = 3x - 2\end{align*}.

Find its volume.

\begin{align*}V &= lwh \\ V &= x(2x+1)(3x-2) \\ V &= (2x^2+x)(3x-2) \\ V &= 2x^2 \cdot 3x + 2x^2 \cdot -2 + x \cdot 3x + x \cdot -2 \\ V &= 6x^3 -4x^2 + 3x^2 -2x \\ V &= 6x^3 -x^2 - 2x\end{align*}

Now let’s apply what we have learned about multiplying binomials to the problem from the introduction.

Real-Life Example Completed

The Community Garden

Here is the problem from the introduction. Reread it and then solve it for the combined area of the two fields.

The students loved walking through the community garden which was adjacent to the town hall. All of the flowers were in full bloom. One of the community workers was there and he presented the students with the following problem. The students suspected that this was the work of their teacher Mr. Travis, but they began to work on solving the problem anyway. Here it is.

A farmer has two rectangular fields. One measures \begin{align*}3x + 7\end{align*} by \begin{align*}2x - 4\end{align*}. The other measures \begin{align*}x^2 + 1\end{align*} by \begin{align*}6x + 5\end{align*}. Find the combined area of the two fields.

Remember you are solving for the combined area of the two fields.

Solution to Real – Life Example

Remember, we need to work on figuring out the area of both rectangles and then the sum of those two areas will give us the total area. Here is how we can solve this problem.

\begin{align*}&(3x+7)(2x-4)+(x^2+1)(6x+5) \\ &= (3x \cdot 2x + 3x \cdot -4 + 7 \cdot 2x + 7 \cdot -4) + (x^2 \cdot 6x +x^2 \cdot 5 + 1 \cdot 6x + 1 \cdot 5) \\ &= (6x^2 -12x +14x -28) + (6x^3 +5x^2 + 6x +5) \\ &= 6x^2 + 2x - 28 +6x^3 + 5x^2 +6x +5 \\ &= 6x^3 + 11x^2 + 8x -23 \end{align*}


Here are the vocabulary words that are found in this lesson.

polynomials with two terms in them.
Perfect Square Trinomial
a trinomial where the first term and the last term are perfect squares because the binomial was to the second power or squared.
firsts, outers, inners, lasts – binomial multiplication

Time to Practice

Directions: Use a table to multiply the following binomials.

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

Directions: Multiply the following binomials vertically.

  1. \begin{align*}(d+2)(4d-1)\end{align*}
  2. \begin{align*}(5x+7)(5x-7)\end{align*}
  3. \begin{align*}(4b^2+3c)(2b-5c^2)\end{align*}

Directions: Multiply the following binomials using the FOIL method:

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

Directions: Use what you have learned to solve each problem.

  1. Find the area of a square whose side measures \begin{align*}5n - 2\end{align*}.
  2. A guard dog is tied to a rope attached to a pole in the ground. The rope measures \begin{align*}x + 5\end{align*}. The dog can walk around the stake to the full length of the rope in all directions. Find how much area she can guard.
  3. A company is designing a cardboard box that will measure \begin{align*}l = z, w = z - 3,\end{align*} and \begin{align*}h = 2z + 4\end{align*}. What will its volume be?
  4. Use the same box measurements from problem 13 to find the box’s surface area.

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Difficulty Level:
At Grade
Date Created:
Jan 14, 2013
Last Modified:
Apr 11, 2016
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