NCERT Class 9 Maths Chapter 2 Notes Polynomials - Download PDF

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Polynomials Class 9 Notes

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Polynomials

Polynomials are an algebraic expression that has one or more variables with non-negative exponents. A polynomial can be represented by $P(x)$ where $P$ is called a polynomial in $x$ and the exponent of $x$ is called the highest degree of $x$. As shown in the figure below.

Example: 2x + 1, 3xy, 3x³ + 2x² + x + 1, x + y + z etc.

Terms

An expression that is a part of the polynomial expression is called a term.

Example: 4x² + 2x + 2
Where 4x², 2x and 2 are the terms of a polynomial expression.

Coefficient

A factor that multiples a variable or a term is called a coefficient. A coefficient can be a whole number, a fraction or a decimal, and these numbers can be negative or positive.

Types of Polynomials

Based on the degrees and terms that a polynomials have, polynomials can be classified into two categories:
1. Number of terms
2. Degree of polynomials

Based On the Number Of Terms

A polynomial expression can have one or more terms, and according to the number of terms, polynomials are as follows -
1. A polynomial with only one term is called a monomial.
Example: $4x$, $7pq$, $9p^2$

2. A polynomial with only two unlike terms is called a binomial.
Example: $4x + 1$, $7y^2$ + $3$

3. A polynomial with only three unlike terms is called a trinomial.
Example: $4y^2$ + $2y$ - $2$

Based On the Degree Of Polynomials

1. A polynomial with 0 as a degree is called a zero or constant polynomial.
Example: $2x^0$, $5y^0$

2. A polynomial with 1 as a degree is called a linear polynomial.
Example: $3x$ + $1$, $4y$ + $5$

3. A polynomial with 2 as a degree is called a quadratic polynomial.
Example: $5x^2$ + $3x$ + $1$

4. A polynomial that has 3 as a degree is called a cubic polynomial.
Example: $7x^3$ + $5x$ - $2$

Polynomials in One Variable

A polynomial expression that has one type of variable in the entire expression is known as a polynomial in one variable.
Example: 8x² + 4x + 2, 5y + 2, 6z.

Zeros of A Polynomial

The value of the variable that makes polynomials equal to zero is called the zeros of a polynomial.

Example: For the polynomial $p(y)$ = $x^2$ - $9$,
the zeros are $x = 3$ and $x = -3$,
because $p(3)$ = $3^2$ - $9$ = $0$ and p(-3) = (-3)² - 9 = 0

Finding Zeros of Polynomials for Linear Equations

The standard form of a linear equation is P(x) = ax + b, and for this equation, the formula for finding the zeros is as follows:
Zero of Polynomial (x) = $\frac{-\text{Constant}}{\text{ Coefficent of x}}$

Example: Find the zero of a polynomial P(x) = 2y - 8.
Here,
Constant = 8
Coefficient = 2

Zero of Polynomial (y) = $\frac{-\text{Constant}}{\text{ Coefficent of y}}$

Zero of Polynomial (y) = $\frac{-8}{2}$

Zero of Polynomial (y) = $-4$

Finding Zeros of Polynomials for Quadratic Equations

The standard form of a quadratic equation is ax² + bx + c = 0, where a, b and c are constants. Therefore, the formula for finding zeros of quadratic equations is:

$X$ = $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Example: Find the zeros of a polynomial of the quadratic equation 5x² + 4x + 1 = 0.
After comparing the question equation with the standard quadratic equation ax² + bx + c = 0, we get,
a = 5
b = 4
c = 1
And,

$X$ = $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

After putting the values in the formula, we get,

$X$ = $\frac{-4 \pm \sqrt{4^2 - 4 × 5 × 1}}{2 × 5}$

$X$ = $\frac{-4 \pm \sqrt{16 - 20}}{10}$

$X$ = $\frac{-4 \pm \sqrt{-4}}{10}$

$X$ = $\frac{-4 \pm {(-2)}}{10}$

Therefore, $x$ = $\frac{-3}{5}$ and $x$ = $\frac{-1}{5}$

Relationship Between Zeros And Coefficients Of a Polynomial

For quadratic Polynomials, the standard quadratic equation is:
ax² + bx + c and let α and β be the roots of the standard quadratic equation, then,

Sum of the zeroes (α + β) = $\frac{-b}{a}$ or $\frac{\text{-Coeffiecient of x}}{\text{Coeffiecinet of x}^2}$

Product of the zeroes (αβ) = $\frac{c}{a}$ or $\frac{\text{Constant term}}{\text{Coefficient of x}^2}$

Example: x² + 7x + 10
After comparing the question equation with the standard quadratic equation ax² + bx + c = 0, we get,
b = 7
a = 1
c = 10

Sum of the zeroes (α + β) = $\frac{-7}{1}$

Product of the zeroes (αβ) = $\frac{10}{1}$

For Cubic Polynomials:
Let α, β and γ be the roots of a cubic polynomial ax³ + bx² + cx + d, then

α + β + γ = $\frac{-b}{a}$

αβ + βγ + γα = $\frac{c}{a}$

αβγ = $\frac{-d}{a}$

Example: Find zeros for the polynomial equation P(x) = 3x³ + 2x² + x + 2

After comparing the equation with standard polynomial equation ax³ + bx² + cx + d we get,
a = 3
b = 2
c = 1
d = 2

α + β + γ = $\frac{-b}{a}$

αβ + βγ + γα = $\frac{c}{a}$

αβγ = $\frac{-d}{a}$

After putting in the values, we get,

α + β + γ = $-\frac{2}{3}$

αβ + βγ + γα = $\frac{1}{3}$

αβγ = $-\frac{2}{3}$

Remainder Theorem

It is a fundamental method that is used to determine the remainder of the polynomials when a polynomial is divided by a linear expression. If P(x) is a polynomial with degree greater than or equal to 1 and if the polynomial is divided by (x-a), then P(a) is called the remainder. Here, (x-a) is called the divisor of the polynomial P(a) is equal to 0. This is applied in factorising polynomials.

The general formula is as follows:

${P(x)}$ = $(x - a) Q(x) + P(a)$

Where,
P(x) = Polynomial equation
Q(x) = Quotient
P(a) = Remainder
(x - a) = Divisor

Example: If P(x) is 6x² + 5x + 1 is divided by x - 2 and the quotient is 6x + 17, then the remainder will be 35.
Here, x - 2 is a divisor, and we know that,
x - a = 0,
Therefore, x - 2 = 0
x = 2
By placing the values in the P(x), we get,
= 6 × (2)² + 5 × 2 + 1
= 24 + 10 + 1
= 35

Factor Theorem

When the polynomials are completely divisible, the factor theorems are used. It is used for factoring polynomials or for finding the roots.

Proof Using Remainder Theorem

Let P(x) be a polynomial which is divided by (x - a), then P(a) = 0.
According to the remainder theorem,
P(x) = (x - a) × Q(x) + P(a)
Since, P(a) = 0, then
P(x) = (x - a) × Q(x) + P(a)
It is proved that (x - a ) is a factor of the polynomial P(x).

Factorisation of Polynomials

The process of defining the polynomials as the product of two or more terms is known as the factorisation of polynomials.

Example: Find the roots of the polynomial P(x) = 6x² + 17x + 5 using the factorisation method.

P(x) = 6x² + 17x + 5

⇒ 6x² + 15x + 2x + 5 = 0

⇒ 3x(2x + 5) + 1(2x + 5) = 0

⇒ (2x + 5), (3x + 5)

x = $\frac{-5}{3}$, $\frac{-5}{2}$,

Algebraic Identities

In polynomials, some algebraic identities are given that are valid for certain values. Algebraic identities are equations and identities are as follows:

1. (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx
2. (x + y)³ = x³ + y³ + 3xy(x + y)
3. (x – y)³ = x³ – y³ – 3xy(x – y)
4. x³ + y³ + z³ – 3xyz = (x + y + z) (x² + y² + z² – xy – yz – zx)

Polynomials: Previous Year Question and Answer

Question 1:
If $\frac{1}{x}+x=4$, then find $\frac{1}{x^2}+x^2$.

Solution:
$\frac{1}{x}+x=4$
squaring both sides, we get,
⇒ $(\frac{1}{x}+x)^2=4^2$
⇒ $\frac{1}{x^2}+2×\frac{1}{x}×x+x^2=16$
⇒ $\frac{1}{x^2}+x^2=16-2$
⇒ $\frac{1}{x^2}+x^2=14$
Hence, the correct answer is 14.

Question 2:
If 4x² + y² = 40 and xy = 6, then find the value of 2x + y.

Solution:
Given: 4x² + y² = 40 and xy = 6
(2x + y)² = 4x² + y² + 2 × 2x × y
⇒ (2x + y)² = 40 + 4 × 6
⇒ (2x + y)² = 64
$\therefore$ 2x + y = 8
Hence, the correct answer is 8.

Question 3:
If $a-\frac{1}{a-5}=10$, then the value of $(a-5)^3-\frac{1}{(a-5)^3}$ is:

Solution:
$a-\frac{1}{a-5}=10$ .............(1)
Subtract 5 in equation (1) on both sides
$(a-5)-\frac{1}{a-5}=10-5$
⇒ $(a-5)-\frac{1}{a-5}=5$
Now, take $(a - 5) = x$
So, the equation becomes
⇒ $x-\frac{1}{x}=5$
Cubing both sides
⇒ $x^3-\frac{1}{x^3}-3(x-\frac{1}{x}) = 125$
⇒ $x^3-\frac{1}{x^3} = 125+3(5)$
⇒ $x^3-\frac{1}{x^3} = 140$
Hence, the correct answer is 140.

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