We Distribute, Yet Things Multiply

Start with a shopping bill, not with algebra. Suppose you are buying supplies for a class project: 6 notebooks at ₹40 each and 6 pens at ₹15 each. One student works out the two bills separately, 6 × 40 = 240 rupees for the notebooks and 6 × 15 = 90 rupees for the pens, and adds them to get 240 + 90 = 330 rupees. A second student says that one notebook and one pen together cost 40 + 15 = 55 rupees, and there are 6 such pairs, so the bill is 6 × 55 = 330 rupees. Both students are right, and the agreement is not luck. It is a rule of arithmetic called the distributive law.

In symbols the rule says a(b + c) = ab + ac. Read it slowly. On the left you add first and then multiply once. On the right you multiply twice and add at the end. The multiplier a is shared out, or distributed, to both terms inside the bracket, and neither term is allowed to be left out. That is exactly what the title of this chapter is pointing at: you distribute one multiplier, and the number of small products you have to work out multiplies.

The clearest reason why the law is true is a picture. Draw a rectangle whose breadth is a and whose length is b + c. Its area is a(b + c). Now cut it with a single vertical line so that the length splits into a piece of length b and a piece of length c. The two smaller rectangles have areas ab and ac. Cutting a sheet of paper does not change how much paper there is, so the whole must equal the sum of the parts. A picture is made of lengths, so what it settles is the law for positive numbers; the rule is then taken to hold for negative numbers as well, and every check in this chapter agrees with that.

        b      +      c
   +---------+----------+
 a |    ab   |    ac    |
   +---------+----------+
   whole area = a(b + c) = ab + ac

Plain numbers behave the same way, because every number can be split. Take 4 × 13. Since 13 = 10 + 3, the law gives 4 × 13 = 4 × 10 + 4 × 3 = 40 + 12 = 52, which is the answer you already know. Subtraction is covered too, because trimming a strip off the rectangle trims away its area: a(b - c) = ab - ac. So 8 × 19 = 8 × (20 - 1) = 160 - 8 = 152.

One warning before you go any further. Multiplication distributes over addition, but addition does not distribute over multiplication. Test that claim rather than believing it. On one side, 2 + (3 × 4) = 2 + 12 = 14. On the other side, (2 + 3) × (2 + 4) = 5 × 6 = 30. The two answers disagree, so no such law exists. The distributive law runs in one direction only, and you should be able to say which direction without stopping to think.

Once you trust the law, a surprising number of multiplications stop needing paper at all. The method never changes: split one factor into a friendly part and a small correction, multiply twice, then add or subtract.

Take 7 × 98. The number 98 is uncomfortable, but 100 is easy and 98 = 100 - 2. So 7 × 98 = 7 × (100 - 2) = 700 - 14 = 686. Multiplying by 100 is instant, multiplying by 2 is instant, and a single subtraction finishes the job. You could instead have split 98 as 90 + 8, giving 630 + 56 = 686. The answer is the same, which is reassuring, but the arithmetic was harder. Any split is allowed, so choose the one that leaves you the least work.

Numbers just above a round figure are handled the same way, with a plus instead of a minus. For 25 × 104, write 104 = 100 + 4, so 25 × 104 = 25 × 100 + 25 × 4 = 2500 + 100 = 2600.

Money questions are where this pays off. Twelve bus tickets at ₹35 each cost 35 × 12 = 35 × 10 + 35 × 2 = 350 + 70 = 420 rupees. Forty-five chairs at ₹99 each cost 99 × 45 = 45 × (100 - 1) = 4500 - 45 = 4455 rupees. Eighteen sets of stickers at ₹25 each cost 18 × 25 = 25 × (20 - 2) = 500 - 50 = 450 rupees.

Here is something worth noticing. The column method you learned in earlier classes is the distributive law wearing a uniform. When you multiply 47 by 6 in a column, taking the 7 first and then the 4 tens, you are really computing 6 × (40 + 7) = 240 + 42 = 282. The place value split is done for you by the way the digits are arranged.

You may also split both factors. Then every part of the first factor must meet every part of the second, so two pieces times two pieces give four products. For 23 × 14, split 23 into 20 + 3 and 14 into 10 + 4.

          10     +     4
     +--------+---------+
  20 |   200  |    80   |
     +--------+---------+
   3 |    30  |    12   |
     +--------+---------+
   200 + 80 + 30 + 12 = 322

So 23 × 14 = 322. Keep that four cell grid in your mind, because the algebra later in this chapter is the very same grid with letters written in the margins instead of numbers.

Two words first, because they name what you are about to multiply. A monomial is a single term, such as 3x or 2xy or plain 7. A binomial is two terms joined by a plus or a minus sign, such as 2x + 5 or 3a - 7.

Letters change nothing about the rule. In 3x(2x + 5), the monomial 3x sits outside the bracket and must reach both terms inside it. Work through the two products separately, and handle numbers with numbers and letters with letters.

First product: 3x × 2x. The numbers give 3 × 2 = 6, and the letters give x × x = x2, so the product is 6x2. Second product: 3x × 5 = 15x. Putting them together, 3x(2x + 5) = 6x2 + 15x. Those two terms cannot be joined into one, because x2 and x are unlike terms. An expanded answer with two terms is a finished answer.

The reason x × x = x2 is the index rule xm × xn = xm+n: multiplying powers of the same letter adds the indices. So x2 × x3 = x5, and a plain x counts as x1.

The area picture still works, and it is worth drawing once. A rectangle of breadth 3x whose length is split into 2x and 5 has two parts.

         2x        +      5
   +------------+-----------+
3x |    6x²     |    15x    |
   +------------+-----------+
   total = 6x² + 15x

Signs need care. In 4a(3a - 7) the minus travels with the 7, so the answer is 12a2 - 28a. When the monomial itself is negative, every term inside flips sign: -5m(2m + 3) = -10m2 - 15m. Notice that the second term became negative even though it was positive inside the bracket. A negative multiplied by a positive is negative, and there is no exception for algebra.

Two letters are no harder. For 2xy(3x + 4y), the first product is 2xy × 3x = 6x2y and the second is 2xy × 4y = 8xy2, so the expansion is 6x2y + 8xy2.

Always check. Put x = 2 into the original 3x(2x + 5): the monomial becomes 6, the bracket becomes 4 + 5 = 9, and the value is 6 × 9 = 54. Now put x = 2 into the expansion 6x2 + 15x: that is 6 × 4 + 15 × 2 = 24 + 30 = 54. The two values agree, so the expansion has survived a real test. This check takes about ten seconds and catches almost every sign slip.

Now both factors are brackets, as in (x + 3)(x + 5). The move is to treat the whole second bracket as a single object for one step. Distribute it over x + 3, and then distribute again inside each piece.

Step one: (x + 3)(x + 5) = x(x + 5) + 3(x + 5). Step two: x(x + 5) = x2 + 5x and 3(x + 5) = 3x + 15. Step three, collect: x2 + 5x + 3x + 15 = x2 + 8x + 15, because 5x and 3x are like terms.

The same four products appear as the four cells of a two by two grid, which is why the area model is worth keeping.

          x      +      5
    +---------+----------+
  x |   x²    |    5x    |
    +---------+----------+
  3 |   3x    |    15    |
    +---------+----------+
   x² + 5x + 3x + 15 = x² + 8x + 15

Check it with x = 2. The original gives (2 + 3)(2 + 5) = 5 × 7 = 35. The expansion gives 4 + 16 + 15 = 35. Agreed.

Minus signs simply travel with their numbers. For (x + 7)(x - 4), the four products are x2, -4x, +7x and -28, and collecting the middle two gives x2 + 3x - 28. Test at x = 2: the original is 9 × (-2) = -18, and the expansion is 4 + 6 - 28 = -18.

A coefficient in front of a letter changes nothing about the method. For (2a + 3)(a + 5), the products are 2a2, 10a, 3a and 15, giving 2a2 + 13a + 15. At a = 2 the original is 7 × 7 = 49 and the expansion is 8 + 26 + 15 = 49. When both brackets carry minus signs, as in (3p - 2)(2p - 5), the four products are 6p2, -15p, -4p and +10, since a negative times a negative is positive. The answer is 6p2 - 19p + 10. At p = 2 the original is 4 × (-1) = -4 and the expansion is 24 - 38 + 10 = -4.

One shortcut is worth memorising because it appears so often. When each bracket starts with the same letter standing alone, with no number written in front of it, (x + p)(x + q) = x2 + (p + q)x + pq. Add the two numbers for the middle coefficient and multiply them for the last term. For (x + 3)(x + 5) that is 3 + 5 = 8 and 3 × 5 = 15, which matches what you found the long way. That condition about the number in front really matters. In (2a + 3)(a + 5) both brackets do start with the same letter, but the 2 in front of the first a puts the shortcut out of action: adding and multiplying the numbers would give a2 + 8a + 15, while the true answer worked out above is 2a2 + 13a + 15. With a number in front, go back to the four products. Also keep count of those products: two terms times two terms always give four, and three terms times two terms give six. If you write down fewer than four before you start collecting like terms, one has been skipped.

Some products turn up so often that it is worth expanding them once and remembering the shape of the answer. Every one of them comes from the same four cell expansion you have just practised, so none of them has to be learned blindly.

Start with a square. (a + b)2 means (a + b)(a + b). The four products are a2, ab, ab and b2. The two middle terms are like terms, so they combine into 2ab and the result is (a + b)2 = a2 + 2ab + b2. Test it with a = 3 and b = 4: the left side is 7² = 49, and the right side is 9 + 24 + 16 = 49.

That test also settles a very frequent slip. It is tempting to write that (a + b)2 equals a2 + b2, but with a = 3 and b = 4 that would give 9 + 16 = 25, and the true value is 49. The middle term 2ab is not optional, and it is the whole reason squaring a sum is interesting.

With a minus sign, (a - b)2 = a2 - 2ab + b2. The last term is positive because -b × -b = +b2. Test with a = 7 and b = 2: the left side is 5² = 25 and the right side is 49 - 28 + 4 = 25.

The third pattern is the prettiest. (a + b)(a - b) gives the products a2, -ab, +ab and -b2. The middle two cancel exactly, leaving (a + b)(a - b) = a2 - b2. Test with a = 9 and b = 4: the left side is 13 × 5 = 65 and the right side is 81 - 16 = 65.

These three make mental arithmetic even faster. To square 102, write it as 100 + 2, so 102² = 10000 + 2 × 100 × 2 + 4 = 10000 + 400 + 4 = 10404. To square 98, write it as 100 - 2, so 98² = 10000 - 400 + 4 = 9604. For a product of two numbers sitting equally far on either side of a round number, use the third pattern: 53 × 47 = (50 + 3)(50 - 3) = 2500 - 9 = 2491, and 41 × 39 = (40 + 1)(40 - 1) = 1600 - 1 = 1599.

They work with letters too. (x + 4)2 = x2 + 8x + 16, and (2y - 3)2 = 4y2 - 12y + 9, where the first term is 2y × 2y = 4y2 and the middle term is twice the product 2y × 3. That product is 6y, so twice it is 12y, and the minus sign in the bracket makes it -12y. Check the second one at y = 1: the left side is (-1)² = 1 and the right side is 4 - 12 + 9 = 1.

Everything so far has been the law read from left to right, turning one bracket into several terms. Read it from right to left and it becomes just as useful: ma + mb = m(a + b). Spotting a shared multiplier and pulling it out in front is called factorising by taking out a common factor.

Look at 6x + 9. Both terms can be divided by 3, since 6x = 3 × 2x and 9 = 3 × 3. So 6x + 9 = 3(2x + 3). Check by expanding back: 3 × 2x = 6x and 3 × 3 = 9, which returns the expression you started with. Expanding your own answer costs a few seconds and proves that you have not changed the expression. It does not on its own prove that you have taken out everything you could, so after expanding, look at the bracket once more and ask whether its terms still share anything.

To take out as much as possible in one go, find the highest common factor of all the terms. Take the HCF of the number parts, then take each letter that appears in every term, raised to its lowest power. For 12a2 + 18a, the HCF of 12 and 18 is 6, and the letter a appears in both terms with lowest power 1, so the common factor is 6a. Dividing gives 12a2 ÷ 6a = 2a and 18a ÷ 6a = 3, so 12a2 + 18a = 6a(2a + 3).

Notice that 3a(4a + 6) is also equal to the same expression, but it is only half done, because 4a and 6 still share a factor of 2. A fully factorised answer leaves a bracket with nothing left to take out.

Two more of the same kind. For 4a2b + 8ab2, the numbers give HCF 4, the letter a has lowest power 1 and so does b, so the common factor is 4ab. Dividing, 4a2b ÷ 4ab = a and 8ab2 ÷ 4ab = 2b, giving 4ab(a + 2b). For 15x2y - 25xy2, the common factor is 5xy and the answer is 5xy(3x - 5y).

One trap deserves its own line. In 5x + 5, the whole of the second term is the common factor, so what stays behind is 1, not nothing: 5x + 5 = 5(x + 1). Writing 5(x) would lose a whole term, and expanding back would catch it at once.

Sometimes no single factor divides all four terms, yet the expression still factorises in pairs. In 2x + 2y + ax + ay, the first pair gives 2(x + y) and the second pair gives a(x + y). Now (x + y) is itself the common factor, so 2x + 2y + ax + ay = (x + y)(2 + a). This is called factorising by grouping. Another one to try: pq - pr + q - r becomes p(q - r) + 1(q - r), which is (q - r)(p + 1).

Distribution is not a classroom trick. It is how a real situation turns into an expression, and how an expression turns back into a number when you finally know the value of the letter.

Shopping first. If one book costs ₹(3y + 20), then 15 books cost 15(3y + 20) = 45y + 300 rupees. Suppose y turns out to be 12. Using the expression, the cost is 45 × 12 + 300 = 540 + 300 = 840 rupees. Working directly, one book costs 3 × 12 + 20 = 56 rupees and 15 books cost 15 × 56 = 840 rupees. The agreement is the point: the expression is a machine that stores the answer for every possible price.

Now shapes. A rectangle of breadth 3 and length (x + 5) has area 3x + 15 square units. A rectangle of length (x + 2) and breadth (x + 6) has area x2 + 8x + 12 square units. Check at x = 4: the sides are 6 and 10, giving 60, and the expression gives 16 + 32 + 12 = 60.

A borders question uses the square pattern nicely. A square lawn has side x metres, and a path 1 metre wide runs all around it on the outside. The outer square then has side (x + 2) metres, because the path adds 1 metre at each end, so the outer area is x2 + 4x + 4 square metres. The path alone is the outer area minus the lawn, which is 4x + 4 square metres. For a lawn of side 12 metres, the path covers 4 × 12 + 4 = 52 square metres, and you can confirm that by subtracting: 14 × 14 - 12 × 12 = 196 - 144 = 52.

Two sign traps are worth practising until they feel automatic. A minus in front of a bracket changes the sign of everything inside it. So 5 - 2(x - 3) = 5 - 2x + 6 = 11 - 2x, because -2 × -3 = +6. Check at x = 1: the original is 5 - 2(-2) = 9, and the answer is 11 - 2 = 9. Similarly 7 - (x + 4) = 7 - x - 4 = 3 - x, since the invisible multiplier in front of the bracket is -1.

Finally, build the checking habit. After expanding or factorising, substitute a number into the question and into your answer, and see whether the two values match. Avoid 0 and 1 when you can, because 0 wipes out most of the terms and 1 hides wrong coefficients, so a sign error can slip through. Numbers like 2 or 3 test far more of your working. This one habit will do more for your accuracy than any amount of rereading.