Algebra 1 · A.APR.A.1 · A.APR.C.4

Special products

When the two brackets are the same, or differ only in one sign, what shape does their product always take?

You need

Multiplying two binomialsAlgebra 1

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Proving polynomial identitiesAlgebra 2 Rationalizing a denominatorAlgebra 1 Factoring special patternsAlgebra 1

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Special products

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Why this exists

Multiplying two binomials always works the same way: every term of the first times every term of the second, four products, then collect like terms. Two shapes of product turn up so often that the four products are worth studying once and for all. One is a binomial times itself, such as (x + 5)^2. The other is a sum times the difference of the same two terms, such as (x + 5)(x - 5).

In both, the two middle products are not just like terms but equal in size. In a square they are equal and add up; in a sum times a difference they are opposites and cancel. Because of that, the answer has a fixed shape that can be written down on sight, read backwards later to factor, and even used to do arithmetic such as 104^2 in your head.

Definition

A special product is one of these three products of binomials, where a and b stand for any two terms (numbers, letters, or terms such as 3x or 2y):

the square of a sum, (a + b)^2, which means (a + b)(a + b);
the square of a difference, (a - b)^2, which means (a - b)(a - b);
the product of a sum and a difference of the same two terms, (a + b)(a - b). Two binomials like this are called conjugates of each other.

A trinomial of the form a^2 + 2ab + b^2 or a^2 - 2ab + b^2 is a perfect square trinomial, because it is the square of a binomial. An expression a^2 - b^2 is a difference of two squares.

Intuition

Figure 1The square of side a + b is two squares and two equal rectangles, a^2 + 2ab + b^2.

Take a square of side a + b and cut each side where a ends. Figure 1 shows the four pieces: a square a by a, a square b by b, and two rectangles, each a by b. The two rectangles are the same size, so together they are 2ab:

(a + b)^2 = a^2 + 2ab + b^2

The two rectangles are why the square of a sum is more than the sum of the squares. With numbers, (3 + 4)^2 = 7^2 = 49, while 3^2 + 4^2 = 9 + 16 = 25. The gap is 49 - 25 = 24, which is 2 \cdot 3 \cdot 4: exactly the two 3 by 4 rectangles in the corners of the 7 by 7 square.

Figure 2Take b^2 off the corner of a^2, then move strip B. The same area becomes a rectangle (a + b) by (a - b).

For a sum times a difference, start with a square of side a and take a small square of side b off one corner (Figure 2). What is left has area a^2 - b^2. Cut the L-shape into piece A, which is a wide and a - b tall, and piece B, which is a - b wide and b tall. Turn B a quarter turn and set it beside A. Both pieces are a - b tall now, so together they make one rectangle a + b wide and a - b tall. Moving a piece does not change its area, so

(a + b)(a - b) = a^2 - b^2

Results

For any two terms a and b:

(a + b)^2 = a^2 + 2ab + b^2

(a - b)^2 = a^2 - 2ab + b^2

(a + b)(a - b) = a^2 - b^2

In words: the square of a binomial is the square of the first term, plus or minus twice the product of the two terms, plus the square of the last term. The last term is always added, because (-b)^2 = b^2. A sum times the matching difference is the square of the first term minus the square of the second, with no middle term.

Taking a^2 + b^2 away from the first result gives the size of the gap seen in the intuition:

(a + b)^2 - (a^2 + b^2) = 2ab

So (a + b)^2 equals a^2 + b^2 only when 2ab = 0, that is, when a = 0 or b = 0.

Why it is true

Each result is an ordinary product of two binomials, so the distributive property gives the four products, and only the middle two need a second look.

For the square of a sum,

\begin{aligned} (a + b)(a + b) &= a^2 + ab + ba + b^2\\ &= a^2 + 2ab + b^2 \end{aligned}

because ba = ab by the commutative property of multiplication, so the middle products are like terms and add to 2ab.

For the square of a difference, write a - b as a + (-b) and use the first result with -b in place of b:

\begin{aligned} (a + (-b))^2 &= a^2 + 2a(-b) + (-b)^2\\ &= a^2 - 2ab + b^2 \end{aligned}

since a negative times a negative is positive.

For a sum times a difference,

\begin{aligned} (a + b)(a - b) &= a^2 - ab + ba - b^2\\ &= a^2 - b^2 \end{aligned}

because -ab + ba = -ab + ab = 0. The middle products are opposites, so they cancel.

These steps hold for every value of a and b, so the three results are identities: true for all numbers, positive, negative or zero. The figures only illustrate the case where a and b are positive lengths, with a > b in Figure 2.

Worked examples

Example 1

Squaring a sum and a difference

Expand (h + 7)^2 and (n - 11)^2.

Solution

In (h + 7)^2 the first term is a = h and the last is b = 7. Square the first, add twice the product, add the square of the last:

\begin{aligned} (h + 7)^2 &= h^2 + 2 \cdot h \cdot 7 + 7^2\\ &= h^2 + 14h + 49 \end{aligned}

In (n - 11)^2 the terms are n and 11, joined by a minus, so the middle term is subtracted and the last is still added:

\begin{aligned} (n - 11)^2 &= n^2 - 2 \cdot n \cdot 11 + 11^2\\ &= n^2 - 22n + 121 \end{aligned}

Check at h = 1: 8^2 = 64 and 1 + 14 + 49 = 64. Check at n = 1: (-10)^2 = 100 and 1 - 22 + 121 = 100.

Example 2

A coefficient and two variables

Expand (3x - 2y)^2.

Solution

The pattern works with whole terms, so take a = 3x and b = 2y. The square applies to the whole term, so (3x)^2 = 3^2x^2 = 9x^2 and (2y)^2 = 4y^2 by the exponent rules. Twice the product is 2 \cdot 3x \cdot 2y = 12xy, and it is subtracted because the binomial is a difference:

(3x - 2y)^2 = 9x^2 - 12xy + 4y^2

Check at x = 2, y = 1: (6 - 2)^2 = 16 and 36 - 24 + 4 = 16.

Example 3

Spotting a sum times a difference

Expand (5m + 4)(5m - 4) and (k + 7)(7 - k).

Solution

The first is the sum and the difference of the same two terms, 5m and 4. The middle products -20m and 20m cancel, leaving the square of the first minus the square of the second:

(5m + 4)(5m - 4) = 25m^2 - 16

In the second, the terms are written in a different order. By the commutative property of addition, k + 7 = 7 + k, so this is (7 + k)(7 - k): a sum and a difference of 7 and k, with 7 as the first term. The first term squared comes first and the second is subtracted:

(k + 7)(7 - k) = 49 - k^2

Check at k = 2: 9 \cdot 5 = 45 and 49 - 4 = 45. Which term is subtracted is decided by which one carries the minus sign in the difference, not by where it is written.

Example 4

A border round a quilt

Kestrel sews a square quilt: a center panel of side 12 inches with a border of width b inches on every side. Write the area of the whole quilt as a sum of terms, find the area of the border alone, and find it for b = 3.

Solution

Figure 3Kestrel’s quilt. The border is four strips 12 by b and four corners b by b.

The border adds b on the left and b on the right, so the quilt has side 12 + 2b. Its area is the square of a sum with a = 12 and second term 2b:

\begin{aligned} (12 + 2b)^2 &= 12^2 + 2 \cdot 12 \cdot 2b + (2b)^2\\ &= 144 + 48b + 4b^2 \end{aligned}

The center panel covers 144 square inches, so the border is the rest, 48b + 4b^2 square inches. Figure 3 shows where each term lives: four strips 12 by b give 48b, and four corner squares b by b give 4b^2. At b = 3 the border is 48 \cdot 3 + 4 \cdot 9 = 180 square inches. Check: the quilt is 18 inches wide, and 18^2 - 144 = 324 - 144 = 180.

Example 5

Arithmetic in your head

Work out 104^2, 57^2 and 43 \times 37 without a written multiplication.

Solution

Split each number around a round number that is easy to square. 104 = 100 + 4, so

\begin{aligned} 104^2 &= 100^2 + 2 \cdot 100 \cdot 4 + 4^2\\ &= 10000 + 800 + 16 = 10816 \end{aligned}

57 = 60 - 3, so

\begin{aligned} 57^2 &= 60^2 - 2 \cdot 60 \cdot 3 + 3^2\\ &= 3600 - 360 + 9 = 3249 \end{aligned}

43 and 37 sit 3 either side of 40, so the product is a sum times a difference:

43 \times 37 = (40 + 3)(40 - 3) = 1600 - 9 = 1591

Each time the hard multiplication became squares of round numbers and small numbers.

Example 6

Building a perfect square

What number fills the gap in x^2 + 30x + \underline{\quad} to make a perfect square trinomial? Write the trinomial as a square.

Solution

Match the expression to a^2 + 2ab + b^2. The first term x^2 says a = x. The middle term must be 2ab = 2 \cdot x \cdot b, and it is 30x, so 2b = 30 and b = 15. The missing last term is b^2 = 225:

x^2 + 30x + 225 = (x + 15)^2

In the picture of Figure 1, x^2 is the big square and 30x is two rectangles x by 15; the gap is the 15 by 15 corner that completes the square. With a minus sign the same number works, x^2 - 30x + 225 = (x - 15)^2, since the last term of a perfect square is always added. Adding this number on purpose to solve an equation is completing the square.

Boundaries

The three results are identities for any terms a and b, including negative ones and terms with coefficients. The area pictures are only pictures: they need positive lengths, and Figure 2 also needs a > b.
The sum-times-difference pattern needs the same two terms in both brackets. (x + 3)(x - 5) is not of this shape; its middle products -5x and 3x do not cancel, and it is an ordinary product of two binomials.
An exponent passes through a product but not through a sum. (ab)^2 = a^2b^2 for all a and b, but (a + b)^2 = a^2 + b^2 only when a = 0 or b = 0.
These patterns are for a binomial. Squaring three terms makes nine products, and higher powers of a binomial are not covered here.

Where this goes next

Read from right to left, the results factor: x^2 - 400 = (x + 20)(x - 20) and x^2 + 30x + 225 = (x + 15)^2. That reverse direction is factoring special patterns. Building the missing b^2 on purpose, as in Example 6, is how completing the square turns any quadratic equation into a square that can be undone. The sum times a difference clears square roots: \displaystyle (\sqrt{5} + 2)(\sqrt{5} - 2) = 5 - 4 = 1, which is the idea behind rationalizing a denominator. Identities like these are proved and used in Proving polynomial identities, and the coefficients of higher powers of a + b are the binomial coefficients.

Exercises

Squares of sums and differences

Expand (t + 8)^2.
Expand (y - 3)^2.
Expand (2x + 5)^2.
Expand (4w - 1)^2.
Expand \displaystyle \left(x + \frac{1}{2}\right)^2.

Sums times differences

Expand (p + 10)(p - 10).
Expand (3a - 7)(3a + 7).
Expand (6 + r)(6 - r).
Expand (x^2 + 5)(x^2 - 5).

Two variables

Expand (a + 3b)^2.
Expand (5u - 2v)^2.
Expand (2c + 9d)(2c - 9d).

Arithmetic in your head

Use a special product to work out 203^2 and 89^2.
Use a special product to work out 62 \times 58 and 7.1 \times 6.9.

Problems

A square sandbox of side x feet has a wooden deck 3 feet wide around it on every side. Write the area of the deck alone as a sum of terms, and check it at x = 10 by finding the two areas directly.
Fill each gap to make a perfect square trinomial, and write it as a square: (a) x^2 + 34x + \underline{\quad}; (b) y^2 - 11y + \underline{\quad}; (c) \underline{\quad} + 28k + 49, where the missing term is a positive multiple of k^2.

Reasoning

Show that (a + b)^2 - (a - b)^2 = 4ab for every a and b, and use it to write (n + 3)^2 - (n - 3)^2 in one step.
Explain why the product of two whole numbers that differ by 2 is always one less than the square of the number between them, for example 15 \times 17 = 16^2 - 1.

Answers

t^2 + 16t + 64. Twice the product is 2 \cdot t \cdot 8 = 16t, and 8^2 = 64.
y^2 - 6y + 9. Twice the product is 6y, subtracted; 3^2 = 9 is added.
4x^2 + 20x + 25. (2x)^2 = 4x^2, 2 \cdot 2x \cdot 5 = 20x, 5^2 = 25.
16w^2 - 8w + 1. (4w)^2 = 16w^2, 2 \cdot 4w \cdot 1 = 8w, 1^2 = 1.
\displaystyle x^2 + x + \frac{1}{4}. Twice the product is \displaystyle 2 \cdot x \cdot \frac{1}{2} = x, and \displaystyle \left(\frac{1}{2}\right)^2 = \frac{1}{4}.
p^2 - 100. The middle products -10p and 10p cancel.
9a^2 - 49. The order of the brackets does not matter, since multiplication is commutative.
36 - r^2. Here 6 is the first term and r is the term subtracted.
x^4 - 25. The first term is x^2, and (x^2)^2 = x^4.
a^2 + 6ab + 9b^2. Twice the product is 2 \cdot a \cdot 3b = 6ab, and (3b)^2 = 9b^2.
25u^2 - 20uv + 4v^2. Twice the product is 2 \cdot 5u \cdot 2v = 20uv, subtracted.
4c^2 - 81d^2. (2c)^2 = 4c^2 and (9d)^2 = 81d^2.
203^2 = (200 + 3)^2, which is 40000 + 1200 + 9 = 41209, and 89^2 = (90 - 1)^2, which is 8100 - 180 + 1 = 7921.
62 \times 58 = (60 + 2)(60 - 2), which is 3600 - 4 = 3596, and 7.1 \times 6.9 = (7 + 0.1)(7 - 0.1), which is 49 - 0.01 = 48.99.
The deck is 12x + 36 square feet. The sandbox and deck together make a square of side x + 6, so their area is x^2 + 12x + 36; take away the sandbox’s x^2. At x = 10: 16^2 - 10^2 = 256 - 100 = 156, and 12 \cdot 10 + 36 = 156.
(a) 289, giving (x + 17)^2: 2b = 34, so b = 17. (b) \displaystyle \frac{121}{4}, giving \displaystyle \left(y - \frac{11}{2}\right)^2: 2b = 11, so \displaystyle b = \frac{11}{2}. (c) 4k^2, giving (2k + 7)^2: b^2 = 49 gives b = 7, and 2 \cdot a \cdot 7 = 28k gives a = 2k.
Expand both squares: (a + b)^2 - (a - b)^2 is (a^2 + 2ab + b^2) minus (a^2 - 2ab + b^2). The a^2 and b^2 cancel and 2ab - (-2ab) = 4ab. With a = n and b = 3: (n + 3)^2 - (n - 3)^2 = 12n.
Call the number between them n. The two numbers are n - 1 and n + 1, a sum and a difference of n and 1, so their product is (n + 1)(n - 1) = n^2 - 1. For 15 and 17, n = 16 and 16^2 - 1 = 255 = 15 \times 17.