Factorization is the decomposition of an expression into a product of its factors.

The following are common factorizations.

  1. For any positive integer nn, anbn=(ab)(an1+an2b++abn2+bn1).a^n-b^n = (a-b)(a^{n-1} + a^{n-2} b + \ldots + ab^{n-2} + b^{n-1} ). In particular, for n=2 n=2, we have a2b2=(ab)(a+b) a^2-b^2=(a-b)(a+b).

  2. For n n an odd positive integer, an+bn=(a+b)(an1an2b+abn2+bn1). a^n+b^n = (a+b)(a^{n-1} - a^{n-2} b + \ldots - ab^{n-2} + b^{n-1} ).

  3. a2±2ab+b2=(a±b)2 a^2 \pm 2ab + b^2 = (a\pm b)^2

  4. x3+y3+z33xyz=(x+y+z)(x2+y2+z2xyyzzx) x^3 + y^3 + z^3 - 3 xyz = (x+y+z) (x^2+y^2+z^2-xy-yz-zx)

  5. (ax+by)2+(aybx)2=(a2+b2)(x2+y2) (ax+by)^2 + (ay-bx)^2 = (a^2+b^2)(x^2+y^2). (axby)2(aybx)2=(a2b2)(x2y2) (ax-by)^2 - (ay-bx)^2 = (a^2-b^2)(x^2-y^2).

  6. x2y+y2z+z2x+x2z+y2x+z2y+2xyz=(x+y)(y+z)(z+x) x^2 y + y^2 z + z^2 x + x^2 z + y^2 x + z^2 y +2xyz= (x+y)(y+z)(z+x).

Factorization often transforms an expression into a form that is more easily manipulated algebraically, that has easily recognizable solutions, and that gives rise to clearly defined relationships.

Worked Examples

1. Find all ordered pairs of integer solutions (x,y) (x,y) such that 2x+1=y22^x+ 1 = y^2.

Solution: We have 2x=y21=(y1)(y+1)2^x = y^2-1 = (y-1)(y+1). Since the factors (y1)(y-1) and (y+1)(y+1) on the right hand side are integers whose product is a power of 2, both (y1)(y-1) and (y+1)(y+1) must be powers of 2. Furthermore, their difference is

(y+1)(y1)=2, (y+1)-(y-1)=2,

implying the factors must be y+1=4y+1 = 4 and y1=2y-1 = 2. This gives y=3 y=3, and thus x=3x=3. Therefore, (3,3)(3, 3) is the only solution.

2. Factorize the polynomial

f(a,b,c)=ab(a2b2)+bc(b2c2)+ca(c2a2).f(a, b, c) = ab(a^2-b^2) + bc(b^2-c^2) + ca(c^2-a^2).

Solution: Observe that if a=b a=b, then f(a,a,c)=0f(a, a, c) =0; if b=cb=c, then f(a,b,b)=0f(a, b, b)=0; and if c=a c=a, then f(c,b,c)=0 f(c,b,c)=0. By the Remainder-Factor Theorem, (ab),(bc), (a-b), (b-c), and (ca) (c-a) are factors of f(a,b,c) f(a,b,c). This allows us to factorize

f(a,b,c)=(ab)(bc)(ca)(a+b+c).f(a,b,c) = -(a-b)(b-c)(c-a)(a+b+c).

Note by Arron Kau
7 years, 3 months ago

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Quite helpful as it seems.

Syed Hamza Khalid - 4 years, 3 months ago

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