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# Multiplying Binomials Mentally

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# Multiplying Binomials

Have you ever been to a community garden? Take a look at this dilemma.

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 $3x + 7$ by $2x - 4$ . The other measures $x^2 + 1$ by $6x + 5$ . 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.

### Guidance

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 $(x+5)(x+3)$ .

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 $x \cdot x$ , the second is $5 \cdot x$ , the third is $3 \cdot x$ , and the fourth is $3 \cdot 5$ .

In order to find the total, we will add the four areas: $x^2 + 5x +3x +15$

Now, combine like terms carefully: $x^2 + 8x + 15$ .

Here is one that is a little different.

Multiply $(5x - 8)^2$ .

Remember that the exponent applies to the entire binomial such that $(5x -8)^2 = (5x - 8)(5x - 8)$ .

$5x$ $-8$
$5x$ $25x^2$ $-40x$
$-8$ $-40x$ $64$

$& 25x^2-40x-40x+64 \\&=25x^2-80x +64$

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

$& \qquad \qquad \qquad \quad \text{thousands} \qquad \quad \text{hundreds} \qquad \quad \text{tens} \qquad \quad \text{ones} \\& \quad \quad 73 \ \quad \rightarrow \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 70 \ \quad + \qquad 3 \\& \ \underline{\quad \times 81} \ \quad \rightarrow \ \ \underline{\qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \quad 81 \ \quad + \qquad 1} \\& \quad \quad 73\ \quad \rightarrow \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 70 \ \quad + \qquad 3 \\& \ \underline{+4580} \ \quad \rightarrow \ \ \underline{\ \ \qquad 4000 \qquad + \qquad \ 500 \quad + \ \quad 80 \ \quad + \qquad 0} \\& \quad 4653 \ \quad \leftarrow \ \quad \qquad 4000 \qquad + \ \qquad 500 \quad + \quad 150 \ \quad + \qquad 3$

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.

Take a look at this one.

Multiply $(3x + 2)(5x + 4)$

$& 2^{nd} \text{power} \qquad \ \quad 1^{st} \text{power} \qquad \quad 0 \ \text{power} \\& \qquad \qquad \qquad \qquad \ \ \ \quad 3x \quad + \qquad \ \quad 2 \\& \underline{\qquad \qquad \qquad \qquad \qquad 5x \quad + \qquad \ \quad 4 \;\;} \\& \qquad \qquad \qquad \qquad \ \ \quad 12x \quad + \qquad \ \quad 8 \\& \underline{\qquad 15x^2 \quad + \qquad \ \quad 10x \ \quad + \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\& \qquad 15x^2 \quad + \qquad \ \quad 22x \quad + \qquad \ \quad 8 \quad \rightarrow \quad 15x^2 + 22x + 8$

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

A third way is to use “FOIL”. “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 $(2x+8)(5x-13)$ using the FOIL method.

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

$(2x+8)(5x-13)=10x^2 - 26x + 40x - 104 = 10x^2 + 14x - 104$

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.

Take a look at one more.

Multiply $(5x^3+2x)(7x^2+8)$

$&(5x^3+2x)(7x^2+8) \\&= 5x^3 \cdot 7x^2 + 5x^3 \cdot 8 + 2x \cdot 7x^2 + 2x \cdot 8 \\&= 35x^5 +40x^3 +14x^3 +16x \\&= 35x^5 +54x^3 +16x$

Multiply the following binomials.

#### Example A

$(x+2)(x+4)$

Solution: $x^2+6x+8$

#### Example B

$(x-6)(x+5)$

Solution: $x^2-x-30$

#### Example C

$(x+3)(x-3)$

Solution: $x^2-9$

Now let's go back to the dilemma from the beginning of the Concept.

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.

$&(3x+7)(2x-4)+(x^2+1)(6x+5) \\&= (3x \cdot 2x + 3x \cdot -4 + 7 \cdot 2x + 7 \cdot -4) + (x^2 \cdot 6x +x^2 \cdot 5 + 1 \cdot 6x + 1 \cdot 5) \\&= (6x^2 -12x +14x -28) + (6x^3 +5x^2 + 6x +5) \\&= 6x^2 + 2x - 28 +6x^3 + 5x^2 +6x +5 \\&= 6x^3 + 11x^2 + 8x -23$

### Vocabulary

Binomials
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.
FOIL
firsts, outers, inners, lasts – binomial multiplication

### Guided Practice

Here is one for you to try on your own.

Multiply by using a table.

Multiply $(x-4)(x+6)$

Solution

$x$ $-4$
$x$ $x^2$ $-4x$
$+6$ $6x$ $-24$

$& x^2-4x+6x-24 \\&= x^2+2x-24$

Notice the careful work with the negative and positive signs.

### Practice

Directions: Use a table to multiply the following binomials.

1. $(x+3)(x+5)$
2. $(x-3)(x-5)$
3. $(x+3)(x-3)$
4. $(x+2)(x-8)$
5. $(3x^2+3x)(6x-2)$
6. $(2x-7y)(5x+4y)$
7. $(2x-9)^2$

Directions: Multiply the following binomials vertically.

1. $(d+2)(4d-1)$
2. $(5x+7)(5x-7)$
3. $(4b^2+3c)(2b-5c^2)$

Directions: Multiply the following binomials using the FOIL method:

1. $(p+6)(5p+2)$
2. $(-7y^2-4y)(6y+2)$
3. $(x^3+3x)^2$
4. $(2x+1)(x-4)$
5. $(3x-3)(5x+9$ )
6. $(x+5)^2$