# Polynomial Identities

When we have a sum(difference) of two or three numbers to power of 2 or 3 and we need to remove the brackets we use
polynomial identities(short multiplication formulas):
(x + y)2 = x2 + 2xy + y2
(x - y)2 = x2 - 2xy + y2

Example 1: If x = 10, y = 5a
(10 + 5a)2 = 102 + 2·10·5a + (5a)2 = 100 + 100a + 25a2

Example 2: if x = 10 and y is 4
(10 - 4)2 = 102 - 2·10·4 + 42 = 100 - 80 + 16 = 36

The opposite is also true:
25 + 20a + 4a2 = 52 + 2·2·5 + (2a)2 = (5 + 2a)2

Consequences of the above formulas:

(-x + y)2 = (y - x)2 = y2 - 2xy + x2
(-x - y)2 = (-(x + y))2 = (x + y)2 = x2 + 2xy + y2

Formulas for 3rd degree:

(x + y)3 = x3 + 3x2y + 3xy2 + y3
(x - y)3 = x3 - 3x2y + 3xy2 - y3

Example: (1 + a2)3 = 13 + 3.12.a2 + 3.1.(a2)2 + (a2)3 = 1 + 3a2 + 3a4 + a6

(x + y + z)2 = x2 + y2 + z2 + 2xy + 2xz + 2yz
(x - y - z)2 = x2 + y2 + z2 - 2xy - 2xz + 2yz

### Factor Rules

x2 - y2 = (x - y)(x + y)

x2 + y2 = (x + y)2 - 2xy
or
x2 + y2 = (x - y)2 + 2xy

Example: 9a2 - 25b2 = (3a)2 - (5b)2 = (3a - 5b)(3a + 5b)

x3 - y3 = (x - y)(x2 + xy + y2)
x3 + y3 = (x + y)(x2 - xy + y2)

If n is a natural number

xn - yn = (x - y)(xn-1 + xn-2y +...+ yn-2x + yn-1)

If n is even (n = 2k)

xn + yn = (x + y)(xn-1 - xn-2y +...+ yn-2x - yn-1)

If n is odd (n = 2k + 1)

xn + yn = (x + y)(xn-1 - xn-2y +...- yn-2x + yn-1)

#### More Algebraic Formulas

2(a2 + b2) = (a + b)2 + (a - b)2
(a + b)2 - (a - b)2 = 4ab
(a - b)2 = (a + b)2 - 4ab
a4 + b4 = (a + b)(a - b)[(a + b)2 - 2ab]

#### Problems Involving Polynomial Identities

1) Solve the equation: x2 - 25 = 0
Solution: x2 - 25 = (x - 5)(x + 5)
=> we have to solve the following 2 equations:
x - 5 = 0 or x + 5 = 0
so the equation have two decisions: x = 5 and x = -5

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