Download A factor is a whole number that can be multiplied by another whole

Factors
A factor is a whole number that can be multiplied by another whole
number to produce a particular product. A product is any number that
results from multiplying two or more factors. Since a product is the result of
multiplying two factors together, that means any whole number can be a
product, because at the very least you can generate any number n by
multiplying 1 times that number n itself. For instance the number 3 is the
product of 1 times 3. So the numbers 1 and 3 are both factors of the product
3. The number 2 has to be multiplied by 1.5 to produce the product 3, but
1.5 is not a whole number, so 2 is not a factor of 3. No number larger than a
particular product can ever be a factor of that product, because a number
larger than 3 would have to be multiplied by a value less than 1 to produce
the product 3.
Another way to think of this is that any factor of a number will divide
that number evenly, with no remainder. Thus, for the product 3, the only
factors are 1 and 3.
As another example, the number 6 is the product of 1 times 6. Thus,
the numbers 1 and 6 are both factors of the number 6. In this case, though,
you can also produce the product 6 by multiplying 2 times 3, so both 2 and
3 are also factors of 6. No other whole numbers can be multiplied to produce
the product 6-- if you check to see whether 4 is a factor, you see that you
have to multiply 4 by 1.5, which is not a whole number, so 4 is not a factor
of 6. Similarly, the number 5 would have to be multiplied by 1.2 to produce
the product 6, so 5 is not a factor of 6 either. So for the product 6, we can
say that the factors are 1, 2, 3, and 6.
Determining the factors for a product is not an exact procedure. Instead, it's
more like a process of elimination. For small numbers, like we showed above,
the process is simple. For larger numbers, it's usually fairly easy to find some
factors, but finding all of the factors and knowing that you found all of the
factors can be more difficult. Basically, the process consists of finding some
factors of the product, and breaking down those factors themselves, until you
have a list of factors that cannot be broken down any further. The most
complete approach would be to start with the number 1 and divide the
product systematically by each increasing number to determine which
numbers can be divided into the product evenly (with no remainder). When
you reach a point where the factors begin to repeat, you know there are no
further factors. For example, consider the product 12. We know that 1 is a
factor, because 1 times 12 equals 12. Next we see if 2 is a factor. Since 2 with
a quotient of 6 can divide 12 evenly, then 2 and 6 are also factors. Now we
try 3, and since 3 times 4 equals 12, both 3 and 4 are factors. The next
number in the progression to try is 4, but we have already determined that 4
is a factor-- that is, the factors are beginning to repeat. That means all of the
factors have been found, and it's not necessary to continue trying 5 (which is
not a factor), 6 (which we already determined was a factor when we checked
2), etc. We can say that the factors of 12 are 1, 2, 3, 4, 6, and 12.
Practice Question: List the factors of
A) 36
B) 54
C) 81
Least Common Multiple
Before defining the least common multiple, let's review how we define
multiples. For any number n, a multiple of n is a product obtained when n
is multiplied by any whole number. For instance, if we start with the number
7, then 28 is a multiple of 7 because 7 x 4 = 28. Another multiple of 7 is
42, because 7 x 6 = 42. The number 39 would not be a multiple of 7, as
there is no whole number by which 7 can be multiplied to yield the product
39. However, 39 is a multiple of 3, since 3 x 13 = 39.
Products that are multiples of more than one number are common
multiples for all of their factors. If you think about it, any number that is
not a prime number has to be a common multiple for however many factors
it has. Consider the number 30. We can show that the factors for 30 are 1,
2, 3, 5, 6, 10, 15, and 30. 30 is a multiple of each of these factors (we
usually don't bother including 1 in the list of multiples), so we say that 30 is
a common multiple for the numbers 2, 3, 5, 6, 10, 15, and 30.
Suppose we want to determine the common multiples for two
numbers. We first make out a list of multiples for each number, listing a few
in each case. Then we compare the lists and see if any numbers appear on
both lists. Any number that does appear on both lists is a common multiple
for those two numbers. For example, determine some common multiples for
the numbers 4 and 6. The beginnings of the lists of multiples for each
number look like this:
4
4
8
12
16
20
24
28
32
36
6
12
18
24
30
36
42
48
54
60
To determine some common multiples, simply look at which multiples
appear in both lists. In this case, we can see that 12, 24, and 36 are common
multiples of the numbers 4 and 6. You can never list all of the common
multiples for any two or more numbers, as there are an infinite number of
common multiples, just like each number has an infinite number of
multiples.
Now we finally come to the title topic of this section. The least
common multiple for two or more numbers is the smallest number in the
list of common multiples for the numbers. Once you already have a list of
common multiples generated like we did above, the least common multiple
can be determined by just looking at the list. In the example of the numbers
4 and 6 that we did earlier, the least common multiple of those two numbers
is 12. You can determine the least common multiple for any two or more
numbers by making a list of multiples for each, and seeing what the smallest
number is that shows up on all of the lists-- that number will be the least
common multiple.
Practice Question: Find the least common multiples
A) 48 and 16 B) 63 and 42 C) 55 and 20
Greatest Common Factor
A common Factor is a number that is a factor for two or more products.
We can show that the factors for the number 18 are 1, 2, 3, 6, 9, and 18. The
factors for 32 are 1, 2, 4, 8, 16, and 32. If we compare these lists of factors, we
see that the number 2 appears on both lists. Thus, 2 is a common factor for the
numbers 18 and 32. Since we don't bother listing 1 as a common factor for
numbers (because 1 is a factor for all numbers), for 18 and 32, the only common
factor is 2.
Lets look at another pair of numbers, say 30 and 48. As an exercise,
you can verify that these are the factors:
The factors for 30 are 1, 2, 3, 5, 6, 10, 15, and 30.
The factors for 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48.
Excluding 1 again, the numbers that show up on both factor lists are 2, 3,
and 6.
The common factors for 30 and 48 are 2, 3, and 6.
As you might expect, the greatest common factor is defined as the
largest common factor for two or more numbers. Once the common factors
have been determined, all you need to do is look at the list and see what the
largest number is. For 30 and 48, the largest number on the common
factors list is 6.
The greatest common factor for 30 and 48 is 6.
When you have an equation like -13ab+39ac find the greatest
common factor, which would be 13a and bring it out of the equation. So you
are now left with the equation in simplest form looking like this 13a(b+3c).
Once you have reached this stage you are finished.
Practice Question: Find the greatest common factor
A) 42x y –63xy B) 15a + 21b
Factoring Binomials
To multiply a binomial (a polynomial with two terms like, (x + 3)
by a binomial use FOIL:
First: Multiply the first terms of each binomial.
Outside: Multiply the outside terms (the outer extreme terms of
the expression).
Inside: Multiply the inside terms (the inner extreme terms of the
expression).
Last: Multiply the last terms of each binomial.
Then add them making sure to combine like terms.
Example 1
(x + 3)(x - 5) =
Solution
F: x × x = x
O: -5 × x = -5x
I : 3 × x = 3x
L: +3 × -5 = -15
Therefore, the answer is the sum of F + O + I + L:
x - 5x + 3x - 15
then combine (-5x and +3x)
x - 2x – 15
Example 2
(2a + b)(a - 2b) = 2a -4ab + ab - 2b
=2a - 3ab - 2b
Practice Question: Factor the binomials
A) (x+3) (x+5) B) (y+4) (y+2) C) (a+6)(a+5)
Trinomials
When factoring out trinomials you can put them into two separate
brackets. If you had an equation such as x²+8x+16 you have to remember
that the first term in each bracket must multiply to add up to x². So you would
put in your brackets (x+_) (x+_). The next part is a little trickier. You must
find terms that add up to get 8x and multiply them together to get 16. You
already have the two x’s in each bracket to use.
An easy way to factor out trinomials is to list all the factors that multiply together
to give you your x, and list all the factors that multiply together to give you 16.
Then multiply all the possible combinations together to find which ones add up to
give you 8x.
Examples: x²+10x+25
X²= X*X
25= 1x25, 5x5
(x+1) (x+25) =
x²+26x=25 Wrong answer
(x+5) (x+5)=
~~x²+10x+25~~
So the right numbers to use are (x+5) (x+5)
Trinomials get harder than this when they start to incorporate numbers in front
of the first term or x. Now you have to look more closely at all the factors that
multiply together to find the numbers.
Ex. 4x²+30x+56
4x²= x*4x, 2x*2x
56= 1x56, 2x28, 4x14, 7x8
(x+1) (4x+56)=
4x²+ 60x+56 Wrong answer
(x+2) (4x+28)=
4x²+36x+56 Wrong answer
You also have to remember you can switch around the numbers between brackets
but keep them in their same place, lets use the last example we used.
(4x+2) (x+28)=
4x²+114x+56 Wrong answer
This gives you a whole new answer by using the same numbers rearranged. This
is still obviously not the right answer so you would start looking at some of the
other ones…
(2x+7) (2x+8)=
4x²+30x+56!!!! Right Answer
Practice Questions Factor the trinomials
a) x²+14x+49 b) 3x²+26x+48 c) 5x²+17x+14
Difference of Squares
A binomial is a difference of squares if both terms are perfect squares.
Recall that we may first have to factor out a common factor. After determining
that a binomial is a difference of squares, we factor it into two binomials. The first
binomial being the square root of the first term minus the square root of the
second term. The second binomial being the square root of the first term plus the
square root of the second term. This is represented in the following formula:
A2 - B2 = (A - B)(A + B)
Examples: x2 – 25
First check that the binomial is a difference of squares. x2 and 25 are
both prefect squares so we have a difference of squares. The square root of x2 is x
and the square root of 25 is 5. This gives us:
(x - 5)(x + 5)
16x4 - 100
16x4 is a prefect square as is 100, so we have a difference of squares.
First factor out the common factor. If we do not do this step, our answer will not
be completely factored.
The common factor is 4 giving us 4(4x4 - 25).
The square root of 4x4 is 2x2 and the square root of 25 is 5. This gives us:
4(2x2 - 5)(2x2 + 5)
Practice Questions Find the difference of squares
A) y2- 25 B) a2- 81 C) x2- 36
Difference of Cubes
A binomial is a sum of cubes if both terms are perfect cubes. Recall that we may
first have to factor out a common factor.
If we determine that a binomial is a sum of cubes, we factor it into a binomial and
a trinomial. The binomial being the cube root of the first term plus the cube root
of the second term. The trinomial comes from the binomial. We square the first
term of the binomial, change the sign to subtraction, multiply the two terms
together, and square the second term of the binomial. This is represented in the
following formula
A3 + B3 = (A + B)(A2 - AB + B2)
Examples: 8x3 + 27
First check that we have a sum of cubes.
8x3 and 27 are both perfect cubes and they are being added, so we have a sum of
cubes.
The cube root of 8x3 is 2x and the cube root of 27 is 3.
So, the binomial is (2x + 3).
To get the first term of the trinomial, we square 2x getting 4x2. To get the
second term of the trinomial, we change the sign to - and multiply 2x by 3
getting -6x. To get the third term of the trinomial, we square 3 getting 9.
So, the trinomial is (4x2 - 6x + 9).
The answer is
(2x + 3)(4x2 - 6x + 9)
Examples: 54x7 + 16x
First check that we have a sum of cubes.
Neither 54x7 nor 16x are perfect cubes. First factor out common factor of 2x.
This gives us 2x(27x6 + 8).
Now both 27x6 and 8 are perfect cubes.
The cube root of 27x6 is 3x2 and the cube root of 8 is 2.
So, our binomial is (3x2 + 2).
Now use the binomial to create the trinomial. Square the first term, 3x2, to get
9x4 as the first term of the trinomial. Change the sign from + to - and multiply
3x2 and 2 to get -6x2 as the middle term of the trinomial. Square the second
term, 2, to get 4 as the third term of the trinomial.
So, the trinomial is (9x4 - 6x2 + 4).
The answer is
2x(3x2 + 2)(9x4 - 6x2 + 4)
Practice Questions Find the difference of cubes:
A) x3- 27 B) 8x6 - 125 C) 250x4 - 128x
Factoring Quiz
1.List the factors of the following:
A) 18
B) 65
C) 42
2. What is the least common multiple?
A) 15 and 33
B) 21 and 49
C) 15 and 40
3. What is the greatest common factor?
A) (36x – 90x)
B) (26y2 – 38y)
C) (19a2 – 21a2)
B) (x - 9)(x + 5)
C) (b – 2)(b-3)
B) 6x2 + 21x + 9
C) x2 + 8x + 12
4. Factor the binomials.
A) (h+4)(h+7)
5. Factor the trinomials.
A) x2 + 5x + 6
6. Find the difference of squares.
A) a2 – 49
B) z2 - y2
C) 64x2 – 9
7. Find the difference of Cubes
A) j3 + h3
B) 53 + 73
C) 2x3 – 8x3
Factor Quiz Answers
1. A) 1,2,3,6,9,18
2. A) 3
B) 1,5,13,65
C) 1,2,3,6,7,14,21,42
B) 7 C) 5
3. A) 9x(4 – 10) B) 2y(13y – 19) C) a2(19 – 21)
4. A) h2 + 11h + 21 B) x2 – 4x – 45 C) b2 – 5b + 6
5. A) (x + 3)(x + 2) B) (6x + 3)(x + 3) C) (x + 6)(x + 2)
6. A) (a – 7)(a + 7)
B) (z – y)(z + y)
7. A) (j + h)(j2 - jh + h2)
C) (8x – 3)(8x + 3)
B) (5 + 7)(25 – 35 + 49) C) (2x + 8x)(4x2 - 16x + 84x2)
Appendix:
Practice Question: List the factors of
A) 36
B) 54
C) 81
A) 36= 1x6, 2x18, 3x12, 4x9, 6x6
B) 54= 1x54, 2x27, 3x27, 6x9
C) 81= 1x81, 3x27, 9x9
Practice Question: Find the least common multiples
A) 48 and 16 B) 63 and 42 C) 55 and 20
A) 48 and 16
48= 48, 96, 144, 192
16= 16, 32 48, 64, 80, 96
The least common multiple of 48 and 16 is 96
B) 63 and 42
63= 63, 126, 189
42= 42, 84, 126, 168
The least common multiple of 63 and 42 is 126
C) 55 and 20
55= 55, 110, 165, 220
20= 20, 40, 60, 80, 100, 120, 140, 160, 180, 200, 220
The least common multiple of 55 and 20 is 220
Practice Question: Find the greatest common factor
A) 42x y –63xy B) 15a + 21b
A) 42xy-63xy
42= 1x42, 2x21, 6x7
63= 1x63, 3x21, 7x9
They both have a common factor of xy
7xy(6-9)
B) 15a+21b
15= 1x15, 3x5
21= 1x21, 3x7
3(5a+7b)
Practice Question: Factor the binomials
A) (x+3) (x+5) B) (y+4) (y+2) C) (a+6)(a+5)
A) (x+3) (x+5)
x²+5x+3x+15
x²+8x+15
B) (y+4) (y+2)
y²+2y+4y+8
y²+6y+8
C) (a+6) (a+5)
a²+5a+6a+30
a²+11a+30
Practice Questions Factor the trinomials
a) x²+14x+49 b) 3x²+26x+48 c) 5x²+17x+14
a) x²+14x+49
49= 1x49, 7x7
7+7=14
(x+7) (x+7)
b) 3x²+26x+48
48= 1x48, 2x24, 3x 16, 4x12, 6x8
3x6+8= 26
(3x+8) (x+6)
c) 5x²+17x+14
14= 1x14, 2x7
5x2+7=17
(5x+7) (x+2)
Practice Questions Find the difference of squares
A) y2- 25 B) a2- 81 C) x2- 36
A) y2- 25
(y+5) (y-5)
B) a2- 81
(a+9) (a-9)
C) x2- 36
(x+6) (x-6)
Practice Questions Find the difference of cubes:
A) x3- 27 B) 8x6 - 125 C) 250x4 - 128x
A) x3- 27
27=3x3x3
(x+3) (x2-x3+32)
B) 8x6 - 125
8x6= 2x2+2x2+2x2
125=5x5x5
(2x2 +5) (2x4- 2x25+52)
C) 250x4 - 128x
2x(125x3-64)
125= 5x5x5
64= 4x4x4
2x(5x+4) (5x2-5x4+42)
Websites:
http:// mathforum.org/dr.math/faq/faq/.learn.factor.html
http://www.purplemath.com/modules/specfact.htm
http://www.mcraeclan.com/graemer/math/factoring3a.htm