Download Factoring GCF`s

Factoring GCF’s
What is Factoring?
Factoring is the process that reverses multiplication. Rewriting a polynomial as a product of factors is
called factoring a polynomial.
When we multiply 4 times 5 we obtain the product 20. This is multiplication. Factoring 20 into its
factors of 4 and 5 is factoring.
4 × 5 = 20 is multiplication and 20 = 4 × 5 is factoring.
Likewise ( x + 3)( x + 5) = x 2 + 8 x + 15 is multiplication of polynomials, and the reverse process
x 2 + 8 x + 15 = ( x + 3)( x + 5) is factoring.
There are many ways to factor a polynomial and the first method we will look at involves factoring out a
greatest common factor of two or more terms.
Factors of an Integer:
Factors are numbers that are multiplied together to give a product.
Example: List the factors of 20.
Solution: The following pairs of factors will result in a product of 20.
20
1 x 20
2 x 10
4 x 5
Therefore, the factors of 20 are; 1, 2, 4, 5, 10, 20.
Example: List the factors of 48.
Solution: The following pairs of factors will result in a product of 48.
1
2
3
4
6
48
x
x
x
x
x
48
24
16
12
8
Therefore, the factors of 48 are; 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.
Common Factors of Integers:
A common factor of two or more integers is an integer that is a factor of both integers.
Example: Find the common factors of 18 and 24.
Solution: First, list the factors of each integer, then look for the factors they both have in common.
18
24
1 x 18
1 x 24
2 x 9
2 x 12
3 x 6
3 x 8
4 x 6
The integers 18 and 24 have the following factors in common; 1, 2, 3, and 6. Note that all integers will
have a common factor of 1.
Example: Find the common factors of 36 and 60.
Solution: First, list the factors of each integer, then look for the factors they both have in common.
36
60
1 x 36
1 x 60
2 x 18
2 x 30
3 x 12
3 x 20
4 x 9
4 x 15
6 x 6
5 x 12
6 x 10
The integers 36 and 60 have the following factors in common; 1, 2, 3, 4, 6, 12.
Greatest Common Factor (GCF) of Integers:
The greatest common factor (GCF) of two or more integers is the largest integer that is a factor of each
integer.
Example: Find the greatest common factor of 18 and 24.
Solution: We determined in a previous example that the common factors of 18 and 24 are; 1, 2, 3, and 6.
Because the largest of these is 6, it is the greatest common factor.
Example: Find the greatest common factor of 36 and 60.
Solution: We determined in a previous example that the common factors of 36 and 60 are; 1, 2, 3, 4, 6,
and 12. Because the largest of these is 12, it is the greatest common factor.
Greatest Common Factor of Variables:
The greatest common factor of two or more terms containing the same variable will be the lowest power
of that variable.
Example: Find the greatest common factor of x 3 and x 5 .
Solution: The lowest power of the variable x is x 3 , therefore x 3 is the GCF.
Example: Find the greatest common factor of x 2 y 4 and x 3 y 3 .
Solution: The lowest power of the variable x is x 2 , and the lowest power of the variable y is y 3 ,
therefore x 2 y 3 is the GCF.
Now, let’s put it all together and find the greatest common factor of terms containing integers and
variables.
Example: Find the greatest common factor of 6 x 3 y 2 z 2 , 9 x 2 y 3 z and 12 x 2 y 5 z 3 .
Solution: The greatest common factor of the integers 6, 9, and 12 is 3. The lowest power of the
variable x is x 2 , the lowest power of the variable y is y 2 , and the lowest power of the variable z is z ,
therefore 3x 2 y 2 z is the GCF.
Example: Find the greatest common factor of 10 x 2 y 5 z 2 , 15 x 2 y 3 z 4 and 20 x 5 y 7 z 3 .
Solution: The greatest common factor of the integers 10, 15, and 20 is 5. The lowest power of the
variable x is x 2 , the lowest power of the variable y is y 3 , and the lowest power of the variable z is z 2 ,
therefore 5 x 2 y 3 z 2 is the GCF.
Factoring the Greatest Common Factor from a Polynomial:
To factor the greatest common factor out of a polynomial, we need to determine the greates common
factor of the terms in the polynomial.
Example: Factor out the greatest common factor from the polynomial 3 x + 12 y .
Solution: Both terms of this binomial have a factor of 3 which is the greatest common factor.
3 x + 12 y
3( x ) + 3(4 y )
3( x + 4 y )
This result may be checked by using the distributive property to multiply the greatest common factor by
the polynomial in the parentheses to obtain the original polynomial.
Example: Factor out the greatest common factor from the polynomial 15 xy 2 + 25 z .
Solution: Both terms of this binomial have a factor of 5 which is the greatest common factor.
15 xy 2 + 25 z
5(3 xy 2 ) + 5(5 z )
5(3 xy 2 + 5 z )
This result may be checked by using the distributive property to multiply the greatest common factor by
the polynomial in the parentheses to obtain the original polynomial.
Example: Factor out the greatest common factor from the polynomial 4 x 2 − 3 x .
Solution: Both terms of this binomial have a factor of x which is the greatest common factor.
4 x 2 − 3x
x(4 x) − x(3)
x(4 x − 3)
As always, this result may be checked by multiplying.
Example: Factor out the greatest common factor from the polynomial 12 x 3 − 3 x 2 + 9 x .
Solution: Both terms of this trinomial have a factor of 3x which is the greatest common factor.
12 x 3 − 3x 2 + 9 x
3 x(4 x 2 ) − 3 x ( x ) + 3x (3)
3 x(4 x 2 − x + 3)
As always, this result may be checked by multiplying.
Example: Factor out the greatest common factor from the polynomial 6 x 3 y − 5 x 2 y 2 + 7 x 5 y 3 .
Solution: Both terms of this trinomial have a factor of x 2 y which is the greatest common factor.
6 x 3 y − 5x 2 y 2 + 7 x 5 y 3
x 2 y (6 x ) − x 2 y (5 y ) + x 2 y (7 x 3 y ) 2
x 2 y (6 x − 5 y + 7 x 3 y 2 )
As always, this result may be checked by multiplying.
Example: Factor out the greatest common factor from the polynomial 18 xy 2 − 24 x 2 z + 42 x 2 y 2 .
Solution: The greatest common factor is 6 x .
18 xy 2 − 24 x 2 z + 42 x 2 y 2
6 x(3 y 2 ) − 6 x (4 xz ) + 6 x (7 xy 2 )
6 x(3 y 2 − 4 xz + 7 xy 2 )
As always, this result may be checked by multiplying.
Example: Factor out the greatest common factor from the polynomial ax 3 + bx 2 + cx .
Solution: The greatest common factor is x.
ax 3 + bx 2 + cx
x(ax 2 + bx + c)
As always, this result may be checked by multiplying.
Example: Solve the equation ax + b = cx + d for x in terms of a, b, c, and d.
Solution: To solve for x, first bring together all the terms with an x on the left side and all the terms
without an x on the right. Then factor out and isolate the x.
ax + b = cx + d
ax − cx = d − b
x(a − c) = d − b
d −b
x=
a−c
Example: Solve the equation ab = s 2 − ac for a in terms of b, c, and s.
Solution: To solve for a, first bring together all the terms with an a on the left side and all the terms
without an a on the right. Then factor out and isolate the a.
ab = s 2 − ac
ab + ac = s 2
a (b + c) = s 2
a=
s2
b+c
Factoring Common Binomial Factors:
Some polynomials will have a common binomial factor which may be factored out in the same way that
common monomial factors are factored.
Example: Factor out the common binomial factor from x ( x + 4) + 3( x + 4) .
Solution: The factor ( x + 4) is common to both terms. We may factor out this binomial factor using the
distributive property.
x ( x + 4) + 3( x + 4)
( x + 4)( x + 3)
Example: Factor out the common binomial factor from y ( x − 7) + z ( x − 7) .
Solution: The factor ( x − 7) is common to both terms. We may factor out this binomial factor using the
distributive property.
y ( x − 7 ) + z ( x − 7)
( x − 7)( y + z )
Example: Factor out the common binomial factor from 3 x ( 2 y − 5) + ( 2 y − 5) .
Solution: The factor ( 2 y − 5) is common to both terms. We may factor out this binomial factor using the
distributive property.
3x (2 y − 5) + (2 y − 5)
(2 y − 5)(3 x + 1)
Factoring By Grouping:
If a polynomial has four terms it may be possible to factor by grouping terms together and factoring out
a common binomial factor. The general strategy for factoring by grouping is:
1. Group the first two terms together and the last two terms together.
2. Factor a common monomial factor out of each grouping.
3. Factor out the common binomial factor.
Example: Factor the polynomial 4 y − 20 + xy − 5 x by grouping.
Solution:
4 y − 20 + xy − 5 x
(4 y − 20) + ( xy − 5 x )
4( y − 5) + x ( y − 5)
( y − 5)( 4 + x )
Group first and last two terms together
Factor common monomial factor from each group
Factor out common binomial factor
Example: Factor the polynomial 15 xy − 9 yz + 20 xz − 12 z 2 by grouping.
Solution:
15 xy − 9 yz + 20 xz − 12 z 2
(15 xy − 9 yz ) + ( 20 xz − 12 z 2 ) Group first and last two terms together
Factor common monomial factor from each group
3 y (5 x − 3 z ) + 4 z (5 x − 3 z )
Factor out common binomial factor
(5 x − 3 z )(3 y + 4 z )
When the 3rd term is negative, we must factor a negative value from the last two terms. This will cause
the 4th term to change sign value.
Example: Factor the polynomial xy + 6 y − 4 x − 24 .
Solution: When the third term is negative, the sign value of the 4th term must be changed when
grouping.
xy + 6 y − 4 x − 24
( xy + 6 y ) − ( 4 x + 24)
y ( x + 6) − 4( x + 6)
( x + 6)( y − 4)
Group terms, change sign of 4th term
Factor common monomial factor from each group
Factor out common binomial factor
Example: Factor the polynomial 6 ab + 12 ac − 5bc − 10c 2 .
Solution: When the third term is negative, the sign value of the 4th term must be changed when
grouping.
6ab + 12 ac − 5bc − 10c 2
Group terms, change sign of 4th term
(6ab + 12 ac ) − (5bc + 10c 2 )
Factor common monomial factor from each group
6a (b + 2c ) − 5c(b + 2c )
Factor out common binomial factor
(b + 2c )(6a − 5c )
Example: Factor the polynomial ab − 5b − 2a + 10 .
Solution: When the third term is negative, the sign value of the 4th term must be changed when
grouping.
ab − 5b − 2 a + 10
(ab − 5b) − ( 2a − 10)
b ( a − 5) − 2 ( a − 5)
(a − 5)(b − 2)
Group terms, change sign of 4th term
Factor common monomial factor from each group
Factor out common binomial factor
Applications:
Example: Consider the formula P = nC + nT + D for the total price of a purchase, where P is the total
price of the purchase, n is the number of items purchased, C is the total cost per item, T is the tax on
each item, and D is the total delivery charge. Solve for n in terms of P, C, T, and D.
Solution: The objective is to isolate n. Applying equation solving techniques and factoring we may
obtain:
P = nC + nT + D
P − D = nC + nT
P − D = n(C + T )
P−D
=n
C +T
Example: For an investment earning simple interest, the future value of the investment is represented by
the equation
A = P + Pr t
Where A is the future value or amount, P is the present amount of the investment, r is the annual interest
rate, and t is the time in years. Solve the equation for t.
Solution: The objective is to isolate P. Applying equation solving techniques and factoring we may
obtain:
A = P + Pr t
A = P (1 + rt )
A
= 1 + rt
P
A
− 1 = rt
P
1 A 
 − 1 = t
rP 
Example: The harmonic mean H of two numbers x and y is a type of average. Solve for H given the
following formula for harmonic mean.
Hx + Hy = 2 xy
Solution: Isolate the H by factoring, then solve.
Hx + Hy = 2 xy
H ( x + y ) = 2 xy
2 xy
H=
x+ y