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# Introduction to Polynomials and Zeros of a polynomial Class 9 Notes | EduRev

## Mathematics (Maths) Class 9

Created by: Indu Gupta

## Class 9 : Introduction to Polynomials and Zeros of a polynomial Class 9 Notes | EduRev

The document Introduction to Polynomials and Zeros of a polynomial Class 9 Notes | EduRev is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9

Definition of polynomials:
Polynomials is the algebraic topic. An algebraic expression, which have only non negative powers of the variables is called polynomial.

In an expression, generally we have variables and constant terms.

Example  3 x² + 2 x + 6
Here x is variable and 6 without any variant is called Constant.

Degree:
The degree of a algebraic equation is the highest degree for a term with non-zero coefficient. The degree of a term is sum of the powers of each variable in the term.
Example
4x⁷-5
In this expression we have only one term and the power of the term is 7, so the expression has deg 7

5 x³ y³ - 4 x² y² + 7 x y
Here, it has 3 terms and power of the first term is 3+3=6, the power of the second term is 2+2=4, and the power of the last term is 1+1=2, here the highest power is 6, so the expression has deg 3+3=6 We can do basic arithmetic operations with two algebraic equation. That is we can add, subtract, multiply or divide any two polynomial.

For adding any two polynomials we have to combine the like terms.

Add 4 x² + 7 x - 6, x² - 3 x - 2
For adding these two poly-nomials we have to combine the like terms. Here the like terms are 4x² and x², 7x and -3x , -6 and -2
If we combine 4x² and x² we will get 5x²
If we combine 7x and -3x we will get 4x
If we combine -6 and 2 we will get -4
So the final answer is 5x² + 4x - 4

Subtraction:
(2 x³ - 2 x² + 4 x - 3)- (x³ + x² - 5 x + 4)

Step 1: In the first step we are going to multiply the negative with inner terms.

= 2x³ -2x² + 4x -3 - x³-x²+5x-4

Step 2: In the second step we have to combine the like terms

= 2x³ - x³ -2x²-x² + 4x + 5x - 3 - 4

Step 3: After combining the like terms we will get the answer

= x³ + x² + 9x - 7

Multiplication:
there are two formats for this: horizontal and vertical, like in addition.

The simplest case of multiplication of polynomials is multiplication of monomials.

For instance:
Simplify: ( 5 x² )(-2x ³)

For multiplying these two monomials  we have to just multiply the numbers and add the powers using the exponent rule.

So (-6x²)(3x³) = -18 x² ⁺ ³= -18 x⁵

Division:
Division of polynomials involves two cases, the first one is simplification,which is reducing the fraction and the second one is long division.

Polynomial in One Variable
The degree of polynomials in one variable is the highest power of the variable in the algebraic expression. For example, in the following equation: x2+2x+4. The degree of the equation is 2 .i.e. the highest power of variable in the equation.

Multivariable polynomial
For a multivariable polynomial, it the highest sum of powers of different variables in any of the terms in the expression. Take following example,  x5+3x4y+2xy3+4y2-2y+1.  It is a multivariable polynomial in x and y, and the degree of the polynomial is 5 – as you can see the degree in the terms x5 is 5, x4y  it is also 5 (4+1) and so the highest degree among these individual terms is 5.

A polynomial of two variable x and y, like axrys   is the algebraic sum of several terms of the prior mentioned form, where r and s are possible integers. Here, the degree of the polynomial is r+s where r and s are whole numbers.

Note: Exponents of variables of a polynomial .i.e. degree of polynomials should be whole numbers.

How to find the Degree of a Polynomial?

There are 4 simple steps are present to find the degree of a polynomial:-

Example: 6x5+8x3+3x5+3x2+4+2x+4

Step 1: Combine all the like terms that are the terms of the variable terms.

(6x5+3x5)+8x3+3x2+2x+(4+4)

Step 2: Ignore all the coefficients

x5+x3+x2+x+x0

Step 3: Arrange the variable in descending order of their powers

x5+x3+x2+x+x0

Step 4: The largest power of the variable is the degree of the polynomial

deg(x5+x3+x2+x+x0) = 5

Zeros of a polynomial:
Here we are going to see how to find zero of a polynomials.
If the value of a polynomial is zero for some value of the variable then that value is known as zero of the polynomial.

Definition:
Let p(x) be a polynomial in x. If p(x) = 0, then we say that a is a zero of the polynomial p(x).
Let us consider the example given below.

Zeros of linear polynomial
Example 1:Find the zeros of the following linear polynomial
p(x)  =  2x + 3

Solution:
p(x)  =  2x + 3
Now we have to think about the value of x, for which the given function will become zero.
For that let us factor out 2
p(x)  =  2 (x + 3/2)
Instead of "x" , if we apply -3/2 p(x) will become zero.
Hence -3/2 is the zero of the given linear polynomial.

Example 2: Find the zeros of the following linear polynomial
p(x)  =  4x - 1

Solution :
p(x)  =  4x - 1
Now we have to think about the value of x, for which the given function will become zero.
For that let us factor out 4
p(x)  =  4 (x - 1/4)
By applying the value 1/4 instead of x, the function p(x) will become zero.
Hence 1/4 is the zero of the given linear polynomial.

For a quadratic equation, there will be two zeros. In order to find those zeros, we may use the method called factoring.
Example 3: Find the zeros of the quadratic equation x² + 17 x + 60 by factoring.
Solution :
p(x)  =  x² + 17 x + 60
p(x)  =  x² + 12x + 5x + 60
p(x)  =  x (x + 12) + 5 (x + 12)
p(x)  =  (x + 5) (x + 12)
If x =  -5
p(x)  =  (-5 + 5) (-5 + 12)  =  0
If x =  -12
p(x)  =  (-12 + 5) (-12 + 12)  =  0
Hence the zeros are -5 and -12.

Zeros of cubic polynomial
For a cubic equation, there will be three zeros. In order to find those zeros, we may use the methods
(i)  Factor theorem
(ii)  Synthetic division

Example 4: Find the zeros of the following polynomial 4 x³ - 7 x + 3
Solution :
Let p (x) = 4 x³ - 7 x + 3 x = 1
p (1) = 4 (1)³ -7 (1) + 3
= 4 - 7 + 3
= 7 - 7
= 0
So we can decide (x - 1) is a factor. To find other two factors, we have to use synthetic division. So, the factors are (x - 1) and (4 x² - 4 x - 3). By factoring this quadratic equation we get  (2 x +
3) (2 x - 1)
Hence the required three factors are (x - 1) (2 x + 3) (2 x - 1)

Remainder theorem and Factor theorem :
Remainder theorem :
If a polynomial P(x) is divided by (x-a), the remainder is P(a).
Using division algorithm, we have
P(x)  =  Q(x)(x-a) + P(a)
Here, Q(x) is the quotient when P(x) is divided by (x-a).

Factor theorem:
A polynomial P(x) would have a factor (x-a), if and only if P(a)  =  0.

Remainder theorem and Factor theorem - Examples
Example 1: Using Remainder theorem, find the remainder when the polynomial 3x³- 2x² + 6x - 7 is divided by (x-2).
Solution :
Step 1 :
Let P(x)  =  3x³- 2x² + 6x - 7.
If x - 2 = 0, then x  =  2.
Step 2 :
To know the remainder when P(x) divided by (x-2), aaaaa     plug x = 2 in P(x).
Remainder  =  P(2)
Remainder  =  3(2)³- 2(2)² + 6(2) - 7
Remainder  =  24 - 8 + 12 - 7
Remainder  =  21

Example 2: Using Remainder theorem, find the remainder when the polynomial  7x⁴ - x²- 3x + 9 is divided by (x-6).
Solution :
Step 1 :
Let P(x)  =  7x⁴ - x²- 3x + 9
If x - 6 = 0, then x  =  6.
Step 2 :
To know the remainder when P(x) divided by (x-6), aaaaa     plug x = 6 in P(x).
Remainder  =  P(6)
Remainder  =  7(6)⁴ - (6)² - 3(6) + 9
Remainder  =  9072 - 36 - 18 + 9
Remainder  =  9027

Example 3: Using Remainder theorem, find the remainder when the polynomial  x³+ 3x² - 5x + 2 is divided by (x+5).
Solution :
Step 1 :
Let P(x)  =  x³+ 3x² - 5x + 2

If x + 5 = 0, then x  =  - 5.
Step 2 :
To know the remainder when P(x) divided by (x+5), aaaaa     plug x = - 5 in P(x).
Remainder  =  P(-5)
Remainder  =   (-5)³+ 3(-5)² - 5(-5) + 2
Remainder  =  -125 + 3(25) + 25 + 2
Remainder  =  -125 + 75 + 25 + 2
Remainder  =  -23

Example 4: Using Factor theorem, check whether (x-2) is a factor of the polynomial x⁴ - 3x³ + 2x² + 8x - 16.
Solution :
Step 1 :
Let P(x)  =  x⁴ - 3x³ + 2x² + 8x - 16
If x - 2 = 0, then x  =  2.
Step 2 :
Using Factor theorem, to check whether (x-2) is factor of P(x), plug x = 2 in P(x).
P(2)  =  (2)⁴ - 3(2)³ + 2(2)² + 8(2) - 16
P(2)  =  16 - 24 + 8 + 16 - 16
P(2)  =  0
Because P(2) = 0, by Factor theorem, (x-2) is a factor of the polynomial P(x).

Example 5: Using Factor theorem, check whether (x+4) is a factor of the polynomial x² - 8x + 16.
Solution :
Step 1 :
Let P(x)  =  x² - 8x + 16
If x + 4 = 0, then x  =  -4.
Step 2 :
Using Factor theorem, to check whether (x+4) is factor of P(x), plug x = -4 in P(x).
P(-4)  =  (-4)² + 8(-4) + 16
P(-4)  =  16 - 32 + 16
P(-4)  =  0
Because P(-4) = 0, by Factor theorem, (x+4) is a factor of the polynomial P(x).

Example 6: Using Factor theorem, check whether (x-3) is a factor of the polynomial x³ - 2x² + 5x + 6.
Solution :
Step 1 :
Let P(x)  =  x³ - 2x² + 5x + 6
If x - 3 = 0, then x  =  3.
Step 2 :
Using Factor theorem, to check whether (x-3) is factor of P(x), plug x = 3 in P(x).
P(3)  =  (3)³ - 2(3)² + 5(3) + 6
P(3)  =  27 - 2(9) + 15 + 6
P(3)  =  27 - 18 + 15 + 6
P(3)  =  30 ≠ 0
Because P(3) ≠ 0, by Factor theorem, (x-3) is not a factor of the polynomial P(x).

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## Mathematics (Maths) Class 9

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