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FACTORING - THE MATRIX METHOD
When you have to factor a polynomial into its factors, there are three ways to proceed.
1. Work it out in your head – a simple trial-and-error approach.
2. Set up a matrix and systematically work out the combination of integers that will work for you
3. Apply the formula
The Simple Approach
Example: x2 + 5x + 6 = 0
The first term is 1x2, the middle term is + 5x, and the last term is + 6.
Usually, the factors of the first term are easy: first term: x  x.
Work out all possible factors of the last term: 2  3; 1  6
From the FOIL, you want the "O" and the "I" to add/subtract [depending on the sign of the last term: ( +
) add and ( - ) subtract] to equal the middle term.
+2 +3 = 5
+1 + 6 = 7
Only the first one works
Now write the factors for the last terms like this: (x 2)(x 3).
Check the signs – they are all positive, therefore the factoring arrangement becomes:
x2 + 5x + 6 = 0 = (x + 2)(x + 3) = 0
The roots of the equation then become x = -2 and x = -3
To recap:
 2x
 3x
5x (middle term )
therefo re :
x  2x  3  0
x  2  0 or x  3  0
x  2 or x  -3
Let’s try a problem where we probably will need to develop the entire matrix of possibilities:
2. a 2  8a  20  0
We have three choices :
a
1a
20 
a
20a
2 a 10 
a
10a
4 a
5
5a
1a
2a
4a
19a
8a
1a
Notice that we are looking for the difference (the last term above is negative "-20" ).
 8a is what we need (10a - 2a  8a).
a - 2a  10  0
a  2  0 or a  10  0
a  2 or a  10
3. x 2  2 x  24
Now we have four choices :
x
1x 24 
x
24 x
2 x 12 
12 x
x
3 x 8
8x
x
4  x 6 
6x
1x
2x
3x
4x
23 x
10 x
5x
2x
Notice that we are looking for the difference (the last term above is negative "-24" ).
- 2x is what we need (-6x  4x  -2x), so :
x  6x  4  0
x  6  0 or x  4  0
x  6 or x  -4
4. 3x2  6x  72 = 0
Look at all the combinations we have:
(3x 1)(1x 72)
(3x 2)(1x 36)
(3x 3)(1x 24)
(3x 4)(1x 18)
(3x 6)(1x 12)
(3x 8)(1x 9)
(1x 1)(3x 72)
(1x 2)(3x 36)
(1x 3)(3x 24)
(1x 4)(3x 18)
(1x 6)(3x 12)
(1x 8)(3x 9)
Unfortunately, we have to try them all. Fortunately, we can simplify the equation:


3 x 2  2 x  24  0
3x  6 x  4  0
(See problem #3.)
x  6 or x  -4
The full matrix approach
How to set up the matrix for equations having a x2 coefficient greater than 1
Example 4 : 3x2 + 13x – 10 = 0
Here it would be helpful to set up the entire matrix
X-Factor
3
3
3
3
a*X
30
3
15
6
Factor a
10
1
5
2
Factor b
1
10
2
5
The ‘X-Factor’ is the coefficient of x2
b*X
1
10
2
5
Mixture ++
31
13
17
11
+-
-+
--
29
-7
13
1
-29
7
-13
-1
-31
-13
-17
-11
In columns 2 and 3 we place all possible fator combinations of the last term. Note that we list each possible combination
TWICE – in the second reiteration we merely reverse the two factors
In column 4 we multiply the X-facor by column 2
In the ‘Mixture’ column we compute all possible sums of the numbers in columns 4 and 5, attributing to the numbers every
combination of + or – value. Note that the order is the same: + + ; + - ; - + ; - This is important, because in picking the ‘right’ combination from the mixture column, we also must give credence to whether
the product (the coefficent of the third term) is positive or negative. Since, in this example, the third term coefficient is –10,
we must insist on picking numbers only from the 2 nd and 3rd interior listing of the Mixture column. Had we been looking for a
positive product, we could consider only the first and the fourth listings.
Note that in the ‘Mixture’ column, only ONE number comes out the way we need it. We are looking for the value of + 13, but
we can consider only the interior listings 2 and 3
Hence, we go with the “2 5” choice (the third option from the matrix). Now we need to put them in the proper place.
Remember that in our matrix, it is the factor we list in column 2 that gets multiplied by the X-Factor. In this case it is the
number “2” that gets multiplied by 3. This means that the number “2” must go in the OTHER parenthesis than the “3x”
Consequently, the factor of 3x2 + 13x – 10 = (3x –5)(x – 2) = 0
This is complex, but once you learn the steps and remember them, this approach will always work, and mechanically, it will
take you only seconds to work out the matrix.
Here are some examples to work on:
Example 5: 4x2 – 33x + 35 = 0
Example 6: 3x2 + 11x – 40 = 0
Example 7: 2x2 – 15x – 18 = 0
Example 8: 7x2 – 121x – 34 = 0
(Answers at the end of this section)
Factors and Prime Factors
Every positive, non-prime integer has at least two prime factors.
21 is a product of two prime numbers, 3 and 7
38 is a product of two prime numbers, 2 and 19
If a question asks for the ‘prime factorization’ of 38, the answer is “ 2 and 19” If the question is ‘how many prime factors
does 38 have’, the answer is ‘two’
If a question asks about the factors of 38, the answer is ‘four’ positive integer factors . Any integer, whether prime or not, has
at least two positive integer factors, namely the number 1 and the number itself.
Now, let us kick this up a notch. Take the integer 12. What is its prime factors? The answer is ‘two’ – namely 2 and 3.
12 can be written as (23)(3)
On the other hand, if the question is about [positive integer] ‘factors’ then we maust include the following: 1, 2, 3, 4, 6, 12
In other words, the integer 12 has six positive integer factors. Remember, when you deoineate them, always to ‘bookend’ the
list with the number ‘1’ and the factored number itself.
Next level
How many [positive integer] factors does 84 have?
Applying the basic rule we just learned, we start with 1 and 84, and then all we have got to do, is fill in all the numbers inbetween that are factors of 84. A smart way to do that, is to go up the ladder – ask yourself is 84 divisible by 2 if yes, then add
2 and 42 to the list. Next, ask if 84 is divisible by 3. If yes, then add 3 and 84/3 = 28 to the list
In summary, we obtain the following list as the factors of 84: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84
Let us see if there is an easy way to hone in on his. 84 can be written as (2)(2)(3)(7)
If you treat this like a combination problem, you can show that the number of distinct possible SINGLE factors = 3
The number of distinct possible factors involving the mere product of TWO of its inherent factors,, is (4C2)/2 + 1=4
The number of distinct possible factors involving the mere product of THREE of its inherent factors,, is (4C3)=3
The number of distinct possible factors involving the mere product of FOUR of its inherent factors,, is (4C4) = 1
These add up to 11 – but we must also add the number ‘1’ – as you know, 1 is not a prime number, so we could not include it
in the above analysis. Another way to say this, is that we also obtain a factor of 84 by utilizing zero of its prime facors –
namely the factor ‘1’
In summary, the integer 84 has 12 positive integer factors.
Wouldn’t it be neat if there were a FORMULA for all of this – it sure would save us a lot of time and heartache on the GMAT.
You are in luck – there is:
All you have to do, is commit the formula to memory
Remember that 84 can be written as (22)(31)(71) - here we give you the formula:
If Q consists of (P1)x(P2)y(P3)z then Q has (x +1)(y +1)(z +1) factors (where P1 …P3 are distinct prime
factors of Q)
Under this formula, 84 would have to have (2 + 1)(1 + 1)(1 +1) factors. That is to say, 3 times 2 times 2 = 12
Therefore, if the GMAT question asks you to figure out how many factor the integer 1001 has – not to worry. Apply the
formula.
The integer 1001 is often found on the GMAT because it is the product of 7 times 11 times 13.
Hence, 1001 would have (1 + 1) (1 + 1) (1 + 1) factors, or 8 factors.
They would be: 1, 7, 11, 13, 711, 1113, 713, 71113
Answers:
Example 5: 4x2 – 33x + 35 = 0
(4x – 5)(x – 7) = 0
Example 6: 3x2 + 11x – 40 = 0
(3x – 5)(x + 8) = 0
Example 7: 2x2 – 15x – 18 = 0
(2x – 3)(x – 6) = 0
Example 8: 7x2 – 121x – 34 = 0
(7x – 2)(x – 17) = 0