Algebraic expressions like monomials, binomials, trinomials, quadrinomials, etc., can be identified according to the number of their terms. A polynomial contains one or more than one term with exponents of variables as only whole numbers. A polynomial does not consist of variables with negative exponents. So all possible algebraic expressions are not polynomials. Let’s learn about multiplying a polynomial by a polynomial.

In polynomials, we can apply fundamental operations: addition, subtraction, multiplication, and division. So we can multiply a polynomial by a polynomial. Generally, we apply the distributive property for multiplication. This article will discuss how to multiply a polynomial by a polynomial and the steps involved.

Polynomials

The algebraic expressions with the exponents of the variables as whole numbers are polynomials.
Polynomials are algebraic expressions involving variables and constants with whole-number exponents of the variables.

Monomial

It is a polynomial that includes only one term. For example, \(5 x,-6 x y, 3 a^{2} c, 9,-10\) etc., are monomials.

Binomial

A polynomial including two unlike terms is called a binomial. For example \((2 x+3 y),(6-3 x),\left(y^{2}-x y^{2}\right)\) etc., are binomials.

Trinomial

A polynomial containing three unlike terms is called a trinomial. For example, \((a+y+c),(x+2 a+6 z),\left(a^{3}-y^{2}-z^{3}\right)\) etc., are trinomials.

Multiplication of a Polynomial by a Polynomial

It is a way of multiplying two polynomials. When the first polynomial’s terms are multiplied by the second polynomial, a third polynomial is obtained. There are various ways to multiply polynomials, depending on the type of polynomial we are using. There are different rules for multiplying polynomials based on the type of polynomial. To multiply polynomials, multiply the coefficient by a coefficient and multiply the variable by a variable.

Multiplication of a Polynomial by a Polynomial using Distributive Property

We use the distributive property to multiply a polynomial with a polynomial. Let’s say a polynomial has to be multiplied by another polynomial. By using the distributive property, the above result can be written as:

\((a+b)(c+d)=a c+a d+b c+b d\)

Multiplication of a Polynomial by a Polynomial using Box Method

Two polynomials can be multiplied using the box method. It has a box with terms written on it, with their corresponding products written inside of it.
For example, 

Multiply \((y+7)\) by \((y+3)\) to get the answer.

Solution: \((y+7)\) and \((y+3)\) are two polynomials written in horizontal and vertical directions. Please take a look at the sign with its corresponding term. We have shown what we get after multiplying the corresponding terms in the figure below:

So, in the box method, two binomials are multiplied in a box, as described above. 

\(\left(y^{2}+3 y+7 y+21\right)\) is now the sum. 

As a result, the final product will be \(\left(y^{2}+10 y+21\right)\)

Example 1: Multiply \((x+2)\) and \((x+3).\)

Considering two polynomials \((x+2), (x+3),\) the polynomials contain the same variable.

Step 1: In this case, \((x+2)\) and \((x+3)\) are the polynomials with two terms. We will multiply \((x+2)\) by \((x+3)\) using the distributive property.

We will multiply each term of \((x+2)\) by each term of \((x+3).\)

The first term of the polynomial \((x+2)\) is \(x\) and \((x+3)\) is \(x.\) Now, \(x \times x=x^{2}\) (using the law of the exponents)

Step 2: Now, multiplying \(x\) by \(3,\) we have \(3x.\)

Step 3: Similarly, the second term of the polynomial \((x+2)\) is \(2.\) Now multiply \(2\) by the first term of \((x+3)\) that is \(x\) we have, \(2x.\)

Step 4: At last, we multiply \(2\) by the second term of \((x+3)\) that is \(3\) and, we have \(6.\)

Hence, the answer is \(x^{2}+3 x+2 x+6=x^{2}+5 x+6\)

Example 2: Multiply polynomials \(\left(2 a+a^{2}\right),\left(a+3 a^{2}\right)\)

Considering two polynomials \(\left(2 a+a^{2}\right),\left(a+3 a^{2}\right)\) the polynomials contain the same variable. In this case, we need to apply the law of exponents in each step.

Step 1: In this case, \(\left(2 a+a^{2}\right)\) and \(\left(a+3 a^{2}\right)\) are the polynomials with two terms. We will multiply \(\left(2 a+a^{2}\right)\) by \(\left(a+3 a^{2}\right)\) using the distributive property.
We will multiply each term of \(\left(2 a+a^{2}\right)\) by each term of \(\left(a+3 a^{2}\right)\).
The first term of the polynomial \(\left(2 a+a^{2}\right)\) is \(2a\) and \(\left(a+3 a^{2}\right)\) is \(a.\) Now, \(2 a \times a=2 a^{2}\) (using the law of the exponents)

Step 2: Now multiplying \(2a\) by \(3 a^{2}\) we have, \(6 a^{3}\).

Step 3: Similarly, the second term of the polynomial \(\left(2 a+a^{2}\right)\) is \(a^{2}\) Now multiply \(a^{2}\) by the first term of \(\left(a+3 a^{2}\right)\) that is a we have, \(a^{3}\)

Step 4: At last, we multiply \(a^{2}\) by the second term of \(\left(a+3 a^{2}\right)\) that is \(3 a^{2}\) and, we have, \(3 a^{4}\)

Hence, the answer is  \(\left(2 a+a^{2}\right)\left(a+3 a^{2}\right)=2 a^{2}+6 a^{3}+a^{3}+3 a^{4}=2 a^{2}+7 a^{3}+3 a^{4}\)

Example 3: Multiply binomial \((4x-3)\) by the trinomial \(\left(5 x^{2}-2 x-1\right)\)

Consider two polynomials \((4 x-3),\left(5 x^{2}-2 x-1\right)\)

Step 1: In this case, \((4x-3)\) is a polynomial with two terms, and \(\left(5 x^{2}-2 x-1\right)\) is a polynomial with three terms. We will multiply \((4x-3)\) by \(\left(5 x^{2}-2 x-1\right)\) using the distributive property.
We will multiply each term of \((4x-3)\) by each term of \(\left(5 x^{2}-2 x-1\right)\)
The first term of the polynomial \((4x-3)\) is \(4x\) and \(\left(5 x^{2}-2 x-1\right)\) is \(5 x^{2}\).
Now, \(4 x \times 5 x^{2}=20 x^{3}\) (using the law of the exponents)

Step 2: Now multiplying \(4x\) by \(-2x\) we have, \(-8 x^{2}\). And \(4x\) by \(-1\) gives \(-4x.\)

Step 3: Similarly, the second term of the polynomial \((4x+-3)\) is \(-3.\) Now multiply \(-3\) by each term of \(\left(5 x^{2}-2 x-1\right)\) we have, \(-15 x^{2}+6 x+3\)

Step 4: Hence, we get, \(20 x^{3}-8 x^{2}-4 x-15 x^{2}+6 x+3=20 x^{3}-23 x^{2}+2 x+3\)

Solved Examples on Multiplying a Polynomial by a Polynomial

Q.1. Multiply \(2 x+x^{3}\) and \(x^{2}+x\)
Ans: Considering two polynomials \(\left(2 x+x^{3}\right),\left(x^{2}+x\right)\)
The polynomials are containing the same variable. In this case, we need to apply the law of exponents in each step.
Step 1: In this case, \(\left(2 x+x^{3}\right)\) and \(\left(x^{2}+x\right)\) are the polynomials with two terms. Let us multiply using the distributive property.
We will multiply each term of \(\left(2 x+x^{3}\right)\) by each term of \(\left(x^{2}+x\right)\)
The first terms of the polynomial \(\left(2 x+x^{3}\right)\) is \(2x\) and the polynomial \(\left(x^{2}+x\right)\) is \(x^{2}\). Now, \(2 x \times x^{2}=2 x^{3}\) (using the law of the exponents)
Step 2: Now multiplying \(2x\) by \(x\) we have, \(2 x^{2}\)
Step 3: Similarly, the second term of the polynomial \(\left(2 x+x^{3}\right)\) is \(x^{3}\). Now multiply \(x^{3}\) by the first term of \(\left(x^{2}+x\right)\) that is \(x^{2}\) we have, \(x^{5}\)
Step 4: At last, we multiply \(x^{3}\) by the second term of \(\left(x^{2}+x\right)\) that is \(x\) and, we have, \(x^{4}\)
Hence, the answer is \(2 x^{3}+2 x^{2}+x^{5}+x^{4}\)

Q.2. Multiply \(5 + x\) and \(4 x^{3}+2 y\)
Ans:  
Step 1: Consider the polynomials \(5+x\) and \(4 x^{3}+2 y\). In this case, \(5\) is a constant, and \(4 x^{3}+2 y\). is a binomial. We will multiply the constant with the coefficient of the first term of the polynomial from the left. It gives \(5×4=20.\) After that, we will write the variable \(\left(x^{3}\right)\) after \(20.\) Hence, we get \(20 x^{3}\) after multiplying \(5\) with the first term of the polynomial.
Step 2: we need to follow the same rule as step \(1\) with the second term of the polynomial, and we get, \(10y\)
Step 3: Similarly, multiplying the second term of \(5+x\) that is \(x\) with \(4 x^{3}+2 y\), we have,
\(4 x^{3}+2 y\),
Hence, the answer is \(20 x^{3}+10 y+4 x^{4}+2 x y\)

Q.3. Multiply \(a + 4\) and \(a – 3\).
Ans:
Step 1: In this case, \((a+4)\) and \((a-3)\) are two binomials. We will multiply \((a+4)\) by \((a-3)\) using the distributive property.
We will multiply each term of \((a+4)\) by each term of \((a-3).\)
The first term of the polynomials \((a+4)\) is \(a\) and \((a-3)\) is \(a.\) Now, \(a \times a=a^{2}\) (using the law of the exponents)
Step 2: Now multiplying a by \(-3\) we have, \(-3a.\)
Step 3: Similarly, the second term of the polynomial \((a+4)\) is \(4.\) Now multiply \(4\) by the first term of \((a-3),\)  we have \(4a.\)
Step 4: At last, we multiply \(4\) by the second term of \((a-3)\) that is \(-3\) and, we have \(-12.\)
Hence, the answer is  \(a^{2}-3 a+4 a-12=a^{2}+a-12 .\)

Q.4. Multiply \(2abc\) and \(a^{2} b+8\)
Ans: Given,  \(2 a b c \) & \(a^{2} b+8\)
We will multiply the coefficients. We get \(2×1=2\) for the first term, and then we will multiply the variables using the exponent rule wherever required. Here, the variable parts are \(abc, a^{2} b\) Multiplying these, we get,\(a b c \times a^{2} b=\left(a \times a^{2}\right) \times(b \times b) \times c=a^{3} b^{2} c\) (as we added the exponents of the same variables as per the exponent rule.)
Next, we will multiply \(2abc\) by the second term of the polynomial, which is \(8.\) It gives \(16abc\)
Hence, the answer is \(2a^{3} b^{2} c+16 a b c\)

Q.5. Multiply \(5 x^{2}+3\) and \(\left( {3x + y + z} \right)\)
Ans:  Consider the polynomials \(5 x^{2}+3, \&(3 x+y+z)\)
Step 1: In this case, \(5 x^{2}+3\) is a polynomial with two terms, and \(3x+y+z\) is a polynomial with three terms. We will multiply the polynomial by each term of the polynomial using the distributive property.
Now, multiplying \(5 x^{2}\) by \((3x+y+z)\), we have, \(15 x^{3}+5 x^{2} y+5 x^{2} z\) 
Step 2: The second term of the polynomial  \(5 x^{2}+3\) is \(3.\) Now multiply \(3\) by \((3x+y+z)\) we have, \(9x+3y+3z\)
Step 3: Adding both the results we get, \(15 x^{3}+5 x^{2} y+5 x^{2} z+9 x+3 y+3 z\)

Summary

This article discussed polynomials, types of polynomials, multiplying one polynomial by another. We also discussed the method of multiplication of a polynomial with a polynomial. At last, we solved some examples of the multiplication of a polynomial by a polynomial.

FAQs

Q.1. How do you define a polynomial?
Ans: The algebraic expressions with the exponents of the variables as whole numbers are polynomials.

Q.2. What are the rules in multiplying polynomials?
Ans: We use the distributive property to multiply a polynomial with another polynomial. Each term of the first polynomial is multiplied with each term of the second, and if any like terms are there, they are added to get the final answer. For example (a+b)(c+d) = ac + ad + bc + bd.

Q.3. What are two methods for multiplying polynomials?
Ans: There are two methods for multiplying polynomials. These are the distributive method for multiplication and the box method for multiplication.

Q.4. What is the importance of multiplying polynomials?
Ans: Polynomials are multiplied using the exponent rules and the distributive property to simplify the final product. This method helps simulate real-world situations. When learning how to solve algebraic equations involving polynomials, it is crucial to understand the multiplication of polynomials.

Q.5. How do you simplify expressions by multiplying polynomials?
Ans: After multiplying polynomials, we will group the like terms and unlike terms and then add the like terms to simplify.

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