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4.3: Factor Special Products

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    Learning Objectives

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

    • Factor perfect square trinomials
    • Factor differences of squares
    • Factor sums and differences of cubes
    Be Prepared

    Before you get started, take this readiness quiz.

    1. Simplify: \((3x^2)^3\).

    2. Multiply: \((m+4)^2\).

    3. Multiply: \((x−3)(x+3)\).

    We have seen that some binomials and trinomials result from special products—squaring binomials and multiplying conjugates. If we recognize these kinds of polynomials, we can use the special products patterns to factor them much more quickly.

    Factor Differences of Squares

    One special product we are familiar with is the Product of Conjugates pattern. We use this to multiply two binomials that were conjugates. Here’s an example:

    We have open parentheses 3x minus 4 close parentheses open parentheses 3x plus 4. This is of the form a minus b, a plus b. We rewrite as open parentheses 3x close parentheses squared minus 4 squared. Here, 3x is a and 4 is b. This is equal to 9 x squared minus 16.

    A difference of squares factors to a product of conjugates (in this context, \(a-b\) is said to be conjugate to \(a+b\), and vice versa.)

    Difference of Squares Pattern

    If \(a\) and \(b\) are real numbers,

    a squared minus b squared equals a minus b, a plus b. Here, a squared minus b squared is difference of squares and a minus b, a plus b are conjugates.

    Remember, “difference” refers to subtraction. So, to use this pattern we must make sure we have a binomial in which two squares are being subtracted.

    Example \(\PageIndex{1}\)

    Factor \(64y^2−1\).

    Solution

    Step 1 is to check if the binomial 64 y squared minus 1 fits the pattern. For that we check the following: Is this a difference? Yes. Are the first and last terms perfect squares? Yes.
    Step 2 is to write both terms as squares, So, we have open parentheses 8y close parentheses squared minus 1 squared.
    Step 3 is to write the product of conjugates 8y minus 1, 8y plus 1.
    Step 4 is to check. We multiply to get the original binomial

    Try It \(\PageIndex{2}\)

    Factor \(121m^2−1\).

    Answer

    \((11m−1)(11m+1)\)

    Try It \(\PageIndex{3}\)

    Factor \(81y^2−1\).

    Answer

    \((9y−1)(9y+1)\)

    Factor Difference of Squares

    \(\begin{array} {llll} \textbf{Step 1.} &\text{Does the binomial fit the pattern?} &\qquad &\hspace{5mm} a^2−b^2 \\ &\text{Is this a difference?} &\qquad &\hspace{2mm} \text{____−____} \\ &\text{Are the first and last terms perfect squares?} & & \\ \textbf{Step 2.} &\text{Write them as squares.} &\qquad &\hspace{3mm} (a)^2−(b)^2 \\ \textbf{Step 3.} &\text{Write the product of conjugates.} &\qquad &(a−b)(a+b) \\ \textbf{Step 4.} &\text{Check by multiplying.} & & \end{array}\)

    It is important to remember that sums of squares do not factor into a product of binomials. There are no binomial factors that multiply together to get a sum of squares. After removing any GCF, the expression \(a^2+b^2\) can not be factored any further!

    The next example shows variables in both terms.

    Example \(\PageIndex{4}\)

    Factor \(144x^2−49y^2\).

    Solution

    \(\begin{array} {lll} &\quad &144x^2−49y^2 \\ \text{Is this a difference of squares? Yes.} &\quad &(12x)^2−(7y)^2 \\ \text{Factor as the product of conjugates.} &\quad &(12x−7y)(12x+7y) \\ \text{Check by multiplying.} &\quad &(12x−7y)(12x+7y) \\ \text{Check by multiplying.} &\quad & \\ &\quad & \\ &\quad & \\ \hspace{14mm} (12x−7y)(12x+7y) &\quad & \\ \hspace{21mm} 144x^2−49y^2\checkmark &\quad & \end{array}\)

    Try It \(\PageIndex{5}\)

    Factor \(196m^2−25n^2\).

    Answer

    \((16m−5n)(16m+5n)\)

    Try It \(\PageIndex{6}\)

    Factor \(121p^2−9q^2\).

    Answer

    \((11p−3q)(11p+3q)\)

    As always, we should look for a common factor first whenever we have an expression to factor. Sometimes a common factor may “disguise” the difference of squares and we won’t recognize the perfect squares until we factor the GCF.

    Also, to completely factor the binomial in the next example, we’ll factor a difference of squares twice!

    Example \(\PageIndex{7}\)

    Factor \(48x^4y^2−243y^2\).

    Solution

    \(\begin{array} {ll} &48x^4y^2−243y^2 \\ \text{Is there a GCF? Yes, }3y^2\text{—factor it out!} &3y^2(16x^4−81) \\ \text{Is the binomial a difference of squares? Yes.} &3y^2\left((4x^2)^2−(9)^2\right) \\ \text{Factor as a product of conjugates.} &3y^2(4x^2−9)(4x^2+9) \\ \text{Notice the first binomial is also a difference of squares!} &3y^2((2x)^2−(3)^2)(4x^2+9) \\ \text{Factor it as the product of conjugates.} &3y^2(2x−3)(2x+3)(4x^2+9) \end{array}\)

    The last factor, the sum of squares, cannot be factored.

    \(\begin{array} {l} \text{Check by multiplying:} \\ \hspace{10mm} 3y^2(2x−3)(2x+3)(4x^2+9) \\ \\ \\ \hspace{15mm} 3y^2(4x^2−9)(4x^2+9) \\ \hspace{20mm} 3y^2(16x^4−81) \\ \hspace{19mm} 48x^4y^2−243y^2\checkmark\end{array}\)

    Try It \(\PageIndex{8}\)

    Factor \(2x^4y^2−32y^2\).

    Answer

    \(2y^2(x−2)(x+2)(x^2+4)\)

    Try It \(\PageIndex{9}\)

    Factor \(7a^4c^2−7b^4c^2\).

    Answer

    \(7c^2(a−b)(a+b)(a^2+b^2)\)

    The next example has a polynomial with 4 terms. So far, when this occurred we grouped the terms in twos and factored from there. Here we will notice that the first three terms form a perfect square trinomial.

    Example \(\PageIndex{10}\)

    Factor \(x^2−6x+9−y^2\).

    Solution

    Notice that the first three terms form a perfect square trinomial.

       
    Factor by grouping the first three terms.  
    Use the perfect square trinomial pattern. \((x-3)^{2}-y^{2}\)
    Is this a difference of squares? Yes.  
    Yes—write them as squares. .
    Factor as the product of conjugates. .
      \((x-3-y)(x-3+y)\)

    we may want to rewrite the solution as \((x−y−3)(x+y−3)\).

    Try It \(\PageIndex{11}\)

    Factor \(x^2−10x+25−y^2\).

    Answer

    \((x−5−y)(x−5+y)\)

    Try It \(\PageIndex{12}\)

    Factor \(x^2+6x+9−4y^2\).

    Answer

    \((x+3−2y)(x+3+2y)\)

    Factor Perfect Square Trinomials (optional discussion)

    Some trinomials are perfect squares. They result from multiplying a binomial times itself. We squared a binomial using the Binomial Squares pattern in a previous chapter.

    In open parentheses 3x plus 4 close parentheses squared, 3x is a and 4 is b. Writing it as a squared plus 2ab plus b squared, we get open parentheses 3x close parentheses squared plus 2 times 3x times 4 plus 4 squared. This is equal to 9 x squared plus 24x plus 16.

    The trinomial \(9x^2+24x+16\) is called a perfect square trinomial. It is the square of the binomial \(3x+4\).

    In this chapter, we will start with a perfect square trinomial and factor it into its as many factors as possible. We could factor this trinomial using the methods described in the last section, since it is of the form \(ax^2+bx+c\). But if we recognize that the first and last terms are squares and the trinomial fits the perfect square trinomials pattern, we will save ourselves a lot of work. Here is the pattern—the reverse of the binomial squares pattern.

    Perfect Square Trinomial Pattern

    If \(a\) and \(b\) are real numbers

    \[a^2+2ab+b^2=(a+b)^2,\nonumber\]

    \[a^2−2ab+b^2=(a−b)^2.\nonumber\]

    To make use of this pattern, we have to recognize that a given trinomial fits it. Check first to see if the leading coefficient is a perfect square, \(a^2\). Next check that the last term is a perfect square, \(b^2\). Then check the middle term—is it the product, \(2ab\)? If everything checks, we can easily write the factors.

    The sign of the middle term determines which pattern we will use. When the middle term is negative, we use the pattern \(a^2−2ab+b^2\), which factors to \((a−b)^2\).

    The steps are summarized here.

    Factor Perfect Square Trinomials

    \(\begin{array} {lllll} \textbf{Step 1.} &\text{Does the trinomial fit the pattern?} &\quad &\hspace{7mm} a^2+2ab+b^2 &\hspace{7mm} a^2−2ab+b^2 \\ &\text{Are the first and last terms perfect squares?} &\quad & &\\ &\text{Write them as squares.} &\quad &\hspace{5mm}(a)^2\hspace{16mm} (b)^2 &\hspace{6mm}(a)^2\hspace{16mm} (b)^2 \\ &\text{Check the middle term. Is it }2ab? &\quad &\hspace{12mm} {\,}^{\searrow}{\,}_{2·a·b}{\,}^{\swarrow} &\hspace{12mm} {\,}^{\searrow}{\,}_{2·a·b}{\,}^{\swarrow} \\ \textbf{Step 2.} &\text{Write the square of the binomial.} &\quad &\hspace{13mm} (a+b)^2 &\hspace{13mm} (a−b)^2 \\ \textbf{Step 3.} &\text{Check by multiplying.} & & & \end{array}\)

    Remember the first step in factoring is to look for a greatest common factor. Perfect square trinomials may have a GCF in all three terms and it should be factored out first. And, sometimes, once the GCF has been factored, we will recognize a perfect square trinomial.

    The above is for consideration, but we will not give any examples of this here.

    Key Concepts

    • Difference of Squares Pattern: If \(a\), \(b\) are real numbers,
      a squared minus b squared is a minus b, a plus b. Here, a squared minus b squared is the difference of squares and a minus b, a plus b are conjugates.
    • How to factor differences of squares
      \(\begin{array} {llll} \textbf{Step 1.} &\text{Does the binomial fit the pattern?} &\qquad &\hspace{5mm} a^2−b^2 \\ &\text{Is this a difference?} &\qquad &\hspace{2mm} \text{____−____} \\ &\text{Are the first and last terms perfect squares?} & & \\ \textbf{Step 2.} &\text{Write them as squares.} &\qquad &\hspace{3mm} (a)^2−(b)^2 \\ \textbf{Step 3.} &\text{Write the product of conjugates.} &\qquad &(a−b)(a+b) \\ \textbf{Step 4.} &\text{Check by multiplying.} & & \end{array}\)
    • Perfect Square Trinomials Pattern: If \(a\) and \(b\) are real numbers,

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

    • How to factor perfect square trinomials
      \(\begin{array} {lllll} \textbf{Step 1.} &\text{Does the trinomial fit the pattern?} &\quad &\hspace{7mm} a^2+2ab+b^2 &\hspace{7mm} a^2−2ab+b^2 \\ &\text{Are the first and last terms perfect squares?} &\quad & &\\ &\text{Write them as squares.} &\quad &\hspace{5mm}(a)^2\hspace{16mm} (b)^2 &\hspace{6mm}(a)^2\hspace{16mm} (b)^2 \\ &\text{Check the middle term. Is it }2ab? &\quad &\hspace{12mm} {\,}^{\searrow}{\,}_{2·a·b}{\,}^{\swarrow} &\hspace{12mm} {\,}^{\searrow}{\,}_{2·a·b}{\,}^{\swarrow} \\ \textbf{Step 2.} &\text{Write the square of the binomial.} &\quad &\hspace{13mm} (a+b)^2 &\hspace{13mm} (a−b)^2 \\ \textbf{Step 3.} &\text{Check by multiplying.} & & & \end{array}\)

    Practice Makes Perfect

    Factor Differences of Squares

    In the following exercises, factor completely using the difference of squares pattern, if possible.

    1. \(25v^2−1\)

    Answer

    \((5v−1)(5v+1)\)

    2. \(169q^2−1\)

    3. \(4−49x^2\)

    Answer

    \((7x−2)(7x+2)\)

    4. \(121−25s^2\)

    5. \(6p^2q^2−54p^2\)

    Answer

    \(6p^2(q−3)(q+3)\)

    6. \(98r^3−72r\)

    7. \(24p^2+54\)

    Answer

    \(6(4p^2+9)\)

    8. \(20b^2+140\)

    9. \(121x^2−144y^2\)

    Answer

    \((11x−12y)(11x+12y)\)

    10. \(49x^2−81y^2\)

    11. \(169c^2−36d^2\)

    Answer

    \((13c−6d)(13c+6d)\)

    12. \(36p^2−49q^2\)

    13. \(16z^4−1\)

    Answer

    \((2z−1)(2z+1)(4z^2+1)\)

    14. \(m^4−n^4\)

    15. \(162a^4b^2−32b^2\)

    Answer

    \(2b^2(3a−2)(3a+2)(9a^2+4)\)

    16. \(48m^4n^2−243n^2\)

    17. \(x^2−16x+64−y^2\)

    Answer

    \((x−8−y)(x−8+y)\)

    18. \(p^2+14p+49−q^2\)

    19. \(a^2+6a+9−9b^2\)

    Answer

    \((a+3−3b)(a+3+3b)\)

    20. \(m^2−6m+9−16n^2\)

    Factor Perfect Square Trinomials

    In the following exercises, factor completely using the perfect square trinomials pattern.

    21. \(16y^2+24y+9\)

    Answer

    \((4y+3)^2\)

    22. \(25v^2+20v+4\)

    23. \(36s^2+84s+49\)

    Answer

    \((6s+7)^2\)

    24. \(49s^2+154s+121\)

    25. \(100x^2−20x+1\)

    Answer

    \((10x−1)^2\)

    26. \(64z^2−16z+1\)

    27. \(25n^2−120n+144\)

    Answer

    \((5n−12)^2\)

    28. \(4p^2−52p+169\)

    29. \(49x^2+28xy+4y^2\)

    Answer

    \((7x+2y)^2\)

    30. \(25r^2+60rs+36s^2\)

    31. \(100y^2−52y+1\)

    Answer

    \((50y−1)(2y−1)\)

    32. \(64m^2−34m+1\)

    33. \(10jk^2+80jk+160j\)

    Answer

    \(10j(k+4)^2\)

    34. \(64x^2y−96xy+36y\)

    35. \(75u^4−30u^3v+3u^2v^2\)

    Answer

    \(3u^2(5u−v)^2\)

    36. \(90p^4+300p^4q+250p^2q^2\)

    Mixed Practice

    In the following exercises, factor completely.

    37. \(64a^2−25\)

    Answer

    \((8a−5)(8a+5)\)

    38. \(121x^2−144\)

    39. \(27q^2−3\)

    Answer

    \(3(3q−1)(3q+1)\)

    40. \(4p^2−100\)

    41. \(16x^2−72x+81\)

    Answer

    \((4x−9)^2\)

    42. \(36y^2+12y+1\)

    43. \(8p^2+2\)

    Answer

    \(2(4p^2+1)\)

    44. \(81x^2+169\)

    45. \(45n^2+60n+20\)

    Answer

    \(5(3n+2)^2\)

    46. \(x^2−10x+25−y^2\)

    Answer

    \((x+y−5)(x−y−5)\)

    47. \(x^2+12x+36−y^2\)

    Writing Exercises

    48. Why was it important to practice using the binomial squares pattern in the chapter on multiplying polynomials?

    Answer

    Answers will vary.

    49. How do you recognize the binomial squares pattern?

    50. Explain why \(n^2+25\neq (n+5)^2\). Use algebra, words, or pictures.

    Answer

    Answers will vary.

    51. Maribel factored \(y^2−30y+81\) as \((y−9)^2\). Was she right or wrong? How do you know?

    Self Check

    a. After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

    b. What does this checklist tell you about your mastery of this section? What steps will you take to improve?

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