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Mathematics LibreTexts

6.4: Special Products

  • Page ID
    18965
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    Learning Objectives

    By the end of this section, you will be able to:

    • Square a binomial using the Binomial Squares Pattern
    • Multiply conjugates using the Product of Conjugates Pattern
    • Recognize and use the appropriate special product pattern

    Note

    Before you get started, take this readiness quiz.

    1. Simplify: ⓐ \(9^2\) ⓑ \((−9)^2\) ⓒ \(−9^2\).
      If you missed this problem, review Exercise 1.5.13.

    Square a Binomial Using the Binomial Squares Pattern

    Mathematicians like to look for patterns that will make their work easier. A good example of this is squaring binomials. While you can always get the product by writing the binomial twice and using the methods of the last section, there is less work to do if you learn to use a pattern.

    \(\begin{array}{ll}{\text { Let's start by looking at }(x+9)^{2} \text { . }}& \\ {\text { What does this mean? }} &{(x+9)^{2}} \\ {\text { It means to multiply }(x+9) \text { by itself. }} & {(x+9)(x+9)}\\ {\text { Then, using FOIL, we get: }} & {x^{2}+9 x+9 x+81}\\ {\text { Combining like terms gives: }} &{x^{2}+18 x+81} \\ \\ {\text { Here's another one: }} & {(y-7)^{2}} \\ {\text { Multiply }(y-7) \text { by itself. }} & {(y-7)(y-7)} \\ {\text { Using FOIL, we get: }} & {y^{2}-7 y-7 y+49} \\ {\text { And combining like terms: }} & {y^{2}-14 y+49} \\ \\ {\text { And one more: }} & {(2 x+3)^{2}} \\ {\text { Multiply. }} & {(2 x+3)(2 x+3)} \\ {\text { Use FOL: }} & {\text { 4x }+6 x+6 x+9} \\ {\text { Combine like terms. }} & {4 x^{2}+12 x+9}\end{array}\)

    Look at these results. Do you see any patterns?

    What about the number of terms? In each example we squared a binomial and the result was a trinomial.

    \[(a+b)^{2}=\underline{\qquad}+\underline{\qquad}+\underline{\qquad}\]

    Now look at the first term in each result. Where did it come from?

    This figure has three columns. The first column contains the expression x plus 9, in parentheses, squared. Below this is the product of x plus 9 and x plus 9. Below this is x squared plus 9x plus 9x plus 81. Below this is x squared plus 18x plus 81. The second column contains the expression y minus 7, in parentheses, squared. Below this is the product of y minus 7 and y minus 7. Below this is y squared minus 7y minus 7y plus 49. Below this is the expression y squared minus 14y plus 49. The third column contains the expression 2x plus 3, in parentheses, squared. Below this is the product of 2x plus 3 and 2x plus 3. Below this is 4x squared plus 6x plus 6x plus 9. Below this is 4x squared plus 12x plus 9.

    The first term is the product of the first terms of each binomial. Since the binomials are identical, it is just the square of the first term!

    \[(a+b)^{2}=a^{2}+\underline{\qquad}+\underline{\qquad}\]

    To get the first term of the product, square the first term.

    Where did the last term come from? Look at the examples and find the pattern.

    The last term is the product of the last terms, which is the square of the last term.

    \[(a+b)^{2}=\underline{\qquad}+\underline{\qquad}+b^{2}\]

    To get the last term of the product, square the last term.

    Finally, look at the middle term. Notice it came from adding the “outer” and the “inner” terms—which are both the same! So the middle term is double the product of the two terms of the binomial.

    \[(a+b)^{2}=\underline{\qquad}+2ab+\underline{\qquad}\]

    \[(a+b)^{2}=\underline{\qquad}-2ab+\underline{\qquad}\]

    To get the middle term of the product, multiply the terms and double their product.

    Putting it all together:

    BINOMIAL SQUARES PATTERN

    If a and b are real numbers,

    \[\begin{array}{l}{(a+b)^{2}=a^{2}+2 a b+b^{2}} \\ {(a-b)^{2}=a^{2}-2 a b+b^{2}}\end{array}\]

    No Alt Text

    To square a binomial:

    • square the first term
    • square the last term
    • double their product

    A number example helps verify the pattern.

    \(\begin{array}{ll} & {(10+4)^{2}} \\{\text { Square the first term. }}& {10^{2}+\underline{\qquad}+\underline{\qquad}} \\ {\text { Square the last term. }} & {10^{2}+\underline{\qquad}+\frac{1}{4^{2}}} \\ {\text { Double their product. }} & {10^{2}+2 \cdot 10 \cdot 4+4^{2}} \\ {\text { Simplify. }} & {100+80+16} \\ {\text { Simplify. }} & {196}\end{array}\)

    To multiply \((10+4)^2\) usually you’d follow the Order of Operations.

    \[\begin{array}{c}{(10+4)^{2}} \\ {(14)^{2}} \\ {196}\end{array}\]

    The pattern works!

    Exercise \(\PageIndex{1}\)

    \(\text { Multiply: }(x+5)^{2}\)

    Answer
    x plus 5, in parentheses, squared. Above the expression is the general formula a plus b, in parentheses, squared.
    Square the first term. x squared plus blank plus blank. Above the expression is the general form a squared plus 2 a b plus b squared.
    Square the last term. x squared plus blank plus 5 squared.
    Double the product. x squared plus 2 times x times 5 plus 5 squared. Above this expression is the general formula a squared plus 2 times a times b plus b squared.
    Simplify. x squared plus 10 x plus 25.

    Exercise \(\PageIndex{2}\)

    Multiply: \((x+9)^{2}\)

    Answer

    \(x^{2}+18 x+81\)

    Exercise \(\PageIndex{3}\)

    Multiply: \((y+11)^{2}\)

    Answer

    \(y^{2}+22 y+121\)

    Exercise \(\PageIndex{4}\)

    Multiply: \((y-3)^{2}\)

    Answer
    y minus 3, in parentheses, squared. Above the expression is the general formula a minus b, in parentheses, squared.
    Square the first term. y squared minus blank plus blank. Above the expression is the general form a squared plus 2 a b plus b squared.
    Square the last term. y squared minus blank plus 3 squared.
    Double the product. y squared minus y times y times 3 plus 3 squared. Above this expression is the general formula a squared plus 2 times a times b plus b squared.
    Simplify. y squared minus 6 y plus 9.

    Exercise \(\PageIndex{5}\)

    Multiply: \((x-9)^{2}\)

    Answer

    \(x^{2}-18 x+81\)

    Exercise \(\PageIndex{6}\)

    Multiply: \((p-13)^{2}\)

    Answer

    \(p^{2}-26 p+169\)

    Exercise \(\PageIndex{7}\)

    Multiply: \((4 x+6)^{2}\)

    Answer
    4 x plus 6, in parentheses, squared. Above the expression is the general formula a plus b, in parentheses, squared.
    Use the pattern. 4 x squared plus 2 times 4 x times 6 plus 6 squared. Above this expression is the general formula a squared plus 2 times a times b plus b squared.
    Simplify. 16 x squared plus 48 x plus 36.

    Exercise \(\PageIndex{8}\)

    Multiply: \((6 x+3)^{2}\)

    Answer

    \(36 x^{2}+36 x+9\)

    Exercise \(\PageIndex{9}\)

    Multiply: \((4 x+9)^{2}\)

    Answer

    \(16 x^{2}+72 x+81\)

    Exercise \(\PageIndex{10}\)

    Multiply:\((2 x-3 y)^{2}\)

    Answer
    contains 2 x minus 3 y, in parentheses, squared. Above the expression is the general formula a plus b, in parentheses, squared.
    Use the pattern. 2 x squared minus 2 times 2 x times 3 y plus 3 y squared. Above this expression is the general formula a squared minus 2 times a times b plus b squared.
    Simplify. 4 x squared minus 12 x y plus 9 y squared.

    Exercise \(\PageIndex{11}\)

    Multiply: \((2 c-d)^{2}\)

    Answer

    \(4 c^{2}-4 c d+d^{2}\)

    Exercise \(\PageIndex{12}\)

    Multiply: \((4 x-5 y)^{2}\)

    Answer

    \(16 x^{2}-40 x y+25 y^{2}\)

    Exercise \(\PageIndex{13}\)

    Multiply: \(\left(4 u^{3}+1\right)^{2}\)

    Answer
    4 u cubed plus 1, in parentheses, squared. Above the expression is the general formula a plus b, in parentheses, squared.
    Use the pattern. 4 u cubed, in parentheses, squared, plus 2 times 4 u cubed times 1 plus 1 squared. Above this expression is the general formula a squared plus 2 times a times b plus b squared.
    Simplify. 16 u to the sixth power plus 18 u cubed plus 1.

    Exercise \(\PageIndex{14}\)

    Multiply: \(\left(2 x^{2}+1\right)^{2}\)

    Answer

    \(4 x^{4}+4 x^{2}+1\)

    Exercise \(\PageIndex{15}\)

    Multiply: \(\left(3 y^{3}+2\right)^{2}\)

    Answer

    \(9 y^{6}+12 y^{3}+4\)

    Multiply Conjugates Using the Product of Conjugates Pattern

    We just saw a pattern for squaring binomials that we can use to make multiplying some binomials easier. Similarly, there is a pattern for another product of binomials. But before we get to it, we need to introduce some vocabulary.

    What do you notice about these pairs of binomials?

    \[(x-9)(x+9) \qquad(y-8)(y+8)\qquad (2x-5)(2x+5)\]

    Look at the first term of each binomial in each pair.

    This figure has three products. The first is x minus 9, in parentheses, times x plus 9, in parentheses. The second is y minus 8, in parentheses, times y plus 8, in parentheses. The last is 2x minus 5, in parentheses, times 2x plus 5, in parentheses

    Notice the first terms are the same in each pair.

    Look at the last terms of each binomial in each pair.

    This figure has three products. The first is x minus 9, in parentheses, times x plus 9, in parentheses. The second is y minus 8, in parentheses, times y plus 8, in parentheses. The last is 2x minus 5, in parentheses, times 2x plus 5, in parentheses.

    Notice the last terms are the same in each pair.

    Notice how each pair has one sum and one difference.

    This figure has three products. The first is x minus 9, in parentheses, times x plus 9, in parentheses. Below the x minus 9 is the word “difference”. Below x plus 9 is the word “sum”. The second is y minus 8, in parentheses, times y plus 8, in parentheses. Below y minus 8 is the word “difference”. Below y plus 8 is the word “sum”. The last is 2x minus 5, in parentheses, times 2x plus 5, in parentheses. Below the 2x minus 5 is the word “difference” and below 2x plus 5 is the word “sum”.

    A pair of binomials that each have the same first term and the same last term, but one is a sum and one is a difference has a special name. It is called a conjugate pair and is of the form (a−b),(a+b).

    CONJUGATE PAIR

    A conjugate pair is two binomials of the form

    \[(a-b),(a+b)\]

    The pair of binomials each have the same first term and the same last term, but one binomial is a sum and the other is a difference.

    There is a nice pattern for finding the product of conjugates. You could, of course, simply FOIL to get the product, but using the pattern makes your work easier.

    Let’s look for the pattern by using FOIL to multiply some conjugate pairs.

    \[\begin{array}{cc}{(x-9)(x+9)} & {(y-8)(y+8)} & (2x-5)(2x+5)\\ {x^{2}+9 x-9 x-81} & {y^{2}+8 y-8 y-64} & {4 x^{2}+10 x-10 x-25} \\ {x^{2}-81} & {y^{2}-64} & {4 x^{2}-25}\end{array}\]

    This figure has three columns. The first column contains the product of x minus 9 and x plus 9. Below this is the expression x squared plus 9x minus 9x minus 81. Below this is x squared minus 81. The second column contains the product of y minus 8 and y plus 8. Below this is the expression y squared plus 8y minus 8y minus 64. Below this is y squared minus 64. The third column contains the product of 2x minus 5 and 2x plus 5. Below this is the expression 4x squared plus 10x minus 10x minus 25. Below this is 4x squared minus 25.

    Each first term is the product of the first terms of the binomials, and since they are identical it is the square of the first term.

    \[\begin{array}{c}{(a+b)(a-b)=a^{2}-}\underline{\qquad} \\ {\text { To get the}\textbf{ first term, square the first term. }}\end{array}\]

    The last term came from multiplying the last terms, the square of the last term.

    \[\begin{array}{c}{(a+b)(a-b)=a^{2}-b^{2}} \\ {\text { To get the}\textbf{ last term, square the last term. }}\end{array}\]

    What do you observe about the products?

    The product of the two binomials is also a binomial! Most of the products resulting from FOIL have been trinomials.

    Why is there no middle term? Notice the two middle terms you get from FOIL combine to 0 in every case, the result of one addition and one subtraction.

    The product of conjugates is always of the form \(a^2-b^2\). This is called a difference of squares.

    This leads to the pattern:

    PRODUCT OF CONJUGATES PATTERN

    If a and b are real numbers,

    This figure is divided into two sides. On the left side is the following formula: the product of a minus b and a plus b equals a squared minus b squared. On the right side is the same formula labeled: a minus b and a plus b are labeled “conjugates”, the a squared and b squared are labeled squares and the minus sign between the squares is labeled “difference”. Therefore, the product of two conjugates is called a difference of squares.

    The product is called a difference of squares.

    To multiply conjugates, square the first term, square the last term, and write the product as a difference of squares.

    Let’s test this pattern with a numerical example.

    \(\begin{array}{ll} & (10-2)(10+2)\\ {\text { It is the product of conjugates, so the result will be the }} \\ {\text { difference of two squares. }} & \underline{\qquad} - \underline{\qquad}\\ {\text { Square the first term. }}& 10^2 - \underline{\qquad} \\ {\text { Square the last term. }} & 10^2 - 2^2\\ {\text { Simplify. }} & 100 -4\\ {\text { Simplify. }} & 96\\ {\text { What do you get using the Order of Operations? }} \\ \\ & (10-2)(10+2) \\ & (8)(12) \\ & 96 \end{array}\)

    Notice, the result is the same!

    Exercise \(\PageIndex{16}\)

    Multiply: \((x-8)(x+8)\)

    Answer

    First, recognize this as a product of conjugates. The binomials have the same first terms, and the same last terms, and one binomial is a sum and the other is a difference.

    It fits the pattern. The product of x minus 8 and x plus 8. Above this is the general form a minus b, in parentheses, times a plus b, in parentheses.
    Square the first term, x. x squared minus blank. Above this is the general form a squared minus b squared.
    Square the last term, 8. x squared minus 8 squared.
    The product is a difference of squares. x squared minus 64.

    Exercise \(\PageIndex{17}\)

    Multiply: \((x-5)(x+5)\)

    Answer

    \(x^{2}-25\)

    Exercise \(\PageIndex{18}\)

    Multiply: \((w-3)(w+3)\)

    Answer

    \(w^{2}-9\)

    Exercise \(\PageIndex{19}\)

    Multiply: \((2 x+5)(2 x-5)\)

    Answer

    Are the binomials conjugates?

    It is the product of conjugates. The product of 2x plus 5 and 2x minus 5. Above this is the general form a minus b, in parentheses, times a plus b, in parentheses.
    Square the first term, 2x. 2 x squared minus blank. Above this is the general form a squared minus b squared.
    Square the last term, 5. 2 x squared minus 5 squared.
    Simplify. The product is a difference of squares. 4 x squared minus 25.

    Exercise \(\PageIndex{20}\)

    Multiply: \((6 x+5)(6 x-5)\)

    Answer

    \(36 x^{2}-25\)

    Exercise \(\PageIndex{21}\)

    Multiply: \((2 x+7)(2 x-7)\)

    Answer

    \(4 x^{2}-49\)

    The binomials in the next example may look backwards – the variable is in the second term. But the two binomials are still conjugates, so we use the same pattern to multiply them.

    Exercise \(\PageIndex{22}\)

    Find the product:\((3+5 x)(3-5 x)\)

    Answer
    It is the product of conjugates. The product of 3 plus 5 x and 3 minus 5 x. Above this is the general form a plus b, in parentheses, times a minus b, in parentheses.
    Use the pattern. 3 squared minus 5 x squared. Above this is the general form a squared minus b squared.
    Simplify. 9 minus 25 x squared.

    Exercise \(\PageIndex{23}\)

    Multiply: \((7+4 x)(7-4 x)\)

    Answer

    \(49-16 x^{2}\)

    Exercise \(\PageIndex{24}\)

    Multiply: \((9-2 y)(9+2 y)\)

    Answer

    \(81-4 y^{2}\)

    Now we’ll multiply conjugates that have two variables.

    Exercise \(\PageIndex{25}\)

    Find the product: \((5 m-9 n)(5 m+9 n)\)

    Answer
    This fits the pattern. 5 m minus 9 n and 5 m plus 9 n. Above this is the general form a plus b, in parentheses, times a minus b, in parentheses.
    Use the pattern. 5 m squared minus 9 n squared. Above this is the general form a squared minus b squared.
    Simplify. 25 m squared minus 81 n squared.

    Exercise \(\PageIndex{26}\)

    Find the product: \((4 p-7 q)(4 p+7 q)\)

    Answer

    \(16 p^{2}-49 q^{2}\)

    Exercise \(\PageIndex{27}\)

    Find the product: \((3 x-y)(3 x+y)\)

    Answer

    \(9 x^{2}-y^{2}\)

    Exercise \(\PageIndex{28}\)

    Find the product: \((c d-8)(c d+8)\)

    Answer
    This fits the pattern. The product of c d minus 8 and c d plus 8. Above this is the general form a plus b, in parentheses, times a minus b, in parentheses.
    Use the pattern. c d squared minus 8 squared. Above this is the general form a squared minus b squared.
    Simplify. c squared d squared minus 64.

    Exercise \(\PageIndex{29}\)

    Find the product: \((x y-6)(x y+6)\)

    Answer

    \(x^{2} y^{2}-36\)

    Exercise \(\PageIndex{30}\)

    Find the product: \((a b-9)(a b+9)\)

    Answer

    \(a^{2} b^{2}-81\)

    Exercise \(\PageIndex{31}\)

    Find the product:\(\left(6 u^{2}-11 v^{5}\right)\left(6 u^{2}+11 v^{5}\right)\)

    Answer
    This fits the pattern. The product of 6 u squared minus 11 v to the fifth power and 6 u squared plus 11 v to the fifth power. Above this is the general form a plus b, in parentheses, times a minus b, in parentheses.
    Use the pattern. 6 u squared, in parentheses, squared, minus 11 v to the fifth power, in parentheses, squared. Above this is the general form a squared minus b squared.
    Simplify. 36 u to the fourth power minus 121 v to the tenth power.

    Exercise \(\PageIndex{32}\)

    Find the product:\(\left(3 x^{2}-4 y^{3}\right)\left(3 x^{2}+4 y^{3}\right)\)

    Answer

    \(9 x^{4}-16 y^{6}\)

    Exercise \(\PageIndex{33}\)

    Find the product:\(\left(2 m^{2}-5 n^{3}\right)\left(2 m^{2}+5 n^{3}\right)\)

    Answer

    \(4 m^{4}-25 n^{6}\)

    Recognize and Use the Appropriate Special Product Pattern

    We just developed special product patterns for Binomial Squares and for the Product of Conjugates. The products look similar, so it is important to recognize when it is appropriate to use each of these patterns and to notice how they differ. Look at the two patterns together and note their similarities and differences.

    COMPARING THE SPECIAL PRODUCT PATTERNS

    \(\begin{array}{ll }{\textbf { Binomial Squares }}&{\textbf { Product of Conjugates }} \\ {(a+b)^{2}=a^{2}+2 a b+b^{2}} & {(a-b)(a+b)=a^{2}-b^{2}} \\ {(a-b)^{2}=a^{2}+2 a b+b^{2}} \\ {\text { - Squaring a binomial }}& {\text { - Multiplying conjugates }} \\ {\text { - Product is a trinomial }} & {\text { - Product is a binomial }} \\ {\text { - Inner and outer terms with FOIL are the same. }} &{\text { - Inner and outer terms with FOIL are opposites. }} \\ {\text { - Middle term is double the product of the terms. }} &{\text { - There is no middle term. }} \end{array}\)

    Exercise \(\PageIndex{34}\)

    Choose the appropriate pattern and use it to find the product:

    1. \((2 x-3)(2 x+3)\)
    2. \((8 x-5)^{2}\)
    3. \((6 m+7)^{2}\)
    4. \((5 x-6)(6 x+5)\)
    Answer

    1. \((2x−3)(2x+3)\) These are conjugates. They have the same first numbers, and the same last numbers, and one binomial is a sum and the other is a difference. It fits the Product of Conjugates pattern.

    This fits the pattern. The product of 2 x minus 3 and 2 x plus 3. Above this is the general form a plus b, in parentheses, times a minus b, in parentheses.
    Use the pattern. 2 x squared minus 3 squared. Above this is the general form a squared minus b squared.
    Simplify. 4 x squared minus 9.
    2. \((8 x-5)^{2}\) We are asked to square a binomial. It fits the binomial squares pattern.
    8 x minus 5, in parentheses, squared. Above this is the general form a minus b, in parentheses, squared.
    Use the pattern. 8 x squared minus 2 times 8 x times 5 plus 5 squared. Above this is the general form a squared minus 2 a b plus b squared.
    Simplify. 64 x squared minus 80 x plus 25.
    3. \((6 m+7)^{2}\) Again, we will square a binomial so we use the binomial squares pattern.
    6 m plus 7, in parentheses, squared. Above this is the general form a plus b, in parentheses, squared.
    Use the pattern. 6 m squared plus 2 times 6 m times 7 plus 7 squared. Above this is the general form a squared plus 2 a b plus b squared.
    Simplify. 36 m squared plus 84 m plus 49.
    4. \((5 x-6)(6 x+5)\)This product does not fit the patterns, so we will use FOIL.
    \(\begin{array}{ll} & (5 x-6)(6 x+5)\\ {\text { Use FOIL. }} & {30 x^{2}+25 x-36 x-30} \\ {\text { Simplify. }} & {30 x^{2}-11 x-30}\end{array}\)

    Exercise \(\PageIndex{35}\)

    Choose the appropriate pattern and use it to find the product:

    1. \((9b−2)(2b+9)\)
    2. \((9p−4)2\)
    3. \((7y+1)^2\)
    4. \((4r-3)(4r+3)\)
    Answer
    1. FOIL; \(18 b^{2}+77 b-18\)
    2. Binomial Squares; \(81 p^{2}-72 p+16\)
    3. Binomial Squares; \(49 y^{2}+14 y+1\)
    4. Product of Conjugates; \(16 r^{2}-9\)

    Exercise \(\PageIndex{36}\)

    Choose the appropriate pattern and use it to find the product:

    1. \((6x+7)^2\)
    2. \((3x−4)(3x+4)\)
    3. \((2x−5)(5x−2)\)
    4. \((6n−1)^2\)
    Answer
    1. Binomial Squares; \(36 x^{2}+84 x+49\)
    2. Product of Conjugates; \(9 x^{2}-16\)
    3. FOIL; \(10 x^{2}-29 x+10\)
    4. Binomial Squares; \(36 n^{2}-12 n+1\)

    Note

    Access these online resources for additional instruction and practice with special products:

    Key Concepts

    • Binomial Squares Pattern
      • If a,b are real numbers,
        No Alt Text
      • \((a+b)^{2}=a^{2}+2 a b+b^{2}\)
      • \((a-b)^{2}=a^{2}-2 a b+b^{2}\)
      • To square a binomial: square the first term, square the last term, double their product.
    • Product of Conjugates Pattern
      • If a,ba,b are real numbers,
        No Alt Text
      • \((a-b)(a+b)=a^{2}-b^{2}\)
      • The product is called a difference of squares.
    • To multiply conjugates:
      • square the first term square the last term write it as a difference of squares

    Glossary

    conjugate pair
    A conjugate pair is two binomials of the form \((a−b)\) and \((a+b)\); the pair of binomials each have the same first term and the same last term, but one binomial is a sum and the other is a difference.