Polynomial equations mean the relation between numbers and variables are explained in a pattern. Polynomial is one of the major concepts of Mathematics. In Maths, we have studied a variety of equations formed with algebraic expressions. When we talk about polynomials, it is also a form of the algebraic equation and polynomial formula. Let us get started.

**Polynomial Formula**

**What is a Polynomial?**

Polynomial is an algebraic expression, in which the variables have non-negative powers. A polynomial function is an equation, which consists of a single independent variable, where the variable can occur in the equation more than one time with different degrees of the exponent.Â A polynomial expression is the one, which has more than two algebraic terms. As the name suggests, Polynomial is a repetitive addition of a monomial or a binomial.

### Types of Polynomial

Polynomial equation is of four types :

**Monomial:**This type of polynomial contains only one term. For example, x^{2}Â , x, y, 3y, 4z**Binomial:**This type of polynomial contains two terms. For example, x^{2}Â â€“ 10x**Trinomial:**This type of polynomial contains three terms. For example, x^{2}Â â€“ 10x+9**Quadratic Polynomial:**This type of polynomial contains four terms. For example, x^{3}+2x^{2}Â â€“ 10x+7

The generalÂ Polynomial FormulaÂ is,

**F(x) = a _{n}x^{n}Â + bx^{n-1}Â + a_{n-2}x^{n-2}Â + â€¦â€¦.. + rx +s**

- IfÂ nÂ is a natural number: a
^{n}Â – b^{n}Â = (a – b)(a^{n-1}Â + a^{n-2}b +…+ b^{n-2}a + b^{n-1}) - IfÂ nÂ is evenÂ (n = 2a): x
^{n}Â + y^{n}Â = (x + y)(x^{n-1}Â – x^{n-2}y +…+ y^{n-2}x – y^{n-1}) - IfÂ nÂ is odd number: x
^{n}Â + y^{n}Â = (x + y)(x^{n-1}Â – x^{n-2}y +…- y^{n-2}x + y^{n-1})

**Polynomial Identities**

- (x + y)
^{2}= x^{2}Â + 2xy + y^{2} - (x â€“ y)
^{2}= x^{2}Â â€“ 2xy + y^{2} - x
^{2}â€“ y^{2}Â = (x + y)(x â€“ y) - (x + a)(x + b) = x
^{2}+ (a + b)x + ab - (x + y + z)
^{2}= x^{2}Â + y^{2}Â + c^{2}Â + 2xy + 2yz + 2zx - (x + y)
^{3}= x^{3}Â + y^{3}Â + 3xy (x + y) - (x â€“ y)
^{3}= x^{3}Â â€“ y^{3}Â â€“ 3xy (x â€“ y) - x
^{3}+ y^{3}Â = (x + y)(x^{2}Â â€“ xy + y^{2}) - x
^{3}â€“ y^{3}Â = (x â€“ y)(x^{2}Â + xy + y^{2}) - x
^{3}+ y^{3}Â + z^{3Â }â€“ 3xyzÂ = (x + y + z)(x^{2}Â + y^{2}Â + z^{2}Â â€“ xy â€“ yz â€“ zx) - ax
^{2}+ bx + c = 0 then x = \frac{( -b \pm \sqrt{(b^{2}– 4ac)} )}{2a}

## Solved Examples

Q1. Solve the equation: x^{2}Â +16x + 64 = 0

Solution: Factors of x^{2}Â +16x + 64

We the factors of the 64 are 2, 4, 8, 16, 32, 64. The sum of middle term is 8.

Now, x^{2 }+ (8 + 8) x + 8 Í¯ 8

=x^{2 }+8x +8x + 8 Í¯ 8

= x(x + 8) + 8 (x + 8) = (x + 8) (x + 8)

Q.2. Solve x^{3} â€“ 7x^{2} + 12x = 0

Solution: To factorÂ x^{3}âˆ’7x^{2}+12x

= x (x^{2}âˆ’5x+6)

Now, we factorise x^{2}âˆ’7x+12

We the factors of the 12 = 3, 4

Now, x^{2 }âˆ’ (3 + 4) x + 3 Í¯ 4

=x^{2 }-3x -4x + 3 Í¯ 4

= x(x – 3) â€“ 4 (x – 3) = (xâˆ’3) (xâˆ’4)

FactorsÂ of x^{3}âˆ’7x^{2}+12xÂ are Â x (xâˆ’3) (xâˆ’4)

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