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Lesson Explainer: Multiplying Polynomials Mathematics

In this explainer, we will learn how to multiply polynomial expressions together by expanding the parentheses.

In order to multiply any two polynomial expressions together, we should first recall how we multiply a polynomial by a single term (i.e., monomial). We distribute the multiplication over each term in the polynomial. For example,

We can apply this exact same process to multiply any two polynomials. Consider the product (π‘₯+1)(π‘₯+2). This is not quite in the same form since neither factor is a monomial; however, we can treat the factor of (π‘₯+1) as a single term. This means we just multiply each term of π‘₯+2 by the entire factor of (π‘₯+1):

Each term is now the product of a monomial and binomial, so we can expand:

Finally, we collect like terms: π‘₯+π‘₯+2π‘₯+2=π‘₯+(2+1)π‘₯+2=π‘₯+3π‘₯+2.

Hence, we have shown that (π‘₯+1)(π‘₯+2)=π‘₯+3π‘₯+2.

We can simplify this process by noting that we just find the product of all of the terms of each factor and then combine like terms: (π‘₯+1)(π‘₯+2)=(π‘₯Γ—π‘₯)+(π‘₯Γ—1)+(2Γ—π‘₯)+(2Γ—1)=π‘₯+π‘₯+2π‘₯+2=π‘₯+3π‘₯+2.

We can apply this process to find the product of any two polynomials; we distribute one factor over every term in the second factor, expand, and then simplify.

In general, we can expand the product of binomials by finding the sum of the product of each pair of terms from both factors.

Let’s see an example of applying this process to find the product of two binomials.

Example 1: Expanding the Product of Two Binomials and Simplifying

Expand the product (π‘₯+4)(π‘₯+6).

Answer

We are asked to expand the product of two binomials; we can do this by distributing the first factor over every term in the second factor. So, we multiply π‘₯ by (π‘₯+4) and 6 by (π‘₯+4) and add the results:

We can expand each term by distributing the first factor over each binomial:

We can now simplify each term by recalling that π‘₯Γ—π‘₯=π‘₯ and collecting like terms: (π‘₯Γ—π‘₯)+(π‘₯Γ—4)+(6Γ—π‘₯)+(6Γ—4)=π‘₯+4π‘₯+6π‘₯+24=π‘₯+10π‘₯+24.

A special case of multiplying binomials is when we multiply the same binomial by itself, in other words, squaring a binomial. We can use this expansion method to find an expression for the square of a binomial: (π‘Ž+𝑏)=(π‘ŽΓ—π‘Ž)+(π‘ŽΓ—π‘)+(π‘Γ—π‘Ž)+(𝑏×𝑏)=π‘Ž+2π‘Žπ‘+𝑏.

We have the following result.

Property: Squaring a Binomial

We can square a binomial by finding the sum of the product of each pair of terms from both factors. We obtain (π‘Ž+𝑏)=π‘Ž+2π‘Žπ‘+𝑏.

In our next example, we will expand and simplify the square of a binomial.

Example 2: Squaring a Binomial and Simplifying

Expand (2π‘₯+5).

Answer

We are asked to expand the square of a binomial. We can do this by first recalling that squaring means multiplying a number by itself. So, (2π‘₯+5)=(2π‘₯+5)(2π‘₯+5).

We can now expand this product by finding the product of each pair of terms from either factor and adding the results: (2π‘₯+5)(2π‘₯+5)=(2π‘₯Γ—2π‘₯)+(2π‘₯Γ—5)+(5Γ—2π‘₯)+(5Γ—5)=4π‘₯+10π‘₯+10π‘₯+25.

Finally, we combine like terms to obtain 4π‘₯+10π‘₯+10π‘₯+25=4π‘₯+20π‘₯+25.

We can also answer this question by recalling that we can square a binomial using the formula (π‘Ž+𝑏)=π‘Ž+2π‘Žπ‘+𝑏.

Substituting π‘Ž=2π‘₯ and 𝑏=5 into the formula yields (2π‘₯+5)=(2π‘₯)+2(2π‘₯)(5)+5=4π‘₯+20π‘₯+25.

In our next example, we will expand and simplify the product of two binomials with multiple variables and negative coefficients.

Example 3: Expanding the Product of Two Binomials with Mixed Terms and Simplifying

Expand and simplify (βˆ’2π‘₯βˆ’4𝑦)(2π‘₯βˆ’π‘¦).

Answer

We can expand this product by finding the product of each pair of terms from either factor and adding the results: (βˆ’2π‘₯βˆ’4𝑦)(2π‘₯βˆ’π‘¦)=((βˆ’2π‘₯)Γ—2π‘₯)+((βˆ’2π‘₯)Γ—(βˆ’π‘¦))+((βˆ’4𝑦)Γ—2π‘₯)+((βˆ’4𝑦)Γ—(βˆ’π‘¦)).

We can now simplify each term by using the laws of indices: ((βˆ’2π‘₯)Γ—2π‘₯)+((βˆ’2π‘₯)Γ—(βˆ’π‘¦))+((βˆ’4𝑦)Γ—2π‘₯)+((βˆ’4𝑦)Γ—(βˆ’π‘¦))=βˆ’4π‘₯βˆ’8π‘₯𝑦+2π‘₯𝑦+4𝑦.

Finally, we can collect like terms: βˆ’4π‘₯βˆ’8π‘₯𝑦+2π‘₯𝑦+4𝑦=βˆ’4π‘₯+(βˆ’8+2)π‘₯𝑦+4𝑦=βˆ’4π‘₯βˆ’6π‘₯𝑦+4𝑦.

We can apply this process of expanding the product to any polynomial. For example, let’s expand the product ο€Ήπ‘₯+π‘₯𝑦+1(π‘₯βˆ’π‘¦).

We distribute the first factor over each term in the second factor:

It is worth noting that we could have also distributed the second factor over each term in the first factor. It is a personal preference which method to use; however, it is usually easiest to distribute the factor with the most terms over the factor with the least terms.

We can now expand and simplify by using the laws of indices and collecting like terms: π‘₯ο€Ήπ‘₯+π‘₯𝑦+1ο…βˆ’π‘¦ο€Ήπ‘₯+π‘₯𝑦+1=π‘₯+π‘₯𝑦+π‘₯βˆ’π‘₯π‘¦βˆ’π‘₯π‘¦βˆ’π‘¦=π‘₯βˆ’π‘₯𝑦+π‘₯βˆ’π‘¦.

In general, we need to find the sum of the product of each pair of terms from both sets of parentheses.

Let’s now see an example of applying this process to expand and simplify the product of a trinomial and binomial.

Example 4: Multiplying a Binomial by a Trinomial and Simplifying

Expand and simplify (π‘₯βˆ’2𝑦+3)(2π‘₯+𝑦).

Answer

To expand the product of two polynomials, we need to distribute the product of one polynomial over each term in the other. Let’s start by distributing the product of π‘₯βˆ’2𝑦+3 over each term in the binomial: (π‘₯βˆ’2𝑦+3)(2π‘₯+𝑦)=2π‘₯(π‘₯βˆ’2𝑦+3)+𝑦(π‘₯βˆ’2𝑦+3).

We can now expand each of the two terms, distributing the first factor over each trinomial.

For the first term, we have

For the second term, we have

We can now add these expressions together and collect like terms to simplify. We obtain 2π‘₯(π‘₯βˆ’2𝑦+3)+𝑦(π‘₯βˆ’2𝑦+3)=2π‘₯βˆ’4π‘₯𝑦+6π‘₯+π‘₯π‘¦βˆ’2𝑦+3𝑦=2π‘₯βˆ’3π‘₯π‘¦βˆ’2𝑦+6π‘₯+3𝑦.

In our next example, we will expand the cube of a binomial and simplify to identify the equivalent polynomial expression.

Example 5: Cubing a Binomial and Simplifying

Determine which of the following expressions is equivalent to (π‘₯βˆ’π‘¦).

  1. π‘₯βˆ’3π‘₯𝑦+3π‘₯π‘¦βˆ’π‘¦οŠ©οŠ¨οŠ¨οŠ©
  2. π‘₯βˆ’π‘¦οŠ©οŠ©
  3. π‘₯+3π‘₯π‘¦βˆ’3π‘₯π‘¦βˆ’π‘¦οŠ©οŠ¨οŠ¨οŠ©
  4. π‘₯+3π‘₯π‘¦βˆ’3π‘₯𝑦+π‘¦οŠ©οŠ¨οŠ¨οŠ©
  5. π‘₯βˆ’3π‘₯𝑦+3π‘₯𝑦+π‘¦οŠ©οŠ¨οŠ¨οŠ©

Answer

We want to expand the cube of a binomial to identify which of five given polynomials is an equivalent expression. We can do this by successively multiplying the binomial factors. First, we have (π‘₯βˆ’π‘¦)=(π‘₯βˆ’π‘¦)(π‘₯βˆ’π‘¦)(π‘₯βˆ’π‘¦).

We can start off by finding the product of two of the factors. To do this, we take the product of each pair of terms from the first parentheses with the second parentheses and add the results: (π‘₯βˆ’π‘¦)(π‘₯βˆ’π‘¦)=π‘₯βˆ’π‘₯π‘¦βˆ’π‘₯𝑦+𝑦=π‘₯βˆ’2π‘₯𝑦+𝑦.

We can now substitute this expression into our product of three binomials: (π‘₯βˆ’π‘¦)(π‘₯βˆ’π‘¦)(π‘₯βˆ’π‘¦)=ο€Ήπ‘₯βˆ’2π‘₯𝑦+𝑦(π‘₯βˆ’π‘¦).

We can expand the product of two polynomials by distributing one factor over each term in the other factor. We will choose to distribute the first factor over the terms in the second factor: ο€Ήπ‘₯βˆ’2π‘₯𝑦+𝑦(π‘₯βˆ’π‘¦)=π‘₯ο€Ήπ‘₯βˆ’2π‘₯𝑦+π‘¦ο…βˆ’π‘¦ο€Ήπ‘₯βˆ’2π‘₯𝑦+𝑦.

Performing this multiplication for each term separately, we have π‘₯ο€Ήπ‘₯βˆ’2π‘₯𝑦+𝑦=π‘₯βˆ’2π‘₯𝑦+π‘₯𝑦,βˆ’π‘¦ο€Ήπ‘₯βˆ’2π‘₯𝑦+𝑦=βˆ’π‘¦π‘₯+2π‘₯π‘¦βˆ’π‘¦.

We can then add these expressions and combine like terms: π‘₯ο€Ήπ‘₯βˆ’2π‘₯𝑦+π‘¦ο…βˆ’π‘¦ο€Ήπ‘₯βˆ’2π‘₯𝑦+𝑦=π‘₯βˆ’2π‘₯𝑦+π‘₯π‘¦βˆ’π‘¦π‘₯+2π‘₯π‘¦βˆ’π‘¦=π‘₯+(βˆ’2βˆ’1)π‘₯𝑦+(1+2)π‘₯π‘¦βˆ’π‘¦=π‘₯βˆ’3π‘₯𝑦+3π‘₯π‘¦βˆ’π‘¦.

This is answer A.

In our next example, we will use this process of multiplying polynomials to find a simplified expression for a shaded region.

Example 6: Finding the Area of a Rectangular Shape

A rectangle with dimensions (π‘₯βˆ’π‘¦) and (π‘₯+𝑦+1) is cut out from a larger rectangle with dimensions (2π‘₯+𝑦+3) and (2π‘₯+𝑦). Find a simplified expression for the shaded area.

Answer

We can find an expression for the shaded area by finding the area of the larger rectangle and subtracting the area of the smaller rectangle from the area of the larger rectangle. We need to find the product of the dimensions of each rectangle to find their areas.

We can do this by distributing the product of one factor over each term in the other factor.

For the larger rectangle, we have (2π‘₯+𝑦+3)(2π‘₯+𝑦)=2π‘₯(2π‘₯+𝑦+3)+𝑦(2π‘₯+𝑦+3).

We can distribute the monomials over the polynomials to get 2π‘₯(2π‘₯+𝑦+3)=4π‘₯+2π‘₯𝑦+6π‘₯,𝑦(2π‘₯+𝑦+3)=2π‘₯𝑦+𝑦+3𝑦.

We can add these expressions together and collect like terms to find the area of the larger rectangle: (2π‘₯+𝑦+3)(2π‘₯+𝑦)=4π‘₯+2π‘₯𝑦+6π‘₯+2π‘₯𝑦+𝑦+3𝑦=4π‘₯+4π‘₯𝑦+𝑦+6π‘₯+3𝑦.

For the smaller rectangle, we have (π‘₯+𝑦+1)(π‘₯βˆ’π‘¦)=π‘₯(π‘₯+𝑦+1)βˆ’π‘¦(π‘₯+𝑦+1)=π‘₯+π‘₯𝑦+π‘₯βˆ’π‘₯π‘¦βˆ’π‘¦βˆ’π‘¦=π‘₯βˆ’π‘¦+π‘₯βˆ’π‘¦.

We now need to subtract the area of the smaller rectangle from the area of the larger rectangle: 4π‘₯+4π‘₯𝑦+𝑦+6π‘₯+3π‘¦βˆ’ο€Ήπ‘₯βˆ’π‘¦+π‘₯βˆ’π‘¦ο…=4π‘₯+4π‘₯𝑦+𝑦+6π‘₯+3π‘¦βˆ’π‘₯+π‘¦βˆ’π‘₯+𝑦=4π‘₯βˆ’π‘₯+4π‘₯𝑦+𝑦+𝑦+6π‘₯βˆ’π‘₯+3𝑦+𝑦=(4βˆ’1)π‘₯+4π‘₯𝑦+(1+1)𝑦+(6βˆ’1)π‘₯+(3+1)𝑦=3π‘₯+4π‘₯𝑦+2𝑦+5π‘₯+4𝑦.

In our final example, we will expand the product of three binomials to find the values of unknown coefficients.

Example 7: Finding the Unknowns in an Expression by Multiplying Three Binomials

If (2π‘₯βˆ’π‘¦)(2π‘₯βˆ’5𝑦)(3π‘₯+2𝑦)=π‘Žπ‘₯+𝑏π‘₯𝑦+𝑐π‘₯𝑦+π‘‘π‘¦οŠ©οŠ¨οŠ¨οŠ©, what are the values of π‘Ž, 𝑏, 𝑐, and 𝑑?

Answer

On the left-hand side of the equation, we have the product of three binomials and on the right-hand side of the equation, we have a polynomial. We can find a polynomial expression equivalent to the left-hand side of the equation by expanding the product.

Let’s start with expanding the first two products. We find the sum of the products of each pair of terms in either factor to obtain (2π‘₯βˆ’π‘¦)(2π‘₯βˆ’5𝑦)=(2π‘₯Γ—2π‘₯)+(2π‘₯Γ—(βˆ’5𝑦))+((βˆ’π‘¦)Γ—(2π‘₯))+((βˆ’π‘¦)Γ—(βˆ’5𝑦))=4π‘₯βˆ’10π‘₯π‘¦βˆ’2π‘₯𝑦+5𝑦=4π‘₯βˆ’12π‘₯𝑦+5𝑦.

We can now substitute this expression into the product of the three binomials to obtain (2π‘₯βˆ’π‘¦)(2π‘₯βˆ’5𝑦)(3π‘₯+2𝑦)=ο€Ή4π‘₯βˆ’12π‘₯𝑦+5𝑦(3π‘₯+2𝑦).

We can expand this product by distributing the product of one factor over each term in the other factor: ο€Ή4π‘₯βˆ’12π‘₯𝑦+5𝑦(3π‘₯+2𝑦)=3π‘₯ο€Ή4π‘₯βˆ’12π‘₯𝑦+5𝑦+2𝑦4π‘₯βˆ’12π‘₯𝑦+5𝑦=12π‘₯βˆ’36π‘₯𝑦+15π‘₯𝑦+8π‘₯π‘¦βˆ’24π‘₯𝑦+10𝑦=12π‘₯+(βˆ’36+8)π‘₯𝑦+(15βˆ’24)π‘₯𝑦+10𝑦=12π‘₯βˆ’28π‘₯π‘¦βˆ’9π‘₯𝑦+10𝑦.

We are told that this is equal to π‘Žπ‘₯+𝑏π‘₯𝑦+𝑐π‘₯𝑦+π‘‘π‘¦οŠ©οŠ¨οŠ¨οŠ©; for the polynomials to be equal, their coefficients must be equal. Thus, π‘Ž=12, 𝑏=βˆ’28, 𝑐=βˆ’9, and 𝑑=10.

Let’s finish by recapping some of the important points from this explainer.

Key Points

  • We can multiply two polynomials by multiplying every term of one polynomial by the other entire polynomial and summing the results. We can then simplify by combining like terms.
  • The product of any two polynomials can also be found by multiplying every pair of terms from each of the polynomials and summing the results.
  • We can expand a polynomial to a positive integer power by evaluating the products in pairs.

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