Factorization

Quadratic polynomials

Any quadratic polynomial over the complex numbers (polynomials of the form ax2 + bx + c where a, b, and c ∈ \mathbb{C}) can be factored into an expression with the form a(x - \alpha)(x - \beta) \ using the quadratic formula. The method is as follows:

{ax^2 + bx + c = a(x - \alpha)(x - \beta) = a\left(x - \frac{-b + \sqrt{b^2-4ac}}{2a}\right) \left(x - \frac{-b - \sqrt{b^2-4ac}}{2a}\right),}

where α and β are the two roots of the polynomial, found with the quadratic formula.
[edit] Polynomials factorable over the integers

(mx+p)(nx+q),\,\!

where

mn = a,\ pq = c\,\!

and

pn + mq = b. \,

You can then set each binomial equal to zero, and solve for x to reveal the two roots. Factoring does not involve any other formulas, and is mostly just something you see when you come upon a quadratic equation.

Take for example 2x2 − 5x + 2 = 0. Because a = 2 and mn = a, mn = 2, which means that of m and n, one is 1 and the other is 2. Now we have (2x + p)(x + q) = 0. Because c = 2 and pq = c, pq = 2, which means that of p and q, one is 1 and the other is 2 or one is −1 and the other is −2. A guess and check of substituting the 1 and 2, and −1 and −2, into p and q (while applying pn + mq = b) tells us that 2x2 − 5x + 2 = 0 factors into (2x − 1)(x − 2) = 0, giving us the roots x = {0.5, 2}

Note: A quick way to check whether the second term in the binomial should be positive or negative (in the example, 1 and 2 and −1 and −2) is to check the second operation in the trinomial (+ or −). If it is +, then check the first operation: if it is +, the terms will be positive, while if it is −, the terms will be negative. If the second operation is −, there will be one positive and one negative term; guess and check is the only way to determine which one is positive and which is negative.

If a polynomial with integer coefficients has a discriminant that is a perfect square, that polynomial is factorable over the integers.

For example, look at the polynomial 2x2 + 2x - 12. If you substitute the values of the expression into the quadratic formula, the discriminant b2 − 4ac becomes 22 - 4 × 2 × -12, which equals 100. 100 is a perfect square, so the polynomial 2x2 + 2x - 12 is factorable over the integers; its factors are 2, (x - 2), and (x + 3).

Now look at the polynomial x2 + 93x - 2. Its discriminant, 932 - 4 × 1 × -2, is equal to 8657, which is not a perfect square. So x2 + 93x - 2 cannot be factored over the integers.
[edit] Perfect square trinomials
A visual proof of the identity (a+b)2=a2+2ab+b2

Some quadratics can be factored into two identical binomials. These quadratics are called perfect square trinomials. Perfect square trinomials can be factored as follows:

a^2 + 2ab + b^2 = (a + b)^2,\,\!

and

a^2 - 2ab + b^2 = (a - b)^2.\,\!

[edit] Sum/difference of two squares
Main article: Difference of two squares

Another common type of algebraic factoring is called the difference of two squares. It is the application of the formula

a^2 - b^2 = (a+b)(a-b),\,\!

to any two terms, whether or not they are perfect squares. If the two terms are subtracted, simply apply the formula. If they are added, the two binomials obtained from the factoring will each have an imaginary term. This formula can be represented as

a^2 + b^2 = (a+bi)(a-bi). \,\!

For example, 4x2 + 49 can be factored into (2x + 7i)(2x − 7i).
[edit] Factoring by grouping

Another way to factor some polynomials is factoring by grouping. This is done by placing the terms in the polynomial into two or more groups, where each group can be factored by a known method. The results of these factorizations can sometimes be combined to make an even more simplified expression. For example, to factor the polynomial

4x^2+20x+3yx+15y \,

Group similar terms, (4x^2+20x)+(3yx+15y)\,

Factor out Greatest Common Factor, 4x(x+5)+3y(x+5)\,

Factor out binomial (x+5)(4x+3y)\,
[edit] Factoring other polynomials
[edit] Sum/difference of two cubes

Another formula for factoring is the sum or difference of two cubes. The sum can be represented by

a^3 + b^3 = (a + b)(a^2 - ab + b^2),\,\!

and the difference by

a^3 - b^3 = (a - b)(a^2 + ab + b^2).\,\!

For example, x3 − 103 (or x3 − 1000) can be factored into (x − 10)(x2 + 10x + 100).
[edit] See also