Learning Outcomes
- Multiply Polynomials
Key words
- Monomial: a polynomial with one term
- Binomial: a polynomial with two terms
- Trinomial: a polynomial with three terms
The Distributive Property
In the last section we multiplied polynomials by monomials. In this section we will expand our use of the distributive property. Consider the following example.
We distribute the [latex]p[/latex] to get | [latex]x\color{red}{p}+3\color{red}{p}[/latex] |
What if we have [latex]\left(x+7\right)[/latex] instead of [latex]p[/latex] ?
Think of the [latex]x+7[/latex] as the [latex]\color{red}{p}[/latex] above. |
|
Distribute [latex]\left(x+7\right)[/latex] . | |
Distribute again. | [latex]{x}^{2}+7x+3x+21[/latex] |
Combine like terms. | [latex]{x}^{2}+10x+21[/latex] |
Notice that before combining like terms, we had four terms. We multiplied the two terms of the first binomial by the two terms of the second binomial—four multiplications.
Be careful to distinguish between a sum and a product.
[latex]\begin{array}{cccc}\hfill \mathbf{\text{Sum}}\hfill & & & \hfill \mathbf{\text{Product}}\hfill \\ \hfill x+x\hfill & & & \hfill x\cdot x\hfill \\ \hfill 2x\hfill & & & \hfill {x}^{2}\hfill \\ \hfill \text{combine like terms}\hfill & & & \hfill \text{add exponents of like bases}\hfill \end{array}[/latex]
example
Multiply: [latex]\left(x+6\right)\left(x+8\right)[/latex]
Solution
[latex]\left(x+6\right)\left(x+8\right)[/latex] | |
Distribute [latex]\left(x+8\right)[/latex] . | [latex]x\color{red}{(x+8)}+6\color{red}{(x+8)}[/latex] |
Distribute again. | [latex]{x}^{2}+8x+6x+48[/latex] |
Simplify. | [latex]{x}^{2}+14x+48[/latex] |
try it
Now we’ll see how to multiply binomials where the variable has a coefficient.
example
Multiply: [latex]\left(2x+9\right)\left(3x+4\right)[/latex]
Solution
[latex]\left(2x+9\right)\left(3x+4\right)[/latex] | |
Distribute. [latex]\left(3x+4\right)[/latex] | [latex]2x\color{red}{(3x+4)}+9\color{red}{(3x+4)}[/latex] |
Distribute again. | [latex]6{x}^{2}+8x+27x+36[/latex] |
Simplify. | [latex]6{x}^{2}+35x+36[/latex] |
try it
In the previous examples, the binomials were sums. When there are differences, we pay special attention to make sure the signs of the product are correct.
example
Multiply: [latex]\left(4y+3\right)\left(6y - 5\right)[/latex]
Solution
[latex]\left(4y+3\right)\left(6y - 5\right)[/latex] | |
Distribute. | [latex]4y\color{red}{(6y-5)}+3\color{red}{(6y-5)}[/latex] |
Distribute again. | [latex]24{y}^{2}-20y+18y - 15[/latex] |
Simplify. | [latex]24{y}^{2}-2y - 15[/latex] |
try it
Area Model for Multiplying Binomials
Now let’s explore multiplying two binomials visually. For those that use pictures to learn, we can draw an area model to help make sense of the process. We’ll use each binomial as one of the dimensions of a rectangle, and their product as the area.
The model below shows [latex]\left(x+4\right)\left(x+2\right)[/latex]:
Each binomial is expanded into variable terms and constants, [latex]x+4[/latex], along the top of the model and [latex]x+2[/latex] along the left side. The product of each pair of terms is a colored rectangle. Like terms have the same color. The total area is the sum of all of these small rectangles, [latex]x^{2}+2x+4x+8[/latex], If we combine all the like terms, we can write the product, or area, as [latex]x^{2}+6x+8[/latex].
Here is the same example using the distributive property:
Example
Simplify. [latex]\left(x+4\right)\left(x+2\right)[/latex]
Solution
[latex]\left(x+4\right)\left(x+2\right)[/latex]
Distribute the [latex]x[/latex] onto [latex]x+2[/latex], then distribute 4 onto [latex]x+2[/latex].
[latex]x\left(x+2\right)+4\left(x+2\right)[/latex]
[latex]x\left(x\right)+x\left(2\right)+4\left(x\right)+4\left(2\right)[/latex]
Multiply.
[latex]x^{2}+2x+4x+8[/latex]
Combine like terms [latex]\left(2x+4x\right)[/latex].
[latex]x^{2}+6x+8[/latex]
Answer
[latex]\left(x+4\right)\left(2x+2\right)=x^{2}+6x+8[/latex]
Look back at the model above to see where each piece of [latex]x^{2}+2x+4x+8[/latex] comes from. Can you see where you multiply [latex]x[/latex] by [latex]x + 2[/latex], and where you get [latex]x^{2}[/latex] from [latex]x\left(x\right)[/latex]?
When multiplying binomials each term in one binomial is multiplied by each term in the other binomial. Look at the example above: the [latex]x[/latex] in [latex]x+4[/latex] gets multiplied by both the [latex]x[/latex] and the [latex]2[/latex] from [latex]x+2[/latex], and the [latex]4[/latex] gets multiplied by both the [latex]x[/latex] and the [latex]2[/latex].
The following video provides an example of multiplying two binomials using an area model as well as repeated distribution.
The Table Method
The table method is just a simplified form of the area model, where we use the rows and columns of a table rather than rectangles.
We may see a binomial multiplied by itself written as [latex]{\left(x+3\right)}^{2}[/latex] instead of [latex]\left(x+3\right)\left(x+3\right)[/latex]. To find this product, let’s use the table method. We will place the terms of each binomial along the top row and first column of a table, like this:
[latex]x[/latex] | [latex]+3[/latex] | |
[latex]x[/latex] | ||
[latex]+3[/latex] |
Now multiply the term in each column by the term in each row to get the terms of the resulting polynomial. Note how we keep the signs on the terms, even when they are positive, this will help us write the new polynomial.
[latex]x[/latex] | [latex]+3[/latex] | |
[latex]x[/latex] | [latex]x\cdot{x}=x^2[/latex] | [latex]3\cdot{x}=+3x[/latex] |
[latex]+3[/latex] | [latex]x\cdot{3}=+3x[/latex] | [latex]3\cdot{3}=+9[/latex] |
Now we can write the terms of the polynomial from the entries in the table:
[latex]\begin{equation}\begin{aligned}&\;\;\;\;\left(x+3\right)^{2}\\&= x^2 + 3x + 3x + 9 \\ &= x^{2}+6x +9\end{aligned}\end{equation}[/latex]
Do you see how similar this method is to the area model?
Example
Find the product.[latex]\left(3–s\right)\left(1-s\right)[/latex]
Solution
Notice how the binomials are not written in standard form. There is nothing different in the way we find the product. At the end we will reorganize terms so they are in descending order as a matter of convention.
[latex]\left(3–s\right)\left(1–s\right)[/latex]
Use a table this time.
[latex]3[/latex] | [latex]-s[/latex] | |
---|---|---|
[latex]1[/latex] | [latex]3[/latex] | [latex]-s[/latex] |
[latex]-s[/latex] | [latex]-3s[/latex] | [latex]s^2[/latex] |
Notice how the s term is now positive. Collect the terms and simplify.
[latex]\begin{array}{c}\left(3–s\right)\left(1–s\right)\\\text{ }\\=3-3s-s+s^2\\\text{ }\\=3-4s+s^2\end{array}[/latex]
As a matter of convention, we will organize the terms so the one with greatest degree comes first. Pay close attention to the signs on the terms when you reorganize them. The [latex]3[/latex] is positive, so we will use a plus in front of it, and the [latex]4[/latex] is negative so we use a minus in front of it.
[latex]\begin{array}{c}\left(3–s\right)\left(1–s\right)\\\text{ }\\=s^{2}-4s+3\end{array}[/latex]
Answer
[latex]\left(3–s\right)\left(1–s\right)=s^2-4s+3[/latex]
Multiplying Two Binomials Using the Vertical Method
The table method and the area model are just different representations of the Distributive Property. We can use the Distributive Property to find the product of any two polynomials. The Vertical Method is another way to complete the Distributive Property. It is very much like the method we use to multiply whole numbers. Look carefully at this example of multiplying two-digit numbers.
We start by multiplying [latex]23[/latex] by [latex]6[/latex] to get [latex]138[/latex].
Then we multiply [latex]23[/latex] by [latex]4[/latex], lining up the partial product in the correct columns so that we’re really multiplying by 40.
Last, we add the partial products.
Horizontally, this would be written, [latex]23\times 46=23(40+6)=23(40)+23(6)=920+138=10-58[/latex]. So, multiplying vertically is just another representation of the distributive property.
Now we’ll apply the vertical method to multiply two binomials.
example
Multiply using the vertical method: [latex]\left(5x - 1\right)\left(2x - 7\right)[/latex]
Solution
It does not matter which binomial goes on the top, thanks to the commutative property of multiplication.
We line up the columns when we multiply:
Multiply [latex]2x - 7[/latex] by [latex]-1[/latex] . | |
Multiply [latex]2x - 7[/latex] by [latex]5x[/latex] . | |
Add like terms. |
Notice the partial products are the same as the terms in using the Distributive Property.
try it
The Commutative Property of Multiplication tells us that terms can be multiplied in either order. The expression [latex]\left(x+2\right)\left(x+4\right)[/latex] has the same product as [latex]\left(x+4\right)\left(x+2\right)[/latex], [latex]x^{2}+6x+8[/latex]. The order in which we multiply binomials does not matter. What matters is that we multiply each term in one binomial by each term in the other binomial.
Let’s look at another example using the Distributive Property.
Example
Simplify [latex]\left(4x–10\right)\left(2x+3\right)[/latex] using the Distributive Property.
Solution
[latex]\begin{equation}\begin{aligned}&\;\;\;\;(4x–10)(2x+3)\\&=4x(2x)+4x(3)-10(2x)-10(3)\\&=8x^2+12x-20x-30\\&=8x^2-8x-30\end{aligned}\end{equation}[/latex]
Be careful about including the negative sign on the [latex]‑10[/latex], since 10 is subtracted.
Answer
[latex]\left(4x–10\right)\left(2x+3\right)=8x^{2}–8x–30[/latex]
Example
Simplify [latex]\left(x–7\right)^2[/latex]
Solution
Write the product of the binomial:
[latex]{\left(x-7\right)}^2=\left(x–7\right)\left(x–7\right)[/latex]
Let’s use the table method, just because. Note how we carry the negative sign with the [latex]7[/latex].
[latex]x[/latex] | [latex]-7[/latex] | |
[latex]x[/latex] | [latex]x^2[/latex] | [latex]-7x[/latex] |
[latex]-7[/latex] | [latex]-7x[/latex] | [latex]49[/latex] |
Collect the terms, and simplify. Note how we keep the sign on each term.
[latex]\begin{array}{c}x^2-7x-7x+49\\\text{ }\\=x^2-14x+49\end{array}[/latex]
Answer
[latex]x^2-14x+49[/latex]
Caution! It is VERY important to remember that we can’t move the exponent into a grouped sum because of the order of operations!!!!!
INCORRECT: [latex]\left(2+x\right)^{2}\neq2^{2}+x^{2}[/latex]
CORRECT: [latex]\left(2+x\right)^{2}=\left(2+x\right)\left(2+x\right)[/latex]
The video that follows shows another example of using a table to multiply two binomials.
The method you choose to use doesn’t matter as the answers will always be the same.
A common form for products occurs when the binomial is squared. All we have to remember is that squaring a binomial is equivalent to multiplying it by itself.
Example
Simplify: [latex]\left(2x+6\right)^{2}[/latex]
Solution
We will use the Distributive Property. First write the square as the product of two binomials.
[latex]\begin{equation}\begin{aligned}&\;\;\;\;\left(2x+6\right)^{2}\\&=\left(2x+6\right)\left(2x+6\right)\\&=2x(2x)+2x(6)+6(2x)+6(6)\\&=4x^2+12x+12x+36\\&=4x^2+12x+36\end{aligned}\end{equation}[/latex]
Notice that the two [latex]x[/latex]-terms are identical. This will always be the case when squaring a binomial.
Answer
[latex](2x+6)^{2}=4x^{2}+24x+36[/latex]
The next example shows another common form the product of binomials can take, where each of the terms in the two binomials is the same, but the signs in the middle are different.
Example
Multiply the binomials. [latex]\left(x+8\right)\left(x–8\right)[/latex]
Solution
[latex]\begin{equation}\begin{aligned}&\;\;\;\;(x+8)(x-8)\\&=x(x)+x(-8)+8(x)+8(-8)\\&=x^2+8x-8x-64\\&=x^2-64\end{aligned}\end{equation}[/latex]
Notice that the [latex]x[/latex]-term in the answer disappears since the two [latex]x[/latex]-terms are opposites.
Answer
[latex]\left(x+8\right)\left(x-8\right)=x^{2}-64[/latex]
Think About It
There are predictable outcomes when you square a binomial sum or difference. In general terms, for a binomial difference,
[latex]\left(a-b\right)^{2}=\left(a-b\right)\left(a-b\right)[/latex],
the resulting product, after being simplified, will look like this:
[latex]a^2-2ab+b^2[/latex].
The product of a binomial sum will have the following predictable outcome:
[latex]\left(a+b\right)^{2}=\left(a+b\right)\left(a+b\right)=a^2+2ab+b^2[/latex].
The product of a binomial sum and binomial difference of the same two monomial will have the following predictable outcome:
[latex]\left(a+b\right)\left(a-b\right)=a^2-b^2[/latex].
Note that a and b in these generalizations could be integers, fractions, or variables with any kind of constant. You will learn more about predictable patterns from products of binomials in later math classes.
In this section we showed how to multiply two binomials using the distributive property, an area model, by using a table, and the vertical method. All of these methods are variations of the Distributive Property. Practice each method, and decide which one you prefer.
The Distributive Property can be expanded to the product of any two polynomials; each term in the first polynomial must be multiplied into each term in the second polynomial.
Example
Find the product: [latex](3x+4)(2x^2-3x-8)[/latex]
Solution
Distributive Property:
[latex]\begin{equation}\begin{aligned}&\;\;\;\;(\color{red}{3x}\color{blue}{+4})(2x^2-3x-8)\\&=\color{red}{3x}(2x^2)+\color{red}{3x}(-3x)+\color{red}{3x}(-8)\color{blue}{+4}(2x^2)\color{blue}{+4}(-3x)\color{blue}{+4}(-8)\\&=6x^3-9x^2-24x+8x^2-12x-32\\&=6x^3-x^2-36x-32\end{aligned}\end{equation}[/latex]
Answer
[latex](3x+4)(2x^2-3x-8)=6x^3-x^2-36x-32[/latex]
Example
Find the product: [latex](5x^2-x+3)(2x^2+4x-7)[/latex]
Solution
Distributive Property:
[latex]\begin{equation}\begin{aligned}&\;\;\;\;(\color{red}{5x^2}\color{blue}{-x}+\color{green}{3})(2x^2+4x-7)\\&=\color{red}{5x^2}(2x^2)+\color{red}{5x^2}(4x)+\color{red}{5x^2}(-7)\color{blue}{-x}(2x^2)\color{blue}{-x}(4x)\color{blue}{-x}(-7)+\color{green}{3}(2x^2)+\color{green}{3}(4x)+\color{green}{3}(-7)\\&=10x^4+20x^3-35x^2-2x^3-4x^2+7x+6x^2+12x-21\\&=10x^4+20x^3-2x^3-35x^2-4x^2+6x^2+7x+12x-21\\&=10x^4+18x^3-33x^2+19x-21\end{aligned}\end{equation}[/latex]
Answer
[latex](5x^2-x+3)(2x^2+4x-7)=10x^4+18x^3-33x^2+19x-21[/latex]
Try It
Find the product: [latex](x+5)(x^2-3x+4)[/latex]
Try It
Find the product: [latex](2x^2-3x+1)(x^2-4x+3)[/latex]
Summary
Multiplication of binomials and polynomials requires an understanding of the distributive property, rules for exponents, and a keen eye for collecting like terms. Whether the polynomials are monomials, binomials, or trinomials, carefully multiply each term in one polynomial by each term in the other polynomial. Be careful to watch the addition and subtraction signs and negative coefficients. A product is written in simplified form if all of its like terms have been combined.