Factoring
Factoring
Factoring
Factoring
Greatest Common Factor and Factor by Grouping Finding the Greatest Common Factor The Greatest Common Factor, also known as the GCF, has two possible parts to it. The rst is the numerical value the can be factored out of each term. The second part is the variable portion or monomial that can be factored out of each term. The numerical portion can be viewed in this fashion. First determine the lowest coe cient in the expression. Determine if this value will divide into all the other coe cients in the expression. If it does divide into all the other terms then this value is the GCF. If it dies not then list the factors of this value an try these numbers into all the coe cients in the expression. 24x3 36x2 + 48x + 8
The 8 is the lowest value in the expression but it will not divide into 36. The factors of 8 are f1; 2; 4; 8g: The 4 is the value which will go into all the coe cients. 4 6x3 9x2 + 12x + 2
We can now see that this has be reduced as much as possible. Lets take the same equation but modify the variables. 24x5 36x4 + 48x3 + 8x2
Now, determine if the variable x is in every term in this expression. Since that is true the lowest exponent value for this variable x2 .Divide all the terms in this expression by 4x2 : 4x2 6x3 Now lets try some examples. Example 1 25x2 y 2 + 15x3 y + 10xy 3 5xy 5xy + 3x2 + 2y 2 Example 2 128x4 + 96x3 y 2 + 48x2 y 3 + 64x2 y 4 4x2 32x2 + 24xy 2 + 12y 3 + 16y 4 4x2 (4) 8x2 + 6xy 2 + 3y 3 + 4y 4 16x2 8x2 + 6xy 2 + 3y 3 + 4y 4 9x2 + 12x + 2
Example 3 136x8 y 7 z 4 + 112x7 y 8 z 6 + 144x6 y 9 z 5 8x6 y 7 z 4 17x2 + 14xyz 2 + 18y 2 z In this next example that there are two terms in parenthesis. They are treated as if they were one variable in removing the GCF Example 4 18x2 y 2 (a + b) + 24x2 y 3 (a + b) + 12x2 y 2 (a + b) 6x2 y 2 (a + b) (3 (a + b) + 4y (a + b) + 2) Factor by Grouping can you factor the GCF out of the following? x2 + xy + x + y The answer is no. Suppose we factor a portion of this problem using the GCF techniques that we previously discussed. x2 + xy + x + y x (x + y) + 1 (x + y) Now since the factor (x + y) is in both positions you have found what is the GCF. Factor it out as before. x2 + xy + x + y x (x + y) + 1 (x + y) (x + y) (x + 1) Lets see this problem again only in a dierent order. x2 + x + xy + y x (x + 1) + y(x + 1) (x + 1) (x + y) This time the term (x + 1) was the GCF. The solution is the same. One more time let us try this problem in a dierent order. x2 + y + xy + x x2 + y + x (y + 1) This order does not work. This indicates that order can make a dierence in how we do the problem. If we reorder this to one of the two previous ways we will arrive at a conclusion.
2 2 4 3 2
Example 5 ax2 + bx2 + ay 2 + by 2 x2 (a + b) + y 2 (a + b) (a + b) x2 + y 2 Example 6 x3 + x2 + x + 1 x2 (x + 1) + 1 (x + 1) (x + 1) x2 + 1 Example 7 Count the number of terms. ax2 + ax + 4a + bx2 + bx + 4b The number of groups with the number of terms in each group is found by the integer factors of the to# of groups tal number of terms that you have. In this case the number of terms is six. # of terms in each group First look at two groups of three terms ax2 + ax + 4a + bx2 + bx + 4b a x2 + x + 4 + b x2 + x + 4 x2 + x + 4 (a + b) The solution is ax2 + ax + 4a + bx2 + bx + 4b = x2 + x + 4 (a + b) Here is the same problem but you must use three groups of two terms ax2 + bx2 + ax + bx + 4a + 4b x (ax + bx + a) + b (ax + bx + a) x2 (a + b) + x (a + b) + 4 (a + b) (a + b) x2 + x + 4 Example 8 A GCF and Grouping 5k 3 x7 y + 45k 3 x3 y + 20kx7 y + 180kx3 y 5kx3 y k 2 x4 + 9k 2 + 4x4 + 36 5kx3 y k 2 x4 + 9 + 4 x4 + 9 5kx3 y x4 + 9 k2 + 4 2 3 3 2
We have started to develop a set of rules to follow when factoring. General Factoring Rules GCF and Grouping 1.Factor out the Greatest Common Factor. 2. Count the number of terms in the expression. 3. If the number of terms is four or more factor by Grouping. Remember that the integer products of the total number of terms in the expression indicate how many groups with how many terms. 3
Factoring Trinomials
There is a general form of a basic trinomial: axn + bx 2 + c This is a general form when one variable is in the expression. If the lead coe cient a 6= 0 and a = 1 the method is fast and easy. When the leading coe cient a 6= 0 and a 6= 0 the method is more complex and requires patience to apply General General Factoring Rules with Trinomials 1. Factor out the Greatest Common Factor. 2. Count the number of terms in the expression. 3. If the number of terms is four or more factor by Grouping. Remember that the integer products of the total number of terms in the expression indicate how many groups with how many terms. 4. If the number of terms is three then given axn + bx 2 + c (a) If a = 1 then nd the integer factors of c that sum to b: The exponents on variables in the middle term should be one half of the power found in the rst or last term. (b) If a 6= 1 then nd the integer factors of a; c. Next try each pair of factors of a: with a pair of factors of c When the sum of the products of these factors equals bremember to place the factor of c in both ways The exponents on variables in the middle term should be one half of the power found in the rst or last term. Trinomials When a = 1 When the leading coe cient equals one, the problem depends on the following couple of conditions: 1. The exponents of the variables in the middle term must be one half of the variables located in the rst or third position. 2. Find the integer factors of the last coe cient. If the number is positive list both positive and negative pairs. If the value is negative alternating the sign on each product. 3. If any of these products of c sum up to b you have found the solution. Example 9 Given x2 + 5x + 6 The factors of six are: (1; 6) (2; 3) ( 1; 6) ( 2; 3) The sum of (2; 3) adds up to the middle term. (x + 2) (x + 3) x (x + 3) + 2 (x + 3) x2 + 3x + 2x + 6 x2 + 5x + 6 The solution is (x + 2) (x + 3) 4
n n
Example 10 Given : x2 The factors of negative six are: (1; 6) (2; 3) ( 1; 6) ( 2; 3) The sum of (1; 6) equals the middle term. (x + 1) (x x(x x x The solution is (x + 1) (x Example 11 Given x2 The factors of twelve are: ( 1; 12) ( 2; 6) ( 3; 4) (1; 12) (2; 6) (3; 4) The sum of (3; 4) adds up to the middle term. (x
2 2 2
5x
6) 6) 6
6) + 1(x 6x + x 5x 6
6) x 12
4) (x + 3) 4 (x + 3) 4x 12 12
x (x + 3) x + 3x x The solution is (x 4) (x + 3)
2
Example 12 Given : x4 + 5x2 + 6 The factors of six are: (1; 6) (2; 3) ( 1; 6) ( 2; 3) The sum of (2; 3) adds up to the middle term. x2 + 2 x2 + 3
Example 13 Given x4 + 5x2 y 2 + 6y 4 The factors of six are: (1; 6) (2; 3) ( 1; 6) ( 2; 3) The sum of (2; 3) adds up to the middle term. x2 + 2y 2 x2 + 3y 2
x2 x2 + 3y 2 + 2 x2 + 3y 2 x4 + 3x2 y 2 + 2x2 y 2 + 6y 4 x4 + 5x2 y 2 + 6y 4 The solution is x2 + 2y 2 Example 14 Given x4 y 4 + 5x2 y 2 + 6 The factors of six are: (1; 6) (2; 3) ( 1; 6) ( 2; 3) The sum of (2; 3) adds up to the middle term. x2 y 2 + 2 x2 + 3 x2 + 3y 2
x2 y 2 x2 y 2 + 3 + 2 x2 y 2 + 3 x4 y 4 + 3x2 y 2 + 2x2 y 2 + 6 x4 y 4 + 5x2 y 2 + 6 The solution is x2 y 2 + 2 Example 15 Given 12x4 12 x4 The factors of the last coe cient are: ( 1; 12) ( 2; 6) ( 3; 4) (1; 12) (2; 6) (3; 4) The sum of (3; 4) adds up to the middle term. 12 x2 4y 2 x2 + 3y 2 4y 2 x2 + 3y 2 4x2 y 2 12y 4 144y 4 12y 4 12x2 y 2 x2 y 2 144y 4 12y 4 x2 y 2 + 3
x2 + 3y 2 6
Example 16 Given 15x3 y 15xy x2 The factors of the last coe cient are: (1; 24) (2; 12) (3; 8) (4; 6) ; ( 1; 24) ( 2; 12) ( 3; 8) ( 4; 6) The sum of (4; 6) adds up to the middle term. 15xy x 6y 2 x + 4y 2 6y 2 x + 4y 2 6xy 2 24y 4 360xy 5 24y 4 30x2 y 3 2xy 2 360xy 5 24y 4
15xy x x + 4y 2 15xy x2 + 4xy 2 15xy x2 15x34 y The solution is 15xy x 6y 2 x + 4y 2 2xy 2 30x2 y 3
Trinomials when rst coe cient is not 1 When the leading coe cient does not equals one, the problem depends on the following couple of conditions: 5x2 8xy + 3y 2 1. The exponents of the variables in the middle term must be one half of the variables located in the rst or third position. 2. Find the integer factors of the rst and last coe cient. If the number is positive list both positive and negative pairs. If the value is negative alternating the sign on each product. 5x2 8xy + 3y 2
rst term (1; 5) last term ( 1; 3) (1; 3) 3. Place a pair of factors of the rst coe cient in the factoring equation. Next take a pair of factors of the last coe cient and place that into the last position (5x y) (x 3y) = 5x (x = 5x
2
3y) 15xy
y (x
3y)
xy + 3y 2
= 5x2
16xy + 3y 2
This does not work so you must switch the position of ( 1; 3) (5x 3y) (x y) = 5x (x = 5x
2
y) 5xy
3y (x
y)
3xy + 3y 2
= 5x2 7
8xy + 3y 2
This is the original problem The solution therefore is: 5x2 8xy + 3y 2 = (5x 3y) (x y)
4. If All the possible product of the rst and last coe cients have be used with both positions for the last coe cient position, then there is no solution. Factor 12x4 The factors of the rst coe cient are: (1; 12) ; (2; 6) ; (3; 4) The factors of the last coe cient are: (1; 24) (2; 12) (3; 8) (4; 6) ; ( 1; 24) ( 2; 12) ( 3; 8) ( 4; 6) Start from the bottom of the list: 3x2 12x4 4y 4x2 + 6y 24y 2 24y 2 23x2 y 24y 2
16x2 y + 18x2 y
4x2
Notice that both attempts indicate that there is a GCF when there is none You can skip this work if you notice this and try only those choices that prior to multiplying out the term DO NOT HAVE A GCF in the expression. This assumes that you have removed all aspects of a GCF prior to trying the trinomial method. The next attempt is 3x2 + 3y 4x2 8y
Notice that in both groups there appears to be a GCF when there is none. Do not lose your patience, and reverse the choice for the last coe cient. This does not show any GCF in any group so now multiply it out. 3x2 8y 4x2 + 3y 8 4x2 + 3y 32x2 y 24y 2 4x2 + 3y 24y 2
3x2 4x2 + 3y 12x4 + 9x2 y 12x4 The solution is 12x4 23x2 y 23x2 y 8y 8
24y 2 = 3x2
Example 17 Factor 18x2 Factors of the rst coe cient are (1; 18) (2; 9) (3; 6) Factors of the last coe cient are (1; 7) ( 1; 7) (3x 3x (6x 18x 18x Reverse the last terms (3x 3x (6x 18x 18x The solution is 18x2 Example 18 Factor 18x2 Factors of the rst coe cient are (1; 18) (2; 9) (3; 6) Factors of the last coe cient are (1; 24) (2; 12) (3; 8) (4; 6) ; ( 1; 24) ( 2; 12) ( 3; 8) ( 4; 6) (3x (3x (3x (3x 4y) (6x + 6y) 4y) 3y) 2y) y) 3y) (6x + 8y) 2y) (6x + 12y) y) (6x + 24y) 11xy 24y 2 27x + 7 = (3x
2 2 2 2
27x + 7
7) (6x 1) 3x
1) 7 (6x 42x + 7 1)
45x + 7
1) (6x 7) 21x
7) 1 (6x 6x + 7 7)
27x + 7 7)
1) (6x
(3x + 6y) (6x (3x + 8y) (6x (3x + 12y) (6x (3x + 24y) (6x
Each possibility listed above shows a GCF when there is none in the problem. Therefore change the rst term coe cient factors and start again. (2x (2x (2x (2x 4y) (9x + 6y) 4y) 3y) 2y) y) 3y) (9x + 8y) 2y) (9x + 12y) y) (6x + 24y)
(2x + 6y) (9x (2x + 8y) (9x (2x + 12y) (9x (2x + 24y) (6x Only one choice can work: (2x
2 2
24
The solution is 18x2 11xy 24y 2 = (2x 3y) (9x + 8y) Remember to have patience, write out all the products of the rst and last coe cients. Make sure you have removed all of the GCF if there is one in the problem. Look for those pairs of products that do not indicate a GCF in either group. Example 19 12 (a + b) 11 (a + b) 15 (a + b) h i 4 2 (a + b) 12 (a + b) 11 (a + b) 15 (1; 12) (2; 6) (3; 4) The factors of the last coe cient are: (1; 15) ( 3; 5) ( 1; 15) (3; 5) h
2 2 5 3
(a + b)
3 (a + b)
4 (a + b) + 5
h i 2 2 2 (a + b) 3 (a + b) 4 (a + b) + 5 3 4 (a + b) + 5 h i 4 2 2 (a + b) 12 (a + b) + 15 (a + b) 12 (a + b) + 15 h i 4 2 (a + b) 12 (a + b) + 3 (a + b) 15 10
You have the correct location but the sign of the middle term is ho . Just use the other pair of i 2 2 factors (3; 5)which has the sign reversed. The solution is (a + b) (3 (a + b) 5)(4 (a + b) + 3) Example 20 x2 The factors of the last term are (1; 30) (2; 15) (3; 10) (5; 6) ( 1; 30) ( 2; 15) ( 3; 10) ( 5; 6) Notice that not one of these pairs will add to the middle term coe cient. This problem is not s factorable. Example 21 Factor 5a3 b2 ab2 (5a2 The factors of the rst coe cient are (1; 5) The factors of the last coe cient are (1; 3) ( 1: ab2 (5a2 ab2 (5a ab (5a (a ab (5a 5a b The solution is 5a3 b2 Example 22 Factor 144x4 y 5 36y 5 4x4 72x2 y 6 + 36y 7 2x2 y + y 2
3 2 2 2 2
h i 2 2 2 (a + b) 3 (a + b) 4 (a + b) 3 + 5 4 (a + b) 3 h i 4 2 2 (a + b) 12 (a + b) 9 (a + b) + 20 (a + b) + 15 h i 4 2 (a + b) 12 (a + b) + 11 (a + b) 15
3 (a + b) + 5
4 (a + b)
11xy
30y 2
3)
b))
You may try all the factors of the rst term which are: (1; 4) (2; 2) Or the last term which is just (1; 1) ; ( 1; 1) None of these pairs will have a sum of 2 this is as far as this problem will go!!!
11
Special Formulas
Perfect Squares A perfect Square formula is in two versions. We saw this in chapter ve. (a b) = a2
2 2
2ab + b2
(a + b) = a2 + 2ab + b2 .Now the formulas are written backwards for factoring: a2 2ab + b2 = (a b)
2 2
a2 + 2ab + b2 = (a + b)
These type of equations t into rule three of the factoring rules we developed in the previous section for trinomials. Example 23 x2 (x (x Example 24 72x5 168x3 y 3 + 98xy 6 84x2 y 3 + 49y 6 10x + 25 5) (x 5)
2
5)
2x 36x4
The factors of the rst and last coe cients are: 36 ) (1; 36) (2; 18) (3; 12) (4; 9) (6; 6) Try the perfect squares rst: 2x 36x4 2x 6x2 2x 6x2 Example 25 Factor 36k 3 x4 9k 4k 2 x4 9k k 2 4x4 9k 4x4 9k 2x2 9k 2x2 5y 2 5y 2
2
6x2
7y 3
180k 3 x2 y 2 + 225k 3 y 4 + 72km2 x4 20k 2 x2 y 2 + 25k 2 y 4 + 8m2 x4 20x2 y 2 + 25y 4 + 2m2 4x4 k 2 + 2m2 k 2 + 2m2 2x2 5y 2 20x2 y 2 + 25y 4 k 2 + 2m2 12
The Dierence of Squares This is also a formula that we saw in chapter 5. The formula then was (a + b) (a b) = a2 b2
Now for factoring it is reversed:Two Terms Dierence of Squares In order to factor this type of problem the following conditions must be met: 1. (a) There must be a minus sign between the two terms (b) Each exponent must be even (c) Each of the two coe cient must be a perfect square. a2 Example 26 Factor x2 Example 27 Factor 25x4 Example 28 Factor x2 + 25 This has no solution! Why? The last coe cient is a positive twenty ve. This limits the choices to either positive ve or negative ve. Watch what occurs if we assume either one of these is correct: x2 + 25 (x 5) (x 5) x (x 5) 5 (x 5) x2 5x 5x + 25 x2 10x + 25 x2 + 25 (x + 5) (x + 5) x (x + 5) + 5 (x + 5) x2 + 5x + 5x + 25 x2 + 10x + 25 64y 2 5x2 8y 5x2 + 8y 4y 2 2y) (x + 2y) (x b2 = (a + b) (a b)
In either case the fact that the sign of the last coe cient in both factor are positive or negative results in a middle term which does not exist in the original problem. This is why you must have a negative or subtraction sign between the two terms. Example 29 Factor: x64 1 x64 x x x x x x
32 32 32 32 32 32
1 x32 +1
8
+1 x
16
1 x16
4
+1 x
16
1 x8
2
+1 x
16
+1
8
x +1 x +1
4 2 4
1 x4 1 x2 1 1]
+1 x
16 16
+1
8 8
x +1 x +1 13
x +1
+1 x
+1
x +1
x +1
+1
+1
x +1
x + 1 [x + 1] [x
This repeats for several reasons. rst the exponent on the x variable is a perfect factor of two. Second the number 1 is always a perfect square of itself. Example 30 Factor 256x17 y 17 z 7 144x5 y 5 z 13 9z 6 4x6 y 6 + 3z 3 3z 3
81 x2 9 3]
x + 9 [x + 3] [x
Factoring Cubics We also saw the cubic formula in the same section of chapter ve Perfect Cubes (a)
3 3
(b) = ((a)
3
2 2
This is a bit more complex than the other formulas. It requires the following conditions to be met. Sum or Dierence of Cubes. 1 All exponents must be multiples of three 2 Both coe cients must be perfect cubes Example 32 Factor x3 + 27 (x) + (3)
3 3 2 2
((x) + (3)) (x) + (x) (3) + (3) (x + 3) x2 + 3x + 9 Example 33 Factor 8x6 2x2 2x2 2x2 125y 9
3
5y 3 5y 3 5y 3
2x2
+ 2x2
5y 3 + 5y 3
14
Example 34 Factor 128x7 y 6 z 2 + 250xy 3 z 11 2xy 3 z 2 64x6 y 3 + 125z 9 i h 3 3 2xy 3 z 2 4x2 y + 5z 3 h 2xy 3 z 2 4x2 y + 5z 3 4x2 y 2xy 3 z 2 4x2 y + 5z 3 16x4 y 2 Example 35 Factor x6 x2 x2 x2 y6
3
4x2 y
5z 3 + 5z 3
20x2 yz 3 + 25z 6
y2 y2
y2 h
x2
+ x2
y2 + y2
x4 + x2 y 2 + y 4 y] x4 + x2 y 2 + y 4
[x + y] [x
x3 + y 3
(x + y) x2 If you multiply x2 xy + y 2
x2 + xy + y 2 = x4 + x2 y 2 + y 4
15
(a) Dierence of Squares In order to factor this type of problem the following conditions must be met: i. There must be a minus sign between the two terms ii. Each exponent must be even iii. Each of the two coe cient must be a perfect square. a2 (b) Sum or Dierence of Cubes. (a)
3
b2 = (a + b) (a
b)
(b) = (a
b) a2
ab + b2
i. All exponents must be multiples of three ii. Both coe cients must be perfect cubes 6. Examine each piece of you current solution and REPEAT rules three to ve as needed. Example 36 Factor 3x4
2
27x2 9 3]
3x2 x2
3x [x + 3] [x 16
75y 3 25y 2
120x2 y + 25y 2 5y 5y
2
12x
5y
k +1
k 2 + 1 [k + 1] [k
4000ac9
4a 27a3 b6 1000c9 h i 3 3 10c3 4a 3ab2 h 4a 3ab2 10c3 3ab2 4a 3ab2 10c3 Example 39 Factor 36k 4 x8 y 3 33k 4 x5 y 5
+ 3ab2
10c3 + 10c3
11k 4 x3 y 2 11x3 y 2
2 2
11x3 y 2 3x 3x
3 3
15y 4 5y 5y
2 2
4x + 3y 4x + 3y
3
k +1
3x y Example 40 Factor
k + 1 [k + 1] [k
17
Example 41 Factor (x (x ((x (x (x Example 42 Factor 36x2 y 4 z 7 + 49z 9 z 7 36x2 y 4 + 49z 2 Example 43 Factor 8x3 y 3 x3 8y 3 125x3 z 9 125z 9 64w9 y 3 + 1000w9 z 9 8w9 8y 3 125z 9 3) + 125 3) + (5)
3 3 2 2 3
3) + (5)) (x 3 + 5) x2 2) x
2
3)
(x
3) (5) + (5)
6x + 9
5x + 15 + 25
11x + 49
8y 3 125z 9 x3 8w9 h ih i 3 3 3 3 (2y) 5z 3 (x) 2w3 h i 2 2 2y 5z 3 (2y) + (2y) 5z 3 + 5z 3 (x) 2y 5z 3 4y 2 + 10yz 3 + 25z 6 x 2w3
2w3
x2 + 2xw3 + 4w6
18
3 2
22x2 11x
48x = 0 24 = 0
24 ) (1; 24) (2; 12) (3; 8) (4; 6) ( 1; 24) ( 2; 12) ( 3; 8) ( 4; 6) 36x3 2x 18x 2x = 0 x=0
2
22x2 11x
24x = 0 24 = 0 3] = 0 [2x 3] = 0 2x 3 = 0 2x = 3 3 x= 2
81 = 0 x2 9 =0 3] = 0
x + 9 [x + 3] [x
x2 + 9 = 0 x + 3 = 0 x 3 = 0 x2 = 9 x= 3 x=3 No Solution
19
4x2 4] 4] x
2
x+4=0 1 [x + 4] = 0 1 =0 1] = 0 [x 1] = 0 x=1
4] [x + 1] [x [x + 1] = 0 x= 1
20