Download R6. The Least Common Multiple

R6. The Least Common Multiple
The Least (or Lowest) Common Multiple (LCM) of numbers a and b is the
smallest positive integer divisible by both a and b. The most simple way to find
the LCM is to count by a number. Remember when you learned to count by
two’s? 2, 4, 6, 8, . . . ? The LCM can be found by counting by the given numbers
until each number reaches the same number.
Example. Find the LCM of 6, 15, and 9.
6 =⇒ 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90 , 96, . . .
15 =⇒ 15, 30, 45, 60, 75, 90 , 105, . . .
9 =⇒ 9, 18, 27, 36, 45, 54, 63, 72, 81, 90 , . . .
Notice that 90 was the first number that 6, 15 and 9 had in common. Thus,
90 is the LCM. The problem with such is that this method is limited. Consider
the following example.
Example. Find the LCM of x2 − 4, x2 − 5x + 6, and x2 − 9.
Counting by x2 − 4 does not make sense. Set this problem aside and return
to the first example seeking to discover a more global approach to finding the
LCM.
Example. Find the LCM of 6, 15, and 9 with the purpose of finding another
way to find the LCM. Up to this point, most of our work has been possible
because of factoring. Again, it is still the case. Factor, or write the prime
factorization of, 6, 15, and 9. This is seen below.
6
= 2 · 3
15
=
· 3
9
=
· 32
LCM = 2 · 32
· 5
·
· 5 = 90
The LCM is the smallest integer divisible by all three integers: 6, 15, and 9.
It can be found by writing the prime factorization of the three numbers. The
LCM includes the highest power of each of the factors. The number 2 = 21
appears once in 6 but not in the 15 or 9. Thus, one 2 is needed in the LCM.
The number 3 appears once in the 6, once in the 15, and twice in the 9 as 32 ;
thus, two 3’s are needed in the LCM. Finally, the number 5 appears once in 15,
but not in 6 or 9. Thus, one 5 is needed in the LCM, which is illustrated above.
1
Example. Using the technique of finding the LCM by couting factors, return
to the example of find the LCM of x2 − 4, x2 − 5x + 6, and x2 − 9.
Begin the problem by factoring each polynomial.
x2 − 4
x2 − 5x + 6
x2 − 9
LCM
= (x + 2) (x − 2)
=
(x − 2) (x − 3)
=
(x − 3) (x + 3)
= (x + 2) (x − 2) (x − 3) (x + 3)
In the first example, the factors were multiplied to obtain 90. When the LCM is
a product of binomials and trinomials or polynomials, leave the LCM in factored
form.
Example. Find the LCM of 27x3 y 2 and 18y.
Please note that the large spaces between the numbers implies multiplication
between the numbers.
27x3 y 2
18y
LCM
LCM
=
=
2
=
2
= 54x3 y 2
·
·
33
32
33
· x3
· x0
· x3
· y2
· y1
· y2
There is one 2 in the 18 and none in the 27, so a 2 is needed in the LCM. The 27
has three 3’s and the 18 had two 3’s, so three 3’s are needed in the LCM. The
27x3 y 2 has three x’s and the 18y has none, so three x’s are part of the LCM.
And, the 27x3 y 2 has two y’s and the 18y has only one y, so the LCM has two
y’s.
Example. Find the LCM of x5 − 6x4 + 9x3 , 3x3 − 27x, and 2x + 6.
Factor each quadratic. Take a few minutes to factor each polynomial, then
check your results with the following solution.
x5 − 6x4 + 9x3
3x3 − 27x
2x + 6
LCM
or, LCM
=
=
= 2
= 2
=
3
·
3
R-11: 30–42
2
x3
x1
x3
6x3
(x − 3)2
(x − 3)
(x + 3)
(x + 3)
(x + 3) (x − 3)2
(x + 3) (x − 3)2
0.1
Practice Problems
Find the Least Common Multiple (LCM) of the following expressions.
1.
4.
6.
7.
8.
9.
10.
11.
12.
13.
0.2
1.
4.
7.
8.
9.
10.
11.
12.
13.
15; 20
2. 9; 15; 5
12x2 y; 4xy
5. 15ab2 ; 3ab, 10a3 b
5y − 15; y 2 − 6y + 9
6x2 + 60x + 96; 8x + 16
x5 − 4x4 + 4x3 ; 3x3 − 12x, 2x + 4
3x2 − 30x + 75; 4x2 − 100; 2x + 10
x2 − 2x + 1; 3x2 − 3x; 7x2
x2 + 5x − 14; x2 + 8x + 7
2x2 + 9x + 10; 2x2 − x − 15
x2 − 2x − 24; 3x2 + 9x − 12; 2x2 − 4x + 2
Solutions
60
2. 45
12x2 y
5. 30a3 b2
24(x + 8)(x + 2)
6x3 (x + 2)(x − 2)2
12(x + 5)(x − 5)2
21x2 ((x − 1)2
(x + 1)(x − 2)(x + 7)
(x − 3)(x + 2)(2x + 5)
6(x + 4)(x − 6)(x − 1)2
3. 945
6. 5(y − 3)2
3
3. 27; 35; 63