11.2.3: Multiplying Polynomials
- Page ID
- 67621
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- Multiply monomials times polynomials.
- Multiply two binomials.
- Multiply any two polynomials.
Introduction
Multiplying polynomials involves applying the rules of exponents and the distributive property to simplify the product. This multiplication can also be illustrated with an area model, and can be useful in modeling real world situations. Understanding polynomial products is an important step in learning to solve algebraic equations involving polynomials.
Multiplying Monomials
Let's begin by multiplying two simple monomials together. Consider a rectangle whose length is \(\ 2x\) and whose width is \(\ 3x\). To find the area of this rectangle, multiply the length by the width
\(\ \text { Area of rectangle }=(2 x)(3 x)=(2 x)(3 x)=2 \cdot 3 \cdot x \cdot x=6 x^{2}\)
Note that the commutative and associative properties of multiplication are used to rearrange the factors, putting the coefficients together and the variables together.
The area, \(\ 6 x^{2}\), is a product that includes a coefficient (6) and a variable with a whole number exponent (\(\ x^{2}\)). In other words, it's a monomial, too. So the result of multiplying two monomials is another monomial!
Let's try a slightly more complex problem: \(\ -9 x^{3} \cdot 3 x^{2}\)
Multiply. \(\ -9 x^{3} \cdot 3 x^{2}\)
Solution
| \(\ -9 \cdot 3 \cdot x^{3} \cdot x^{2}\) | Use commutative and associative properties of multiplication to rearrange the factors. |
| \(\ -27 \cdot x^{3} \cdot x^{2}\) | Multiply constants. Remember that a positive number times a negative number yields a negative number. |
| \(\ \begin{array}{c} -27 \cdot x^{3+2} \\ -27 \cdot x^{5} \end{array}\) |
Multiply variable terms. Remember to add the exponents when multiplying exponents with the same base. |
\(\ -9 x^{3} \cdot 3 x^{2}=-27 x^{5}\)
That’s it! When multiplying monomials, multiply the coefficients together, and then multiply the variables together. If two variables have the same base, follow the rules of exponents, like this:
\(\ 5 a^{4} \cdot 7 a^{6}=35 a^{10}\)
Find the area of the rectangle:
- \(\ 8 y^{3}\)
- \(\ 15 y^{5}\)
- \(\ 15 y^{10}\)
- \(\ 8 y^{5}\)
- Answer
-
- Incorrect. Multiply the two coefficients to get 15, then multiply the variables. Use the rules of exponents: \(\ y^{3} \cdot y^{2}=y^{3+2}=y^{5}\). The correct answer is \(\ 15 y^{5}\).
- Correct. Multiply \(\ 3 \cdot 5\) and \(\ y^{3} \cdot y^{2}\), using the rules of exponents to get \(\ 15 y^{5}\).
- Incorrect. When you multiply with exponents, if the bases are the same, you add the exponents: \(\ y^{3} \cdot y^{2}=y^{3+2}=y^{5}\). The correct answer is \(\ 15 y^{5}\).
- Incorrect. You correctly multiplied the variables, \(\ y^{3} \cdot y^{2}=y^{3+2}=y^{5}\), but you appear to have added the coefficients rather than multiplied. The correct answer is \(\ 15 y^{5}\).
The Product of a Monomial and a Polynomial
The distributive property can be used to multiply a polynomial by a monomial. Just remember that the monomial must be multiplied by each term in the polynomial. Consider the expression \(\ 2 x\left(2 x^{2}+5 x+10\right)\).
This expression can be modeled with a sketch like the one below.
The model above illustrates the distributive property.
\(\ 2 x\left(2 x^{2}+5 x+10\right)=2 x \cdot 2 x^{2}+2 x \cdot 5 x=2 x \cdot 10=4 x^{3}+10 x^{2}+20 x\)
Here’s an example:
Simplify. \(\ 5 x\left(4 x^{2}+3 x+7\right)\)
Solution
| \(\ 5 x \cdot 4 x^{2}+5 x \cdot 3 x+5 x \cdot 7\) | Distribute the monomial to each term of the polynomial. |
| \(\ 20 x^{3}+15 x^{2}+35 x\) | Multiply. |
\(\ 5 x\left(4 x^{2}+3 x+7\right)=20 x^{3}+15 x^{2}+35 x\)
You may need to rewrite subtraction as adding the opposite.
Simplify. \(\ 7 x^{2}\left(2 x^{2}-5 x+1\right)\)
Solution
| \(\ 7 x^{2}\left[2 x^{2}+(-5 x)+1\right]\) | Rewrite the subtraction as adding the opposite. |
| \(\ 7 x^{2} \cdot 2 x^{2}+7 x^{2} \cdot-5 x+7 x^{2} \cdot 1\) | Distribute the monomial to each term of the polynomial. |
| \(\ 14 x^{4}+(-35) x^{3}+7 x^{2}\) | Multiply. |
| \(\ 7 x^{2}\left(2 x^{2}-5 x+1\right)=14 x^{4}-35 x^{3}+7 x^{2}\) | Rewrite addition of terms with negative coefficients as subtraction. |
Find the product. Watch the signs!
\(\ -3 t^{2}\left(7 t^{3}+3 t^{2}-t\right)\)
- \(\ -21 t^{5}-9 t^{4}+3 t^{3}\)
- \(\ -21 t^{5}+9 t^{4}-3 t^{3}\)
- \(\ -21 t^{6}-9 t^{4}+3 t^{2}\)
- \(\ -21 t^{5}+3 t^{2}-t\)
- Answer
-
- Correct. Rewriting the subtraction as adding the opposite gives \(\ -3 t^{2}\left[7 t^{3}+3 t^{2}+(-t)\right]\). Distributing the monomial \(\ -3 t^{2}\) gives \(\ -3 t^{2} \cdot 7 t^{3}+\left(-3 t^{2}\right) \cdot 3 t^{2}+\left(-3 t^{2}\right) \cdot(-t)\), which is \(\ -21 t^{5}+\left(9 t^{4}\right)+\left(3 t^{3}\right)\). Rewriting addition of terms with negative coefficients as subtraction gives \(\ -21 t^{5}-9 t^{4}+3 t^{3}\).
- Incorrect. The negative must be distributed to all terms along with the \(\ 3 t^{2}\). This changes the sign of the middle and last terms. The correct answer is \(\ -21 t^{5}-9 t^{4}+3 t^{3}\).
- Incorrect. By the laws of exponents, you add (not multiply) exponents when multiplying: \(\ -3 t^{2} \cdot 7 t^{3}+\left(-3 t^{2}\right) \cdot 3 t^{2}+\left(-3 t^{2}\right) \cdot(-t)\) is \(\ -21 t^{5}+\left(-9 t^{4}\right)+\left(3 t^{3}\right)\). The correct answer is \(\ -21 t^{5}-9 t^{4}+3 t^{3}\).
- Incorrect. You must distribute the monomial to all three terms in the polynomial, not just the first one: \(\ -3 t^{2} \cdot 7 t^{3}+\left(-3 t^{2}\right) \cdot 3 t^{2}+\left(-3 t^{2}\right) \cdot(-t)\). The correct answer is \(\ -21 t^{5}-9 t^{4}+3 t^{3}\).
Product of Two Binomials
Now let's explore multiplying two binomials. Once again, you can draw an area model to help make sense of the process. You'll use each binomial as one of the dimensions of a rectangle, and their product as the area.
The model below shows \(\ (x+4)(2 x+2)\):
Each binomial is expanded into variable terms and constants, \(\ x+4\), along the top of the model and \(\ 2 x+2\) along the left side. The product of each pair of terms is a colored rectangle. The total area is the sum of all of these small rectangles, \(\ 2 x^{2}+8 x+2 x+8\). If you combine all the like terms, you can write the product, or area, as \(\ 2 x^{2}+10 x+8\).
You can use the distributive property to determine the product of two binomials.
\(\ (x+4)(2 x+2)\)
Solution
| \(\ x \cdot 2 x+x \cdot 2+4 \cdot 2 x+4 \cdot 2\) | Distribute the \(\ x\) over \(\ 2 x+2\), then distribute 4 over \(\ 2 x+2\). |
| \(\ 2 x^{2}+2 x+8 x+8\) | Multiply. |
| \(\ 2 x^{2}+10 x+8\) | Combine like terms \(\ (8 x+2 x)\). |
\(\ (x+4)(2 x+2)=2 x^{2}+10 x+8\)
Consider the model above to see where each piece of \(\ 2 x^{2}+8 x+2 x+8\) comes from. Can you see where you multiply \(\ x\) by \(\ 2 x+2\), and where you get \(\ 2 x^{2}\) from \(\ x \cdot 2 x\)?
Another way to look at multiplying binomials is to see that each term in one binomial is multiplied by each term in the other binomial. In the example above, the \(\ x\) in \(\ x+4\) gets multiplied by both the \(\ 2 x\) and the 2 from \(\ 2 x+2\), and the 4 gets multiplied by both the \(\ 2x\) and the 2.
Some people use the FOIL method to keep track of which pairs have been multiplied. The letters in FOIL stand for First, Outer, Inner, Last:
First term in each binomial: \(\ (\mathbf{x}+4)(\mathbf{2} \mathbf{x}+2)\mathrm{ x \cdot 2 x=2 x^{2}}\)
Outer terms: \(\ (\mathbf{x}+4)(2 \mathrm{x}+\mathbf 2)\mathrm{ x \cdot 2=2 x}\)
Inner terms: \(\ (\mathrm{x}+{\bf 4})({\bf 2 x}+2) 4 \cdot \mathrm{2 x=8 x}\)
Last terms in each binomial: \(\ (\mathrm{x}+{\bf 4})( 2 \mathrm{x}+{\bf 2}) 4 \cdot 2=8\)
When you add the four results, you get the same answer:
\(\ 2 x^{2}+2 x+8 x+8=2 x^{2}+10 x+8\)
Here is another example, using FOIL this time.
\(\ (4 x-10)(2 x+3)\)
Solution
| \(\ \begin{array}{r} 4 x \cdot 2 x=8 x^{2} \text { (First) } \\ 4 x \cdot 3=12 x \text { (Outer) } \\ -10 \cdot 2 x=-20 x \text { (Inner) } \\ -10 \cdot 3=-30 \text { (Last) } \end{array}\) |
Be careful about including the negative sign on the -10, since 10 is subtracted. |
| \(\ 8 x^{2}+12 x-20 x-30\) | Combine like terms. |
\(\ (4 x-10)(2 x+3)=8 x^{2}-8 x-30\)
Because multiplication is commutative, the terms can be multiplied in either order. The expression \(\ (2 x+2)(x+4)\) has the same product as \(\ (x+4)(2 x+2)\), \(\ 2 x^{2}+10 x+8\). (Work it out and see.) The order in which you multiply binomials does not matter. What matters is that you multiply each term in one binomial by each term in the other binomial.
The last step in multiplying polynomials is to combine like terms. Remember that a polynomial is simplified only when there are no like terms remaining.
Find the product: \(\ (a+10)(2 a-7)\)
- \(\ 2 a^{2}+19 a-70\)
- \(\ 3 a+3\)
- \(\ 2 a^{2}-70\)
- \(\ 2 a^{2}+13 a-70\)
- Answer
-
- Incorrect. The middle term should be \(\ 20 a-7 a\), which is \(\ 13 a\). The correct answer is \(\ 2 a^{2}+13 a-70\).
- Incorrect. Multiply, don’t add, the terms in one binomial by the terms in the other binomial. The correct answer is \(\ 2 a^{2}+13 a-70\).
- Incorrect. Multiply each term in one binomial by the terms in the other binomial. \(\ \text { (a) }(-7)\) and \(\ (10)(2 a)\) are missing. The correct answer is \(\ 2 a^{2}+13 a-70\).
- Correct. Using the FOIL method, you find: \(\ (a+10)(2 a-7)=a \cdot 2 a+a \cdot-7+10 \cdot 2 a+10 \cdot-7=2 a^{2}-7 a+20 a-70=2 a^{2}+13 a-70\).
Product of a Binomial and a Trinomial
Another type of polynomial multiplication problem is the product of a binomial and trinomial. Although the FOIL method cannot be used exactly since there are more than two terms in a trinomial, you still use the distributive property to organize the individual products. Using the distributive property, each term in the binomial must be multiplied by each of the terms in the trinomial. Two examples are shown below.
\(\ (3 x+6)\left(5 x^{2}+3 x+10\right)\)
Solution
| \(\ 3 x\left(5 x^{2}+3 x+10\right)+6\left(5 x^{2}+3 x+10\right)\) | Distribute the trinomial to each term in the binomial. |
| \(\ 3 x \cdot 5 x^{2}+3 x \cdot 3 x+3 x \cdot 10+6 \cdot 5 x^{2}+6 \cdot 3 x+6 \cdot 10\) | Use the distributive property to distribute the monomials to each term in the trinomials. |
| \(\ 15 x^{3}+9 x^{2}+30 x+30 x^{2}+18 x+60\) | Multiply. |
| \(\ 15 x^{3}+\left(9 x^{2}+30 x^{2}\right)+(30 x+18 x)+60\) | Group like terms. |
| \(\ (3 x+6)\left(5 x^{2}+3 x+10\right)=15 x^{3}+39 x^{2}+48 x+60\) | Combine like terms. |
As you can see, multiplying a binomial by a trinomial leads to a lot of individual terms! Some people prefer to set up these problems vertically and gather like terms as they multiply. This method is shown below, using the same problem as above.
\(\ (3 x+6)\left(5 x^{2}+3 x+10\right)\)
Solution
| \(\ \begin{array}{rrr} 3 x +\ \ 6 \\ \times\quad 5 x^{2} +\ 3 x +10 \\ \hline +30 x +60 \end{array}\) |
Set up the problem in a vertical form, and begin by multiplying \(\ 3 x+6\) by \(\ +10\). Place the products underneath, as shown. |
| \(\ \begin{array}{r} 3 x +\ \ 6 \\ \times \quad5 x^{2} +\ \ {\color{blue}3 x} +10 \\ \hline +\ 30 x +60 \\ \ \color{blue}+9 x^{2} +18 x \ \ \ \ \ \ \ \ \ \end{array}\) |
Now multiply \(\ 3 x+6\) by \(\ 3x\). Notice that \(\ \text { (6) }(3 x)=18 x\); since this term is like \(\ 30x\), place it directly beneath it. |
| \(\ \begin{array}{r} 3 x +\ \ 6 \\ \times\quad\quad\quad\quad\quad {\color{green}5 x^{2}} +\ \ 3 x +10 \\ \hline +\ 30 x +60 \\ +\ \ \ 9 x^{2} +18 x \ \ \ \ \ \ \ \ \ \\ \ {\color{green}+15 x^{3} +30 x^{2}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \hline \end{array}\) |
Finally, multiply \(\ 3 x+6\) by \(\ 5 x^{2}\). Notice that \(\ 30 x^{2}\) is placed underneath \(\ 9 x^{2}\). |
| \(\ \begin{array}{r} 3 x +\ \ 6 \\ \times \quad\quad\quad\quad\quad 5 x^{2} +\ \ 3 x +10 \\ \hline \ +30 x +60 \\ \ +\ \ 9 x^{2} +18 x\ \ \ \ \ \ \ \ \ \\ \ +15 x^{3} +30 x^{2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \hline\ +15 x^{3} +39 x^{2} +48 x +60 \end{array}\) |
Now add like terms. |
\(\ 15 x^{3}+39 x^{2}+48 x+60\)
Notice that although the two problems were solved using different strategies, the product is the same. Both the horizontal and vertical methods apply the distributive property to multiply a binomial by a trinomial.
The next example shows multiplication by a binomial and trinomial that each contains subtraction. The example completes the multiplication without rewriting each subtraction as addition of the opposite. Notice where you must be careful with the signs! (If you prefer, you can continue to rewrite subtraction as adding the opposite.)
\(\ (2 p-1)\left(3 p^{2}-3 p+1\right)\)
Solution
| \(\ 2 p\left(3 p^{2}-3 p+1\right)-1\left(3 p^{2}-3 p+1\right)\) | Distribute the trinomial to each term in the binomial. |
| \(\ 2 p \cdot 3 p^{2}+2 p \cdot-3 p+2 p \cdot 1-1\left(3 p^{2}\right)-1(-3 p)-1(1)\) | If you don’t write subtraction as adding the opposite, then be sure to think of it that way. So you are distributing -1 and multiplying each term of the trinomial by -1. |
| \(\ 6 p^{3}-6 p^{2}+2 p-3 p^{2}+3 p-1\) | Multiply. (Notice that the subtracted 1 and the subtracted \(\ 3p\) have a positive product that is added.) |
| \(\ 6 p^{3}-9 p^{2}+5 p-1\) | Combine like terms. |
\(\ 6 p^{3}-9 p^{2}+5 p-1\)
Find the product: \(\ (3 x-2)\left(2 x^{2}+4 x-11\right)\)
- \(\ 6 x^{3}+8 x^{2}-41 x+22\)
- \(\ 6 x^{3}+8 x^{2}-41 x-22\)
- \(\ 6 x^{3}+12 x+22\)
- \(\ 3 x^{3}+8 x^{2}+25 x-22\)
- Answer
-
- Correct. \(\ 3 x\left(2 x^{2}+4 x-11\right)-2\left(2 x^{2}+4 x-11\right)=6 x^{3}+12 x^{2}-33 x-4 x^{2}-8 x+22=6 x^{3}+8 x^{2}-41 x+22\)
- Incorrect. \(\ (-2)(-11)=22\). The correct answer is \(\ 6 x^{3}+8 x^{2}-41 x+22\).
- Incorrect. Each term in the binomial must be multiplied by each term in the trinomial. The correct answer is \(\ 6 x^{3}+8 x^{2}-41 x+22\).
- Incorrect. Multiply the coefficients \(\ (3 x)\left(2 x^{2}\right)\); combine like terms by adding/subtracting coefficients according to their signs. The correct answer is \(\ 6 x^{3}+8 x^{2}-41 x+22\).
Summary
Multiplication of binomials and polynomials requires use of the distributive property as well as the commutative and associative properties of multiplication. 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.