# Ncert Solutions For Class 10 Maths Ex 2.1

## Ncert Solutions For Class 10 Maths Chapter 2 Ex 2.1

DEGREE OF A POLYNOMIAL:-

If p(x) is a polynomial in x, the highest power of x in p(x) is called the degree of polynomial p(x).

And expressions like:- x2+2,1x1$\sqrt{x^{2}}+2\;, \frac{1}{x-1}$etc., are not polynomials.

Example-1:

3x4+2x2+6x = 1, the highest power of the given polynomial equation is 4

Therefore, its degree is 4.

Example-2:

x5y3z + 2xy3+4x2yz2

This equation is of multiple variables (x, y, z) and to find the degree of this equation, we just need to add up the degrees of the variables in each of the terms and also it does not matter whether they are different variables.

Therefore, 1st term (5+3+1), 2nd term (1+3) and 3rd term (2+1+2).

The highest total is 9, of the 1st term,

Therefore, its degree is 9.

Polynomial equation with degree 1 is called a linear polynomial.

Example:     x – 3$\sqrt{3}$ = 0

Polynomial equation with degree 2 is called a quadratic polynomial.

Example:    x2+2x+7=0$x^{2} +\sqrt{2}x+7 = 0$.

Polynomial equation with degree 3 is called a cubic polynomial.

Example:    3x35x2+7x+5=0$3x^{3} – 5x^{2} +\sqrt{7}x + \sqrt{5} = 0$.

Factorization of a quadratic polynomial equation by splitting middle term:-

General form of quadratic polynomial equation:- ax2 + bx + c = 0

Step 1:- Find the product of a (first term) and c (last term).

Step 2:- Then split the product of a and c into two numbers such that their product remains same as (a × c) and their sum or difference is equal to the value of b (middle term).

Step 3:- Now, there will be total four terms in the equation, make two groups with each group having two terms each. Then take whatever is common and equate it to 0, the resultant values of x will be factors of the quadratic equation.

Example:

12 x2 = -11x +15

Solution:

12x 2 + 11x -15 = 0                (product of a and c is 12×15=180)

12x 2 + 20x – 9x -15 = 0      (splitting 180 into 20 and 9 such that, 20 – 9 = 11 and 20×9 = 180).

4x(3x + 5) – 3(3x + 5) = 0

(4x -3)(3x + 5) = 0

4x – 3 = 0 or 3x + 5 = 0

4x = 3 or 3x = – 5

x = 34$\frac{3}{4}$ or x = 53$\frac{-5}{3}$

Therefore, solution of this quadratic equation  is (53$\frac{-5}{3}$, 34$\frac{3}{4}$ ).

This solution is called zeroes or roots of the given polynomial.

Graphical method to find zeroes:-

Total number of zeroes in any polynomial equation = total number of times the curve intersects x-axis.

Example 1:-

In the above graph, the curve intersects x axis at A and B.

Therefore, the equation of the curve will have maximum of 2 zeroes.

Example 2:-

From the above graph, the curve cuts the x-axis at point A

Therefore, number of zeroes of p(x) = 1

Example 3: –

From the above graph, the curve p(x) intersects x-axis at points A, B and C.

Therefore, the number of zeroes of polynomial equation is 3.

Zeroes and Coefficient of a polynomial:

If α and β are zeroes of any quadratic polynomial ax2 + bx + c=0,

Then, Sum of zeroes = α+β = ba$\frac{-b}{a}$

Product of zeroes = αβ = ca$\frac{c}{a}$

If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:

x2 – (α + β)x + αβ = 0