Chapter 4 — Exploring Algebraic Identities

An algebraic expression combines numbers and letters with the signs of operations, such as 3x + 5 or a² − 2ab + b². An equation is a statement that two expressions are equal for some particular values, for example x + 3 = 7 is true only when x = 4. An identity is different. It is an equality that is true for every value of the letters in it. For example, 2(x + 3) = 2x + 6 is true whatever x is, so it is an identity.

The four identities that we use most often are the following:

Why they hold. Expanding the left side by multiplying every term of one bracket by every term of the other proves each one. For Identity I: (a + b)(a + b) = a·a + a·b + b·a + b·b = a² + 2ab + b². For Identity III: (a + b)(a − b) = a² − ab + ba − b² = a² − b², because the two middle terms cancel.

Testing with numbers. A very good habit is to test an identity with numbers. Put a = 5 and b = 2 in Identity I. The left side is (5 + 2)² = 49, and the right side is 25 + 20 + 4 = 49. Put a = 5, b = 2 in Identity III: (7)(3) = 21 and 25 − 4 = 21. If the two sides differ for even one choice of numbers, the claim is not an identity. But remember that a test with numbers can only show an identity is wrong; to show that it is right for all values, you need the algebraic proof.

Using Identity IV. To multiply (x + 3)(x + 5), notice that a = 3 and b = 5. The product is x² + (3 + 5)x + 15 = x² + 8x + 15. For (x − 4)(x + 7), a = −4 and b = 7, so the result is x² + 3x − 28. Keep the signs with the numbers, because this is the most common place to make a mistake.

Where we are going. In this chapter we will see these identities as pictures, use them backwards to factorise expressions, discover new identities, and use them to simplify fractions made from algebraic expressions. The important idea to carry forward is that an identity is a tool: you may use it in either direction, from the product to the expanded form and from the expanded form back to the product.

An algebraic identity can be seen as a statement about areas. Let us picture Identity I, (a + b)² = a² + 2ab + b².

Draw a square whose side is a + b. Its area is (a + b) × (a + b) = (a + b)². Now mark a point on each side so that the side is split into a part of length a and a part of length b. Draw lines across the square through these points, parallel to the sides. The big square is cut into four pieces:

• a square of side a, with area a²,
• a square of side b, with area b²,
• two rectangles, each a by b, with area ab each, which together make 2ab.

The pieces fill the big square exactly, so the total area is a² + 2ab + b². Since this is the same area as (a + b)², we get (a + b)² = a² + 2ab + b². The picture explains why the middle term is 2ab: there are two rectangles of area ab.

A number check. Take a = 5 and b = 2. The big square has side 7 and area 49. The pieces are a square of area 25, a square of area 4, and two rectangles of area 10 each, and 25 + 4 + 10 + 10 = 49. The picture and the algebra agree.

A shortcut for squares. The picture gives a way of squaring numbers mentally. To find 103², write 103 = 100 + 3. Then 103² = 100² + 2 × 100 × 3 + 3² = 10 000 + 600 + 9 = 10 609. You can see in the picture that the big piece is 100² and the other pieces add a little more. For 52², write 50 + 2, giving 2500 + 200 + 4 = 2704.

Making the model yourself. Use graph paper, or fold and cut coloured paper. Take a square of side 7 units, cut off an L-shape, and arrange the pieces to show 5², 2² and two 5 × 2 rectangles. Handling the pieces makes the identity memorable.

A different look. The same square can be used with algebra tiles later. If we let a be a letter x and b be the number 3, then (x + 3)² is a square with a big x-by-x piece, two strips 3 by x, and a small 3 by 3 square, giving x² + 6x + 9. The identity says how the pieces of a square fit.

Now let us picture Identity II, (a − b)² = a² − 2ab + b². Start with a square of side a, so its area is a². Inside it, cut away a strip of width b from one side and another strip of width b from the next side. What remains in the corner is a square of side (a − b).

To count the area of this smaller square, begin with a². Remove the two strips, each with area a × b, so we take away 2ab. But the small b-by-b square in the corner was part of both strips, so we have removed it twice. We must give it back once by adding b². The area of the corner square is therefore a² − 2ab + b². Since this square has side a − b, its area is (a − b)², so (a − b)² = a² − 2ab + b².

A number check. Take a = 7 and b = 2. The square of side 5 has area 25. The formula gives 49 − 28 + 4 = 25. ✓.

The difference of two squares. Identity III says (a + b)(a − b) = a² − b². Picture a large square of side a, and cut a smaller square of side b out of one corner. The shape that is left, an L, has area a² − b². Cut the L along the dotted line and rearrange the pieces:

• One piece is a rectangle of length a and width (a − b).
• The other piece is a rectangle of length (a − b) and width b.

Slide the second rectangle next to the first, rotated so that the sides of length (a − b) line up. Together they form one large rectangle with sides (a + b) and (a − b). Since the area has not changed, a² − b² = (a + b)(a − b).

Using the identity. This is a powerful shortcut. To compute 51 × 49, notice that these are 50 + 1 and 50 − 1. So 51 × 49 = 50² − 1² = 2500 − 1 = 2499. To compute 97 × 103, write 100 − 3 and 100 + 3: the product is 100² − 3² = 10 000 − 9 = 9991. And 998² can be found as (1000 − 2)² = 1 000 000 − 4000 + 4 = 996 004.

Try to see it. With a = 5 and b = 3, a 5 × 5 square with a 3 × 3 corner removed has area 25 − 9 = 16. The cut-and-slide gives a rectangle 8 by 2, whose area is also 16. Experiments with squares of paper show it clearly.

Factorising means writing an expression as a product of simpler expressions. It is the opposite of expanding. Each identity can be read from right to left to give a factorisation. This is the most common use of identities in algebra, so we look at each in turn.

Perfect squares (Identities I and II). If an expression has the form a² + 2ab + b² or a² − 2ab + b², it is a perfect square. To recognise it, check that the first and last terms are perfect squares and that the middle term is twice the product of their square roots.

Example: x² + 6x + 9. The first term is x², the last is 3², and the middle term is 2 × x × 3 = 6x. So x² + 6x + 9 = (x + 3)². Another: 4y² − 12y + 9. Here a = 2y, b = 3 and 2ab = 2 × 2y × 3 = 12y, with a minus sign. So 4y² − 12y + 9 = (2y − 3)².

Difference of two squares (Identity III). Any expression of the form a² − b² factorises as (a + b)(a − b). Example: 4x² − 25 = (2x)² − 5² = (2x + 5)(2x − 5). Another: 49a² − 16b² = (7a + 4b)(7a − 4b). Sometimes the expression must be rewritten first. 18x² − 8 has a common factor of 2: 2(9x² − 4) = 2(3x + 2)(3x − 2).

First step: look for a common factor. Before using any identity, always take out any common factor. For 3x² − 12 the common factor is 3, which gives 3(x² − 4) = 3(x + 2)(x − 2). If you miss it, your factorisation will be incomplete.

Repeated use. Sometimes one identity can be applied twice. x⁴ − 16 = (x²)² − 4² = (x² + 4)(x² − 4) = (x² + 4)(x + 2)(x − 2). The factor x² + 4 cannot be factorised further with real numbers, so we stop there.

Checking. You can always check a factorisation by multiplying the factors back. For (2x + 5)(2x − 5), the product is 4x² − 10x + 10x − 25 = 4x² − 25, which is the original expression. A good habit is to do this check every time.

A factorisation is complete when none of the factors can be factorised further. The factorised form is also useful for finding the values of x that make an expression zero, since a product is zero only if one of its factors is zero.

Algebra tiles are shapes that stand for terms of an expression, so that factorising becomes a matter of arranging tiles into a rectangle. There are three kinds:

• a large square with sides x and x, whose area is x²,
• a long rectangle with sides x and 1, whose area is x,
• a small square with sides 1 and 1, whose area is 1.

The idea is that an expression such as x² + 5x + 6 can be represented by one large square, five long rectangles and six small squares. The area of the whole collection is x² + 5x + 6. If we can arrange all these tiles into one rectangle without gaps or overlaps, then the length and width of that rectangle are the factors of the expression, because area = length × width.

Example: x² + 5x + 6. Place the large square in a corner. Then try to arrange the 5 long rectangles and 6 small squares around it. Put 2 long rectangles below it and 3 long rectangles to the right. This leaves a corner to fill with small squares, a rectangle of 2 rows and 3 columns, which takes exactly 6 small squares. We have built a rectangle whose width is x + 3 and height is x + 2. So x² + 5x + 6 = (x + 2)(x + 3).

Why it works. The rectangle has a total area made of four parts: the large square (x²), a strip of 3 long tiles (3x), another strip of 2 long tiles (2x), and a block of small squares (2 × 3 = 6). Adding gives x² + 3x + 2x + 6 = x² + 5x + 6. This is exactly Identity IV with a = 2 and b = 3.

What the numbers mean. The number of long tiles is 5, the sum of the two numbers 2 and 3. The number of small squares is 6, their product. So to factorise x² + (a + b)x + ab, we look for two numbers a and b that multiply to the constant term and add up to the middle coefficient.

Another example: x² + 7x + 12. We need two numbers with product 12 and sum 7. The numbers 3 and 4 work. With 7 long tiles and 12 small squares, the arrangement is a rectangle with sides (x + 3) and (x + 4). So x² + 7x + 12 = (x + 3)(x + 4).

When the tiles will not fit. Try to build x² + 2x + 2. It needs two numbers with product 2 and sum 2. The pairs of whole numbers with product 2 are 1 and 2, whose sum is 3, so no pair works. The tiles cannot be arranged into a rectangle, which means the expression has no factorisation with whole numbers.

Algebra tiles can be made from cardboard of two colours, one for positive and one for negative terms. Even if you do not have tiles, you can sketch rectangles on paper.

Tiles help us see why factorisation works, but we also need a method that works quickly on paper and for expressions with negative numbers. The method is called splitting the middle term.

When the leading coefficient is 1. For x² + bx + c, find two numbers p and q such that p × q = c and p + q = b. Then x² + bx + c = (x + p)(x + q). Look at the signs:

• If c is positive, p and q have the same sign (both positive when b is positive, both negative when b is negative).
• If c is negative, p and q have opposite signs, and the one with the larger size has the sign of b.

Example: x² − 7x + 12. The product is +12 and the sum is −7, so both numbers are negative: −3 and −4. Then x² − 7x + 12 = (x − 3)(x − 4). Example: x² + x − 12. The product is −12 and the sum is +1, so the numbers are +4 and −3. Then x² + x − 12 = (x + 4)(x − 3).

When the leading coefficient is not 1. For ax² + bx + c, find two numbers p and q such that p × q = a × c and p + q = b. Split the middle term bx into px + qx, and then group.

Example: 2x² + 7x + 3. Here a × c = 6 and b = 7. The numbers 6 and 1 have product 6 and sum 7. So 2x² + 7x + 3 = 2x² + 6x + x + 3. Group the terms: 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3). Check by multiplying: 2x² + 6x + x + 3 = 2x² + 7x + 3. ✓.

Example: 6x² − 5x − 6. Here a × c = −36 and b = −5. We need two numbers with product −36 and sum −5. These are −9 and +4. So 6x² − 9x + 4x − 6 = 3x(2x − 3) + 2(2x − 3) = (3x + 2)(2x − 3). Check: 6x² − 9x + 4x − 6 = 6x² − 5x − 6. ✓.

Finding the numbers. List the factor pairs of a × c, and see which pair has the required sum. For −36, the pairs are (−1, 36), (1, −36), (−2, 18), (2, −18), (−3, 12), (3, −12), (−4, 9), (4, −9), (−6, 6), (6, −6). The pair (4, −9) has sum −5.

Always check. Multiply the factors back to be sure of the result. If the expression has a common factor, take it out first, so that the numbers are smaller. For example, 6x² + 12x + 6 = 6(x² + 2x + 1) = 6(x + 1)².

We now extend the identities to more terms and higher powers. Each can be proved by multiplying out, and each is worth testing with numbers.

The square of a trinomial. (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca. To see this, treat (a + b) as one quantity: (a + b + c)² = ((a + b) + c)² = (a + b)² + 2(a + b)c + c² = a² + 2ab + b² + 2ac + 2bc + c². Test with a = 1, b = 2, c = 3: the left side is 6² = 36, and the right side is 1 + 4 + 9 + 4 + 12 + 6 = 36. ✓.

Cubes of a binomial.

• (a + b)³ = a³ + 3a²b + 3ab² + b³
• (a − b)³ = a³ − 3a²b + 3ab² − b³

Test with a = 3 and b = 1: (3 + 1)³ = 64, and 27 + 27 + 9 + 1 = 64. ✓. With a = 3 and b = 1 for the second: (3 − 1)³ = 8, and 27 − 27 + 9 − 1 = 8. ✓. The coefficients 1, 3, 3, 1 come from a pattern that you will meet again in Pascal’s triangle.

Sum and difference of cubes.

• a³ + b³ = (a + b)(a² − ab + b²)
• a³ − b³ = (a − b)(a² + ab + b²)

Multiply (a + b)(a² − ab + b²) to check: a³ − a²b + ab² + a²b − ab² + b³ = a³ + b³, because the middle terms cancel. Test with a = 2 and b = 1: 8 + 1 = 9, and (3)(4 − 2 + 1) = 3 × 3 = 9. ✓. For the difference: 8 − 1 = 7, and (1)(4 + 2 + 1) = 7. ✓.

Using them to factorise. x³ + 27 = x³ + 3³ = (x + 3)(x² − 3x + 9). And 8y³ − 1 = (2y)³ − 1³ = (2y − 1)(4y² + 2y + 1).

Using them to expand. (2x + 3)³ = (2x)³ + 3(2x)²(3) + 3(2x)(3)² + 3³ = 8x³ + 36x² + 54x + 27.

Watch the signs. In a³ + b³, the second factor has a minus in the middle: a² − ab + b². In a³ − b³, it has a plus: a² + ab + b². A helpful memory is that the sign of the first factor matches the sign of the cube expression, and the middle sign of the second factor is opposite.

Identities save work. They let us find values without carrying out long multiplications, and they let us find some quantities even when we do not know the numbers individually. Here are the main uses.

1. Quick squares and products. 105² = (100 + 5)² = 10 000 + 1000 + 25 = 11 025. 996² = (1000 − 4)² = 1 000 000 − 8000 + 16 = 992 016. 72 × 68 = (70 + 2)(70 − 2) = 4900 − 4 = 4896. 63² − 37² = (63 + 37)(63 − 37) = 100 × 26 = 2600.

2. Expressions in terms of a + b and ab. If a + b = 7 and ab = 10, we can find a² + b² without finding a and b. From (a + b)² = a² + b² + 2ab, we get a² + b² = (a + b)² − 2ab = 49 − 20 = 29. We can also find (a − b)² = (a + b)² − 4ab = 49 − 40 = 9, so a − b = 3 or −3. And a³ + b³ = (a + b)³ − 3ab(a + b) = 343 − 210 = 133. Check with the numbers a = 5 and b = 2: 25 + 4 = 29 ✓, and 125 + 8 = 133 ✓.

3. Finding a sum from a square. If x + 1/x = 3, then squaring both sides, x² + 2 + 1/x² = 9, so x² + 1/x² = 7. This kind of question appears often and uses Identity I.

4. Evaluating expressions at given values. To find the value of x² + 10x + 25 at x = 95, first notice that it equals (x + 5)². At x = 95 this is 100² = 10 000, which is much easier than calculating 95² + 950 + 25 directly.

5. Proving a statement. Show that the product of two consecutive even numbers plus 1 is always a perfect square. Let the numbers be 2n and 2n + 2. Their product plus 1 is 2n(2n + 2) + 1 = 4n² + 4n + 1 = (2n + 1)². So it is always the square of an odd number. For 4 × 6 + 1 = 25 = 5².

6. Geometry and area problems. A square lawn has a side of 20 m with a path of width 1 m around it. The area of the path is (22)² − (20)² = (22 + 20)(22 − 20) = 42 × 2 = 84 m². The identity avoids squaring large numbers.

The skill to build is choosing the right identity. Look at the numbers. Are they near a round number? Try (a + b)² or (a + b)(a − b). Do you know a sum and a product? Try the square identity in reverse. Practising this selection is more valuable than memorising a long list.

Mathematicians do not only use identities; they discover new ones. The method is simple: combine known identities, look for patterns, test with numbers, and then prove.

Discovery 1: adding and subtracting squares. We know (a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b². Add them: (a + b)² + (a − b)² = 2a² + 2b² = 2(a² + b²). Subtract them: (a + b)² − (a − b)² = 4ab. These are new identities. Test with a = 5, b = 2: 49 + 9 = 58 = 2(25 + 4) ✓ and 49 − 9 = 40 = 4 × 10 ✓.

Use of Discovery 1. The second identity gives a way to find products. (a + b)² − (a − b)² = 4ab means ab = [(a + b)² − (a − b)²] ÷ 4. For 25 × 24, take a = 25 and b = 24: (49² − 1²) ÷ 4 = (2401 − 1) ÷ 4 = 600. ✓.

Discovery 2: a product of two binomials. Multiply (a + b)(c + d) = ac + ad + bc + bd. When c = a and d = −b, we get back the difference of squares. When d = a and c = b, we get (a + b)² again. So one rule gives many special cases.

Discovery 3: a pattern in numbers. Look at 1 × 3 + 1 = 4 = 2², 2 × 4 + 1 = 9 = 3², 3 × 5 + 1 = 16 = 4², 4 × 6 + 1 = 25 = 5². The pattern suggests n(n + 2) + 1 = (n + 1)². To prove it, expand: n² + 2n + 1 = (n + 1)². It is true for every n, so we have found an identity.

Discovery 4: from cubes. From (a + b)³ = a³ + 3a²b + 3ab² + b³, rewrite the middle terms as 3ab(a + b). Then a³ + b³ = (a + b)³ − 3ab(a + b). This is the identity we used earlier for finding a³ + b³ from a + b and ab.

The method. Notice the steps that we followed in each discovery: look at special cases, guess a general statement, test it with numbers, and prove it by algebra. A guess is not a proof. For instance, 6 × 6 − 5 × 7 = 1, and 7 × 7 − 6 × 8 = 1; this suggests n² − (n − 1)(n + 1) = 1, which can be proved by expanding. Every identity you meet was once a guess that someone proved.

You can make your own: choose a pattern in numbers, write it in letters, simplify, and see whether the two sides match.

A rational expression is a fraction in which the numerator and the denominator are algebraic expressions, such as (x² − 9) ÷ (x + 3). Just like a fraction of numbers, a rational expression is simplified by cancelling the common factors of the numerator and the denominator. The key step is to factorise first, and this is where identities are useful.

Example 1. Simplify (x² − 9) ÷ (x + 3). Factorise the numerator using a² − b²: x² − 9 = (x + 3)(x − 3). So the expression is (x + 3)(x − 3) ÷ (x + 3). Cancel the common factor (x + 3), leaving x − 3.

Example 2. Simplify (x² + 5x + 6) ÷ (x² − 4). The numerator is (x + 2)(x + 3), and the denominator is (x + 2)(x − 2). Cancelling (x + 2) leaves (x + 3) ÷ (x − 2).

Example 3. Simplify (4a² − 12a + 9) ÷ (2a − 3). The numerator is a perfect square, (2a − 3)². So the expression is (2a − 3)² ÷ (2a − 3) = 2a − 3.

A careful point about restrictions. A fraction is not defined when its denominator is zero. In Example 1, the original expression is not defined at x = −3, because then x + 3 = 0. After cancelling we get x − 3, which looks fine at x = −3, but the original expression still has no value there. So we write: (x² − 9) ÷ (x + 3) = x − 3, for x ≠ −3. In Example 2, the restrictions are x ≠ 2 and x ≠ −2, since the original denominator is (x + 2)(x − 2).

Checking by numbers. Put x = 5 in Example 1. The original gives (25 − 9) ÷ 8 = 16 ÷ 8 = 2, and the simplified form gives 5 − 3 = 2. ✓. In Example 2, at x = 3 we have (9 + 15 + 6) ÷ (9 − 4) = 30 ÷ 5 = 6, and the simplified form (3 + 3) ÷ (3 − 2) = 6 ÷ 1 = 6. ✓.

What you must not do. You can cancel only factors, which are multiplied, and not terms, which are added. In (x + 3) ÷ 3 you cannot cancel the 3, because the 3 in the numerator is a term, not a factor of the whole. But in 3(x + 3) ÷ 3 you can, and the answer is x + 3.

The steps are therefore: factorise numerator and denominator completely, cancel common factors, state any restrictions, and check with a number. This ties together everything we have learnt in this chapter.