Difference between revisions of "Factoring"

Revision as of 03:12, 28 July 2016

Factoring is an essential part of any mathematical toolbox. To factor, or to break an expression into factors, is to write the expression (often an integer or polynomial) as a product of different terms. This often allows one to find information about an expression that was not otherwise obvious.

Differences and Sums of Powers

Using the formula for the sum of a geometric sequence, it's easy to derive the general formula for difference of powers:

$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b + \cdots + ab^{n-2} + b^{n-1})$

If $n=2$ , this creates the difference of squares factorization, $a^2-b^2=(a+b)(a-b)$

This leads to the difference of cubes factorization, $a^3-b^3=(a-b)(a^2+ab+b^2)$

In addition, if $n$ is odd:

$a^n+b^n=(a+b)(a^{n-1} - a^{n-2}b + a^{n-3}b^2 - a^{n-4}b^3 + \cdots + b^{n-1})$

This also leads to the formula for the sum of cubes,

$a^3+b^3=(a+b)(a^2-ab+b^2)$

Another way to discover these factorizations is the following: the expression $a^n - b^n$ is equal to zero if $a = b$ . If one factorizes a product which is equal to zero, one of the factors must be equal to zero, so $a^n - b^n$ must have a factor of $a - b$ . Similarly, we note that the expression $a^n + b^n$ when $n$ is odd is equal to zero if $a = -b$ , so it must have a factor of $a - (-b) = a + b$ . Note that when $n$ is even, $(-b)^n + b^n = 2b^n$ , rather than 0, so this gives us no useful information.

Vieta's/Newton Factorizations

These factorizations are useful for problems that could otherwise be solved by Newton sums or problems that give a polynomial and ask a question about the roots. Combined with Vieta's formulas, these are excellent factorizations that show up everywhere.

$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$

$(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a)$

$(a+b+c)^5=a^5+b^5+c^5+5(a+b)(b+c)(c+a)(a^2+b^2+c^2+ab+bc+ca)$

$(a+b+c)^7=a^7+b^7+c^2+7(a+b)(b+c)(c+a)((a^2+b^2+c^2+ab+bc+ca)^2+abc(a+b+c))$

Other Useful Factorizations

$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
Binomial theorem
Simon's Favorite Factoring Trick
Sophie Germain Identity

Practice Problems

Prove that $n^2 + 3n + 5$ is never divisible by 121 for any positive integer ${n}$ .
Prove that $2222^{5555} + 5555^{2222}$ is divisible by 7. - USSR Problem Book
Factor $(x-y)^3 + (y-z)^3 + (z-x)^3$ .
Factor $x^4 + 1$ into two polynomials with real coefficients.
Given that $a+b+c=0$ , prove that $abc=\dfrac{a^3+b^3+c^3}{3}$ .

Other Resources

More Common Factorizations.

@@ Line 29: / Line 29: @@
 *<math>(a+b+c)^5=a^5+b^5+c^5+5(a+b)(b+c)(c+a)(a^2+b^2+c^2+ab+bc+ca) </math>
+*<math>(a+b+c)^7=a^7+b^7+c^2+7(a+b)(b+c)(c+a)((a^2+b^2+c^2+ab+bc+ca)^2+abc(a+b+c))</math>
 == Other Useful Factorizations ==
 *<math>a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)</math>

Page

Toolbox

Search