Polynomial Formula

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 :

1. Monomial: This type of polynomial contains only one term. For example, x² , x, y, 3y, 4z
2. Binomial: This type of polynomial contains two terms. For example, x² – 10x
3. Trinomial: This type of polynomial contains three terms. For example, x² – 10x+9
4. Quadratic Polynomial: This type of polynomial contains four terms. For example, x³+2x² – 10x+7

The general Polynomial Formula is,

F(x) = a_nxⁿ + bx^n-1 + a_n-2x^n-2 + …….. + rx +s

• If n is a natural number: aⁿ – bⁿ = (a – b)(a^n-1 + a^n-2b +…+ b^n-2a + b^n-1)
• If n is even (n = 2a): xⁿ + yⁿ = (x + y)(x^n-1 – x^n-2y +…+ y^n-2x – y^n-1)
• If n is odd number: xⁿ + yⁿ = (x + y)(x^n-1 – x^n-2y +…- y^n-2x + y^n-1)

Polynomial Identities

1. (x + y)²= x² + 2xy + y²
2. (x – y)²= x² – 2xy + y²
3. x²– y² = (x + y)(x – y)
4. (x + a)(x + b) = x²+ (a + b)x + ab
5. (x + y + z)²= x² + y² + c² + 2xy + 2yz + 2zx
6. (x + y)³= x³ + y³ + 3xy (x + y)
7. (x – y)³= x³ – y³ – 3xy (x – y)
8. x³+ y³ = (x + y)(x² – xy + y²)
9. x³– y³ = (x – y)(x² + xy + y²)
10. x³+ y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx)
11. ax² + bx + c = 0 then x = \frac{( -b \pm \sqrt{(b² – 4ac)} )}{2a}

Solved Examples

Q1. Solve the equation: x² +16x + 64 = 0

Solution: Factors of x² +16x + 64

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

Now, x² + (8 + 8) x + 8 ͯ 8

=x² +8x +8x + 8 ͯ 8

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

Q.2. Solve x³ – 7x² + 12x = 0

Solution: To factor x³−7x²+12x

= x (x²−5x+6)

Now, we factorise x²−7x+12

We the factors of the 12 = 3, 4

Now, x² − (3 + 4) x + 3 ͯ 4

=x² -3x -4x + 3 ͯ 4

= x(x – 3) – 4 (x – 3) = (x−3) (x−4)

Factors of x³−7x²+12x  are  x (x−3) (x−4)