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. 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}, x, y, 3y, 4z**Binomial:**This type of polynomial contains two terms. For example, x^{2}– 10x**Trinomial:**This type of polynomial contains three terms. For example, x^{2}– 10x+9**Quadratic Polynomial:**This type of polynomial contains four terms. For example, x^{3}+2x^{2}– 10x+7

The general Polynomial Formula is,

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

- If n is a natural number: a
^{n}– b^{n}= (a – b)(a^{n-1}+ a^{n-2}b +…+ b^{n-2}a + b^{n-1}) - If n is even (n = 2a): x
^{n}+ y^{n}= (x + y)(x^{n-1}– x^{n-2}y +…+ y^{n-2}x – y^{n-1}) - If n is odd number: x
^{n}+ y^{n}= (x + y)(x^{n-1}– x^{n-2}y +…- y^{n-2}x + y^{n-1})

**Polynomial Identities**

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

## Solved Examples

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

Solution: Factors of x^{2} +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 ͯ 8

=x^{2 }+8x +8x + 8 ͯ 8

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

Q.2. Solve x^{3} – 7x^{2} + 12x = 0

Solution: To factor x^{3}−7x^{2}+12x

= x (x^{2}−5x+6)

Now, we factorise x^{2}−7x+12

We the factors of the 12 = 3, 4

Now, x^{2 }− (3 + 4) x + 3 ͯ 4

=x^{2 }-3x -4x + 3 ͯ 4

= x(x – 3) – 4 (x – 3) = (x−3) (x−4)

Factors of x^{3}−7x^{2}+12x are x (x−3) (x−4)

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