# 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 \$.

Again \$ p = 4 \$ (and thus \$ \ displaystyle - \ frac {p} {2} = - 2 \$). Now \$ q = \$ 5.

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