The aim of factoring is usually to reduce something to "basic building blocks," such as numbers to prime numbers, or polynomials to irreducible polynomials. Factoring integers is covered by the fundamental theorem of arithmetic and factoring polynomials by the fundamental theorem of algebra. Vieta's formulas relate the coefficients of a polynomial to its roots.

The opposite of factorization is expansion. This is the process of multiplying together factors to recreate the original, "expanded" polynomial.

Integer factorization for large integers appears to be a difficult problem. There is no known method to carry it out quickly. Its complexity is the basis of the assumed security of some public key cryptography algorithms, such as RSA.

A matrix can also be factorized into a product of matrices of special types, for an application in which that form is convenient. One major example of this uses an orthogonal or unitary matrix, and a triangular matrix. There are different types: QR decomposition, LQ, QL, RQ, RZ.

Another example is the factorization of a function as the composition of other functions having certain properties; for example, every function can be viewed as the composition of a surjective function with an injective function. This situation is generalized by factorization systems.

Prime factorization of an integer

Factoring a quadratic polynomial

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

where $alpha$ and $beta$ are the two roots of the polynomial, found with the quadratic formula.

Polynomials factorable over the integers

$ax^2+bx+c\; ,!$

would be factored into:

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

$mn\; =\; a,\; ,$

$pq\; =\; c,\; mbox\{\; and\}\; ,$

$pn\; +\; mq\; =\; b.\; ,$

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 $b^2-4ac$ 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.

Perfect square trinomials

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

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

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

Sum/difference of two squares

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

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

For example, $4x^2\; +\; 49$ can be factored into $(2x\; +\; 7i)(2x\; -\; 7i)$.

Factoring other polynomials

Sum/difference of two cubes

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

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

Sum/difference of any two numbers raised to the same power

$a^n\; -\; b^n\; =\; (a\; -\; b)(a^\{n-1\}\; +\; a^\{n-2\}b\; +\; a^\{n-3\}b^2\; +\; ...\; +\; a^2b^\{n-3\}\; +\; ab^\{n-2\}\; +\; b^\{n-1\})\; ,!$

Also, $(a+b)$ is a factor of $a^n\; -\; b^n$ where $n$ is a positive even integer. Such that,

$a^n\; -\; b^n\; =\; (a\; +\; b)(a^\{n-1\}\; -\; a^\{n-2\}b\; +\; a^\{n-3\}b^2\; -\; ...\; -\; a^2b^\{n-3\}\; +\; ab^\{n-2\}\; -\; b^\{n-1\})\; ,!$

Likewise, $(a+b)$ is a factor of $a^n\; +\; b^n$ where $n$ is a positive odd integer. So that,

$a^n\; +\; b^n\; =\; (a\; +\; b)(a^\{n-1\}\; -\; a^\{n-2\}b\; +\; a^\{n-3\}b^2\; -\; ...\; +\; a^2b^\{n-3\}\; -\; ab^\{n-2\}\; +\; b^\{n-1\})\; ,!$

Factoring by grouping

For example, suppose you had the expression

$4x^3sin^2x-312x^2sin^2x+4620xsin^2x-8024sin^2x-3x^3+234x^2-3465x+6018\; ,$

$(4x^3sin^2x-312x^2sin^2x+4620xsin^2x-8024sin^2x)-(3x^3-234x^2+3465x-6018)\; ,$

$4sin^2x(x^3-78x^2+1155x-2006)\; -\; 3(x^3-78x^2+1155x-2006)\; ,$

$(4sin^2x\; -3)(x^3-78x^2+1155x-2006)\; ,$

$(4sin^2x\; -3)(x-59)(x-17)(x-2)\; ,$

$(2sin\; x+sqrt\; 3)(2sin\; x-sqrt\; 3)(x-59)(x-17)(x-2)\; ,$

Other common formulas

Factoring in mathematical logic

See also

• Vieta's formulas
• Program synthesis
• Matrix decomposition
• Unique factorization
• Polynomial expansion, the opposite of factorization
• Factor group
• Factor ring
• FOIL rule
• Deduplication - similar concept in other contexts

External links

• A page about factorization, Algebra, Factoring
• WIMS Factoris is an online factorization tool.
• Polynomial Factoring is a comprehensive tutorial resource on basic factoring of polynomials.