Many algebraic equations require factoring of polynomials in order to develop a solution.

Factoring Polynomials with Common Monomial Factors

In order to factor polynomials with common monomial factors, we must find the highest monomial factor that will divide exacly into each term of the polynomial. For example:

$3x+2bx+9x² = x\left(3+2b+9x\right)$

In the example, the only factor every term has in common is a single $x$.

Factoring the Difference of Two Squares

Since the product of $x+y$ and $x-y$ is $x²-y²$, the factors of $x²-y²$ are $\left(x+y\right)$ and $\left(x-y\right)$. Therefore $x²-y² = \left(x+y\right)\left(x-y\right)$. In order to factor an expression of this type, we must find the square root of each term. We then write two factors: one is the sum of the two square roots; the other is the difference of the two square roots. For example:

$x²-9$  →  $$\sqrt{\textrm{x}²}$$$= ±x$, $$\sqrt{9}$$$= ±3$

$x²-9 = \left(x+3\right)\left(x-3\right) = \left(-x+3\right)\left(-x-3\right)$

Note that there are two possible answers to this, though the result using the positive roots is usually used. Keep in mind, however, that the difference in the signs is often important to the proper solution of an equation.

Factoring Trinomials of the form $x²+bx+c$

A trinomial of the form $x²+bx+c$ is a quadratic where the coefficient of $x²$ is $1$. In order to factor an equation of this form, we must first find two binomials that have the following characteristics:

1) The product of the first terms of both binomials must equal the first term of the trinomial.
2) The product of the last terms of both binomials must be equal to the last term of the trinomial.
3) The algebraic sum of the last terms of the binomials must be equal to the coefficient of the middle term of the trinomial.

For example, let's factor $x²+7x+10$:

1) The product of the first terms of both binomials must equal the first term of the trinomial:

2) The product of the last terms of both binomials must be equal to the last term of the trinomial:
$1\cdot 10=10, 2\cdot 5=10$

3) The algebraic sum of the last terms of the binomials must be equal to the coefficient of the middle term of the trinomial:
$5+2=7$

Our result is then $x²+7x+10 = \left(x+5\right)\left(x+2\right)$

Note

In steps 2 and 3 above, we could have just as likely had fractional or negative values in the mix. This is a simple example, used to show the process. In actuality, factoring trinomials can become a chore of finding common terms. If factoring takes too long, the trinomial expression can be treated as a quadratic equation in order to find the roots.