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.

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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) = anxn + bxn-1 + an-2xn-2 + …….. + rx +s

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

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KUCKOO B

I get a different answer for first example.
I got Q1 as 20.5
median 23 and
Q3 26

Guest
Yashitha

Hi
Same

Guest
virat

yes