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The ideas in this lesson are taken from Multiplication Division 3 ebook. Only a few examples of each problem type are shown; you should make more problems of each kind for the student.

Factoring, part 3
Free lesson plan from HomeschoolMath.net

To find all the prime numbers less than 100 we can use the sieve of Eratosthenes.

1	2	3	4	5	6	7	8	9	10
11	12	13	14	15	16	17	18	19	20
21	22	23	24	25	26	27	28	29	30
31	32	33	34	35	36	37	38	39	40
41	42	43	44	45	46	47	48	49	50
51	52	53	54	55	56	57	58	59	60
61	62	63	64	65	66	67	68	69	70
71	72	73	74	75	76	77	78	79	80
81	82	83	84	85	86	87	88	89	90
91	92	93	94	95	96	97	98	99	100

Cross out every other number starting at 2.
Cross out every third number starting at 3.
(You don't have to check every fourth number. Why?)
Cross out every fifth number starting at 5
(You don't have to check every sixth number. Why?)
Cross out every seventh number starting at 7
(You don't have to check every eighth number. Why?)
(You don't have to check every ninth number. Why?)
(You don't have to check every tenth number. Why?)

The explanation why you don't have to check
numbers bigger than 10 is beyond this text.

The numbers that are not crossed out are primes.

List here all primes between 0 and 30:

____________________________________________________________

We have found how to know if a number is divisible by 2, 5, 10, 3, or 9.
We can use that in factoring.

	260
	/ \
26	×	10

	260
/ \
26	×	10
/ \		/ \
2 × __	×	2 × 5



2	2	6

We know 260 is divisible by 10 and the division is easy to do.

26 is an even number, so it is divisible by 2. To find out the missing divisor, complete the long division on the right.

	135
	/ \
5	×	?

			2	7

5		1	3	5
	-	1	0
			3	5
		-	3	5
				0

135
/ \
5	×	27
\|		/ \
5	×	3	× 9
\|		\|	/ \
5	×	3	× 3 × 3

We know 135 is divisible by 5 (why?).

Dividing 135 by 5 we find out that 135 = 5 × 27.

Here is the complete factorization to prime numbers.

Example problems

1. Factor the following numbers so the factors are prime numbers. Do any needed long divisions in your notebook.

90
/ \

165
/ \

95
/ \

We can use divisibility rules in factoring bigger numbers, too. Look at the examples.

5850

/ \

10 × 585

5850

/ \

10 × 585

/ \ / \

5 × 2 × 5 × ?

5850

/ \

10 × 585

/ \ / \

5 × 2 × 5 × 117

5850 ends in 0, so it is
divisible by 10.

585 is divisible by 5
since it ends in 5.
10 we know is 5 × 2.

Dividing 585 by 5 we find
that 585 = 5 × 117.

5850

/ \

10 × 585

/ \ / \

5 × 2 × 5 × 117

| | | / \

5 × 2 × 5 × 9 × ?

5850

/ \

10 × 585

/ \ / \

5 × 2 × 5 × 117

/ / / / \

5 × 2 × 5 × 9 × 13

5850

/ \

10 × 585

/ \ / \

5 × 2 × 5 × 117

/ / / / \

5 × 2 × 5 × 9 × 13

/ / / / \ \

5 × 2 × 5 × 3 × 3 × 13

117 is divisible by 9
since the sum of its
digits is 1 + 1 + 7 = 9.

The division tells us
that 265 = 5 × 53.

2. Circle or underline the number that the following numbers are divisible by. Use divisibility rules. Then pick the GREATEST divisor of those, do the division, and express the number as a product. Do the long divisions in your notebook. Follow the example given.

Number	Divisible by (circle the numbers)	Division	Expression as a product
260	10 5 3 2	260 10 = 26	260 = 10 × 26
96	10 5 3 2	96 __ = ___	96 = ___ × __
95	10 5 3 2	95 __ = ___	95 = ___ × __
132	10 5 3 2	132 __ = ___	132 = ___ × __

3. Use the previous exercise and factor these numbers. Continue factoring to prime factors.

124

/ \

2 ×

/ / \

2 × ×

260

/ \

10 ×

/ \ / \

/ \

3 ×

/ / \

4. Factor the following numbers in your notebook down to prime factors. You can start factoring by any number that you notice it is divisible by! Try first if the number is divisible by 10, then by 5, then by 2, then by 3 or 9.

306 =
990 =
945 =

Many times primes occur in pairs so that their difference is only two. For example, 5 and 7 are such a pair, so are 11 and 13.

Find all such pairs that are less than 100. Use the sieve of Eratosthenes.
How many such pairs are there?

Factoring, part 3 Free lesson plan from HomeschoolMath.net

Example problems

Factoring, part 3
Free lesson plan from HomeschoolMath.net