Two products that share a factor
Look at 4 × 3 + 4 × 4. Both products have a factor of 4. Draw them as rectangles: 4 rows of 3 squares, and 4 rows of 4 squares. Both rectangles are 4 squares tall, so they fit side by side into one rectangle 4 tall and 3 + 4 = 7 wide.
So 4 × 3 + 4 × 4 = 4 × (3 + 4) = 4 × 7 = 28. This is expanding a bracket read backward: instead of multiplying the 4 into the bracket, we take the shared 4 out of both products.
The colored 4 by 3 rectangle and the white 4 by 4 rectangle share the side 4, so together they are 4 by 7.
Take the common factor outside
The same works with letters. In 4a + 4b, both terms have a factor of 4. Take the 4 outside a bracket, and what is left of each term goes inside: 4a + 4b = 4(a + b).
Writing a sum as a product like this is called factoring. It undoes expanding, so you can always check a factorization by expanding it again: 4(a + b) = 4a + 4b.
The pieces 4a and 4b share the side 4. The other sides, a and b, make the length of the whole rectangle, a + b.
Use the highest common factor
Factor 6x + 9. The highest common factor of 6 and 9 is 3. Write each term as 3 times something: 6x = 3 × 2x and 9 = 3 × 3. Take out the 3: 6x + 9 = 3(2x + 3).
Check by expanding: 3 × 2x + 3 × 3 = 6x + 9. Inside the bracket, 2x and 3 have no common factor left, so the factoring is complete.
6x and 9 share the side 3, and the other sides are 2x and 3.
Letters can be common factors too
Factor . The numbers 12 and 8 have the highest common factor 4, and both terms contain x, so the common factor is 4x. Write each term as 4x times something: and 8x = 4x × 2. So .
Check with a number. When x = 1, is 12 + 8 = 20, and 4x(3x + 2) is 4 × 5 = 20.
Taking out only 4 gives . That is correct, but it is not complete, because and 2x still share a factor of x.
and 8x share the side 4x, and the other sides are 3x and 2.
The usual mistakes
Leaving the factor inside. 8x + 12 is not 4(2x + 12). Divide every term by the 4 you take out: 12 ÷ 4 = 3, so 8x + 12 = 4(2x + 3).
Taking out a number that does not divide every term. 8x + 12 is not 8(x + 12), because 8 does not divide 12. Expanding 8(x + 12) gives 8x + 96.
Worked example: An Order Totaled Quickly by Taking Out a Common Factor
Question A school orders 37 sets. Each set is a calculator at $46 and a case of instruments at $54. (a) Without a calculator, find the total cost of the 37 sets. (b) The delivery bill for n boxes is (24n + 36) dollars, and 12 classes share it equally. Write one class's share as an expression in n, and find the share when n = 10.
1.The calculators cost 37 × 46 dollars and the cases cost 37 × 54 dollars, so the total is 37 × 46 + 37 × 54.
The total cost is 37 × 46 + 37 × 54 dollars. 2.Both products have the factor 37. Take it outside a bracket: 37 × 46 + 37 × 54 = 37(46 + 54).
Both products have the factor 37: 37 × 46 + 37 × 54 = 37(46 + 54). 3.(a) 46 + 54 = 100, so the total cost is 37 × 100 = $3700.
(a) 46 + 54 = 100, so the total cost is 37 × 100 = $3700. 4.The highest common factor of 24 and 36 is 12, so 24n + 36 = 12 × 2n + 12 × 3 = 12(2n + 3).
The highest common factor of 24 and 36 is 12: 24n + 36 = 12(2n + 3). 5.(b) The bill is 12 equal shares of (2n + 3) dollars, so one class pays (2n + 3) dollars. When n = 10 the share is 2 × 10 + 3 = $23. Check: the bill is 24 × 10 + 36 = $276, and 12 × 23 = 276.
(b) One class pays (2n + 3) dollars, which is $23 when n = 10.
Answer: (a) $3700; (b) (2n + 3) dollars, which is $23 when n = 10
Common mistakes
- Writing 37 × 46 + 37 × 54 as 37 × 37 × (46 + 54). The common factor is taken out once, because each of the 37 sets costs 46 + 54 dollars.
- Taking out 6 and stopping at 6(4n + 6). The factorization is correct but not complete, and it shows 6 equal shares, not the 12 shares that the classes need.