Simple Factoring of Quadratics

How do you factor quadratics?

We are going to learn a simple method for factoring quadratics when the leading multiplier is 1 and the quadratic has non-complex integer solutions that are positive.
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Lets expand (x+8)(x+21)
(x+8)(x+21)=x²+29x+168

Now how would we factor something like this if we did not know what the factors are.
If we expand a quadratic, it looks like this:  (x+A)(x+B)=x²+(A+B)x+AB
The second term or the number before x is the addition of 2 roots or A+B
The last term is the multiplication of 2 roots or AB.

First If we factor 168 into all of its roots, we will find all potential factors:

168 factors into these pairs:     1         168
                                                 2         84
                                                 3         56
                                                 4         42
                                                 6         28
                                                 8         21
                                                 12       14
                                                 24         7

Now we see which 2 pairs add up to 29.  The only 2 that add up to 29 are 8 and 21.

Lets do a couple of simpler problems where we do not already know the factors:

Example 1:      x²+12x+32

Lets factor 32:                                                     32
                                                                          2    16
                                                                                2    8
                                                                                     2   4
                                                                                         2   2
                                                      or
                                                    1         32
                                                    2         16
                                                    4          8

So
1 and 32 add up to 33
2 and 16 add up to 18
4 and 8 add up to 12 so this is our solution set so it looks like this factored:
x²+12x+32=(x+4)(x+8)

One more example:
x²+10x+21

 Lets factor 21:                                                 1        21
                                                                         3         7

We see the only 2 number that add up to 10 are 3 and 7 so:
x²+10x+21=(x+3)(x+7)

In a future post we will do numbers with negative roots.

In the Prime of Its Life

The first primes less than 100  and rules to easily find

Here is a common scenario:

You are taking the SAT test and you are asked how many prime numbers there are between 30 and 80.  You know what prime numbers are and you know what prime numbers are but you have limited time and no table to help you.  You decide to skip the question because you do not have enough time to finish it by the end of the test.

This is a very common test question yet it is not trivial because most people have not memorized the prime numbers so here is a quick was to determine the prime numbers. 

1   2   3   4   5   6   7   8   9   10            (4 primes)

11 12 13 14 15 16 17 18 19 20            (4 primes)

21 22 23 24 25 26 27 28 29 30            (2 primes)

31 32 33 34 35 36 37 38 39 40            (2 primes)

41 42 43 44 45 46 47 48 49 50            (3 primes)

51 52 53 54 55 56 57 58 59 60            (2 primes)

61 62 63 64 65 66 67 68 69 70            (2 primes)

71 72 73 74 75 76 77 78 79 80            (3 primes)

81 82 83 84 85 86 87 88 89 90            (2 primes)         

91 92 93 94 95 96 97 98 99 100          (1 prime)

                                                 
 The sieve of Eratosthenesis is a methodology for finding all prime numbers.  Without going into how the sieve of Eratosthenes works, I will say the methodology I present will be similar but not the same and it will be easier to use for numbers between 1 and 100.  

After 2, all even numbers are not prime.  
After 5, all numbers that end in 5 are not prime.
No numbers that end in 0 are prime.

Now here is the trick.  If you are looking at the columns where the numbers end in a 1, 3, 7, or 9, you will see there are quite a few primes.  Of these numbers that end in 1, 3, 7, and 9, all numbers except 1, 3, 49, 77, 91, are divisible by 3.

Okay lets check this out.
21÷3=21              33÷3=11               27÷3=9               39÷3=13
51÷3=14              63÷3=21              57÷3=19              81÷3=27
81÷3=27              93÷3=31              87÷3=29              69÷3=23 

You can easily find any prime on numbers less than 100.  If you are asked if 87 is prime, you will say no it isn't because it is divisible by 3.  If you use the 3 divisibility rule, it is even easier.  

Also one more trick for those taking the SAT or any standardized test, the number of primes in each row after the tens are:
 2, 2, 3, 2, 2, 3, 2, 2, 1 and the ones and 10's each have 4.  Refer to the table above.
If you are asked the question, "How many prime numbers are there between 30 and 80?"  You can answer 16.  2+3+2+2+3+2+2=16