# Factoring Completely

## Key Questions

• You (try to) divide your number by successive primes $2 , 3 , 5 , 7 , 11. . .$

Example :
Say you want to factor $756$
Even number, divide by 2: $756 = 2 \cdot 378$
Even, divide by 2: $756 = 2 \cdot 2 \cdot 189$
Odd, so try 3: $756 = 2 \cdot 2 \cdot 3 \cdot 63$
And again: $756 = 2 \cdot 2 \cdot 3 \cdot 3 \cdot 21$
And again: $756 = 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 7$
7 is a prime, so you're finished.

You write the result as $756 = {2}^{2} \cdot {3}^{3} \cdot 7$

Remark :
You never have to try a prime that is larger than the square root of the number left, because if that would divide evenly, the result is a smaller number, and you would have already tried that.

• For factoring polynomials, "factoring" (or "factoring completely") is always done using some set of numbers as possible coefficient.

We say we are factoring "over" the set.

${x}^{3} - {x}^{2} - 5 x + 5$ can be factored
over the integers as $\left(x - 1\right) \left({x}^{2} - 5\right)$

${x}^{2} - 5$ cannot be factored using integer coefficients. (It is irreducible over the integers.)

over the real numbers ${x}^{2} - 5 = \left(x - \sqrt{5}\right) \left(x + \sqrt{5}\right)$

One more:
${x}^{2} + 1$ cannot be factored over the real numbers, but over the complex numbers it factors as
${x}^{2} + 1 = \left(x - \sqrt{- 1}\right) \left(x + s q r \left(- 1\right)\right)$

Also written: $\left(x - i\right) \left(x + i\right)$