# How to factor x + 1 2 9

## Factoring and quadratic equations

### Theorem of Vieta

With the __Theorem of Vieta__ you can quickly factorize simple quadratic expressions.

$$ \ text {Is} \ quad x ^ 2 + px + q = (x + x_1) (x + x_2), \ quad \ text {then} $$

$$ p = x_1 + x_2 \ quad \ text {and} \ quad q = x_1 \ cdot x_2. $$

Of course, $ p $ and $ q $ can also be negative.

** Important:** If $ p $ and $ q $ are integers and $ x ^ 2 + px + q $ can even be written as $ (x + x_1) (x + x_2) $, $ x_1 $ and $ x_2 $ must be divisors of $ q $ be.

**Example:** $ x ^ 2-6x + $ 5.

Here $ q = 5 $ and $ p = -6 $.

$ q = 5 $ has the divisors $ \ pm 1 $ and $ \ pm 5 $. Then you make a table like this:

$$ \ begin {array} {c | c | c} x_1 & x_2 & p = x_1 + x_2 \ 5 & 1 & 6 \ - 1 & -5 & -6 \ end {array} $$

You omit all combinations that have already occurred in a different order, e.g. $ x_1 = 1 $ and $ x_2 = 5 $.

The second line shows that $ x_1 + x_2 = -1-5 = -6 = p $ is fulfilled.

So the solution is $ x ^ 2-6x + 5 = (x-5) (x-1) $.

Example 1Example 2 calculate yourself

### Square addition

The quadratic addition is a preliminary stage of the $ p $ - $ q $ formula and is used to summarize different powers of a variable.

The square addition to $ x ^ 2 + px $ is $ \ displaystyle x ^ 2 + px + \ big (\ frac {p} {2}) ^ 2 = \ big (x + \ frac {p} {2} \ big) ^ $ 2

**Example: $ x ^ 2-6x + $ 5**

It is $ p = -6 $, $ \ displaystyle \ frac {p} {2} = - 3 $ and therefore $ \ displaystyle \ left (\ frac {p} {2} \ right) ^ 2 = (- 3) ^ 2 = $ 9

So $ x ^ 2-6x + 5 = x ^ 2-6x + 9-9 + 5 = (x-3) ^ 2-4 $

Example 1Example 2Example 3 calculate yourself

### p-q formula

The __quadratic equation__ $ x ^ 2 + px + q = 0 $ is represented by the two numbers $ \ displaystyle x_ {1,2} = - \ frac {p} {2} \ pm \ sqrt {\ Big (- \ frac {p} { 2} \ Big) ^ 2-q} $ solved.

If the expression under the root is zero, there is only one solution $ \ displaystyle x_1 = x_2 = \ frac {p} {2} $.

If the expression under the root is negative, the quadratic equation has no real solution.

**Example 1:** $ x ^ 2 + 4x + 3 = 0 $.

Here $ p = 4 $ (and thus $ \ displaystyle - \ frac {p} {2} = - 2 $) and $ q = 3 $.

Therefore $ x_ {1,2} = - 2 \ pm \ sqrt {(- 2) ^ 2-3} = - 2 \ pm \ sqrt {1} = - 2 \ pm 1 $.

So the solutions are $ x_1 = -2-1 = -3 $ and $ x_2 = -2 + 1 = -1 $.

**Example 2:** $ x ^ 2 + 4x + 4 = 0 $.

Again $ p = 4 $ (and thus $ \ displaystyle - \ frac {p} {2} = - 2 $). Now $ q = $ 4.

Therefore $ x_ {1,2} = - 2 \ pm \ sqrt {(- 2) ^ 2-4} = - 2 \ pm \ sqrt {0} = - 2 $.

There is only one solution here: $ x_1 = x_2 = -2 $

**Example 3: **$ x ^ 2 + 4x + 5 = 0 $.

Therefore $ x_ {1,2} = - 2 \ pm \ sqrt {(- 2) ^ 2-5} = - 2 \ pm \ sqrt {-1} $.

Because the expression under the root is negative, there is no real solution.

Example 1Example 2 calculate yourself

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