# 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)**:

^{2}= x

^{2}+ 2xy + y

^{2}

(x - y)

^{2}= x

^{2}- 2xy + y

^{2}

Example 1: If x = 10, y = 5a

(10 + 5a)^{2} = 10^{2} + 2·10·5a + (5a)^{2} = 100 + 100a + 25a^{2}

Example 2: if x = 10 and y is 4

(10 - 4)^{2} = 10^{2} - 2·10·4 + 4^{2} = 100 - 80 + 16 = 36

The opposite is also true:

25 + 20a + 4a^{2} = 5^{2} + 2·2·5 + (2a)^{2} = (5 + 2a)^{2}

Consequences of the above formulas:

^{2}= (y - x)

^{2}= y

^{2}- 2xy + x

^{2}

(-x - y)

^{2}= (-(x + y))

^{2}= (x + y)

^{2}= x

^{2}+ 2xy + y

^{2}

Formulas for 3rd degree:

^{3}= x

^{3}+ 3x

^{2}y + 3xy

^{2}+ y

^{3}

(x - y)

^{3}= x

^{3}- 3x

^{2}y + 3xy

^{2}- y

^{3}

Example: (1 + a^{2})^{3} = 1^{3} + 3.1^{2}.a^{2} +
3.1.(a^{2})^{2} + (a^{2})^{3} = 1 + 3a^{2} + 3a^{4} + a^{6}

^{2}= x

^{2}+ y

^{2}+ z

^{2}+ 2xy + 2xz + 2yz

(x - y - z)

^{2}= x

^{2}+ y

^{2}+ z

^{2}- 2xy - 2xz + 2yz

### Factor Rules

^{2}- y

^{2}= (x - y)(x + y)

x

^{2}+ y

^{2}= (x + y)

^{2}- 2xy

or

x

^{2}+ y

^{2}= (x - y)

^{2}+ 2xy

Example: 9a^{2} - 25b^{2} = (3a)^{2} - (5b)^{2} =
(3a - 5b)(3a + 5b)

^{3}- y

^{3}= (x - y)(x

^{2}+ xy + y

^{2})

x

^{3}+ y

^{3}= (x + y)(x

^{2}- xy + y

^{2})

If n is a natural number

^{n}- y

^{n}= (x - y)(x

^{n-1}+ x

^{n-2}y +...+ y

^{n-2}x + y

^{n-1})

If n is even (n = 2k)

^{n}+ y

^{n}= (x + y)(x

^{n-1}- x

^{n-2}y +...+ y

^{n-2}x - y

^{n-1})

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

^{n}+ y

^{n}= (x + y)(x

^{n-1}- x

^{n-2}y +...- y

^{n-2}x + y

^{n-1})

#### More Algebraic Formulas

^{2}+ b

^{2}) = (a + b)

^{2}+ (a - b)

^{2}

(a + b)

^{2}- (a - b)

^{2}= 4ab

(a - b)

^{2}= (a + b)

^{2}- 4ab

a

^{4}+ b

^{4}= (a + b)(a - b)[(a + b)

^{2}- 2ab]

#### Problems Involving Polynomial Identities

1) Solve the equation: x^{2} - 25 = 0

**Solution:** x^{2} - 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

#### Related Resources:

Simplifying polynomial expressions - problems with solutions