Haryana State Board HBSE 9th Class Maths Notes Chapter 2 Polynomials Notes.

## Haryana Board 9th Class Maths Notes Chapter 2 Polynomials

**Introduction**

In previous classes, we have learnt about algebraic expressions and various operations on them such as addition, subtraction, multiplication and division. We have also studied the factorization of algebraic expressions. In this chapter, we shall review these concepts and extent them to particular types of expressions known as polynomials. We shall also study the Remainder Theorem and Factor Theorem and their use in the factorization of polynomials.

**Key Words**

→ Polynomial: A polynomial is a algebraic expression in which the variables involved have only non negative integral powers.

→ TermEach individual part of an expression or sum which stands alone or is connected to other parts of an expression by a plus (+) or minus (-) sign e.g., in expression 3x^{2} + 4x + 9, there are three terms, 3x^{2}, 4x and 9.

→ Coefficients of polynomial: Refers to a multiplying factor e.g., in 5x and 7xy; 5 and 7y are the coefficients of x.

→ Algebraic expression: An algebraic expression is a collection of one or more terms, which are seperated from each other by addition (+) or subtraction signs.

→ Standard form of a polynomial: A polynomial is said to be in standard form, if the powers of x are either in increasing or in decreasing order.

→ Remainder theorem: If p(x) is any polynomial of degree greater than or equal to 1 and p(x) is divided by the linear polynomial (x – a) then remainder is p(a).

→ Factor theorem: (x – a) is a factor of the polynomial p(x), if p(a) = 0.

→ Zero of a polynomial: A real number ‘a’ is a zero of a polynomial p(x), if p(a) = 0.

→ Algebraic equation: An algebraic equation is a statement which states that the two expressions are equal.

→ Factors of the polynomial: When a polynomial (an algebraic expression) is expressed as the product of two or more expressions, then each of these expressions is called a factor of the polynomial.

→ Linear factors: If a factors contains no higher power of the variable quantity than the first power is called linear factor e.g., 4x + 5 is a linear factor.

**Basic Concepts**

Polynomial in One Variable:

(a) Polynomial: An expression of the form a_{n}x^{n} + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + ….. + a_{2}x^{n} + a_{1}x + a_{0} is called polynomial, where n is a whole number, a_{0}, a_{1}, a_{2}, a_{3},…, a_{n}, are real numbers and coefficients of the polynomial.

e.g., (i) 4x^{3} + 9x^{2} – 3x + 4 is a polynomial in one variable x.

(ii) y^{4} + 6y^{3} – 4y^{2} + 2y + 7 is a polynomial in one variable y.

(b) Terms and their Coefficients: (i) The polynomial 5x^{3} + 9x^{2} – 3x – 5 has four terms, namely 5x^{3}, 9x^{2}, -3x and -5, and 5, 9, -3 are coefficients of x, x and x respectively.

(ii) The polynomial x^{2} + 3x – 7 has three terms, namely x^{2}, 3x and -7 and 1, 3 are coefficients of x^{2} and x respectively.

(c) Degree of a Polynomial: The highest power of the variable is called the degree of polynomial.

(i) f(x) = 3x^{3} + 4x^{2} – 8x + 10.

Here, highest power term of given polynomial is 3x^{3}, so it’s degree is 3.

(ii) g(x) = 5x^{2} – 4x + 8.

Here, highest power term of given polynomial is 5x^{2}, so it’s degree is 2.

(d) Types of Polynomial (according to degree):

1. Constant Polynomial: A polynomial of degree zero is called a constant polynomial.

eg., 3, -7, \(\frac{5}{9}\), etc. are all constant polynomials.

2. Linear Polynomial: A polynomial of degree 1 is called a linear polynomial.

e.g., (i) 3x + 7 is a linear polynomial in one variable x.

(ii) 4y – 8 is a linear polynomial in one variable y.

(iii) ax + b, a ≠ 0 is a linear polynomial in one variable x.

3. Quadratic Polynomial: A polynomial of degree 2 is called a quadratic polynomial.

e.g., (i) 11x^{2} + 6x – \(\frac{3}{2}\) is a quadratic polynomial in x.

(ii) y^{2} + 6y – 8 is a quadratic polynomial in y.

(iii) ax^{2} + bx + c, a ≠ 0 is a quadratic polynomial in x.

4. Cubic Polynomial: A polynomial of degree 3 is called a cubic polynomial.

e.g., (i) 7x^{3} + 5x^{2} + 6 is a cubic polynomial in x.

(ii) 3x^{2} – 4xy^{2} + 7 is a cubic polynomial in x and y.

(iii) ax^{3} + bx^{2} + cx + d, a ≠ 0 is a cubic polynomial in x.

5. Biquadratic Polynomial: A polynomial of degree 4 is called a biquadratic polynomial.

e.g., (i) x^{4} – 5x^{3} + 3x^{2} – 6x + 4 is a biquadratic polynomial in x.

(ii) ax^{4} + bx^{3} + cx^{2} + dx + e, a ≠ 0 is a biquadratic polynomial in x.

(e) Types of Polynomial (According to terms):

(i) Monomial: A polynomial having one term is called a monomial.

e.g., 5, 2x, 4xy, 3x^{2} are all monomials.

(ii) Binomial: A polynomial having two terms is called a binomial.

eg., 4x + 3, 5x^{2} + 9, 7x^{2} + 7x are all binomials.

(iii) Trinomial: A polynomial having three terms is called a trinomial.

e.g., 7x^{3} + 4x^{2} + 5, x^{2} + 5x + 3 are all trinomials.

(iv) Zero Polynomial: A polynomial consisting of one term, namely zero only, is called a zero polynomial.

The degree of a zero polynomial is not defined.

Zeroes of a Polynomial: Consider the polynomial:

p(x) = 2x^{3} – 3x^{2} + 4x + 5.

If we replace x by 1 everywhere in p(x), we get

p(1) = 2 × (1)^{3} – 3 × (1)^{2} + 4 × 1 + 5

= 2 – 3 + 4 + 5

= 11 – 3 = 8.

So, we can say that the value of p(x) at x = 1 is 8.

Similarly,

P(0) = 2 × (0)^{3} – 3 × (0)^{2} + 4 × 0 + 5

= 0 – 0 + 0 + 5 = 5.

So, we can say that the value of p(x) at x = 0 is 5 and

p(-2) = 2 × (-2)^{3} – 3 × (-2)^{2} + 4 × (-2) + 5

= -16 – 12 – 8 + 5

= -31.

So, we can say that the value of p(x) at x = -2 is -31. Now, consider another polynomial

f(x) = x^{2} – 4

f(2) = (2)^{2} – 4 = 4 – 4 = 0

f(-2) = (-2)^{2} – 4 = 4 – 4 = 0.

We can say, that 2 and -2 are the zeroes of the polynomial.

In general, a real number ‘a’ is said to be a zero of a polynomial p(x), if the value of the polynomial p(x) at x = a is 0 i.e., p(a) = 0.

Now, consider general linear polynomial p(x) = ax + b, a ≠ 0.

For zero of the polynomial p(x) = 0.

0 = ax + b

⇒ ax = -b

⇒ x = \(\frac{-b}{a}\)

∴ \(\frac{-b}{a}\) is the zero of the polynomial.

(v) A zero of polynomial need not to be zero.

(vi) Number of zeroes is equal to the degree of polynomial.

1. Remainder Theorem: Let p(x) be any polynomial of degree ≥ 1 and let a be any real number. If the polynomial p(x) is divided by linear polynomial (x – a), then the remainder is p(a).

Proof:

Suppose that when p(x) is divided by (x – a), then the quotient is q(x) and the remainder is r(x), then we have

p(x) = (x – a).q(x) + r(x)

where degree of r(x) < degree (x – a) or r(x) = 0. Since, degree of (x – a) = 1, degree of r(x) = 0 i.e., constant.

Let r(x) = r, then we have

p(x) = (x – a).q(x) + r ……..(i)

Putting x = a in (i), we get

p(a) = (a – a).q(a) + r

p(a) = 0·q(a) + r

p(a) = 0 + r

p(a) = r

Hence, the remainder is p(a) when p(x) is divided by (x – a).

Hence proved.

Corollary I. When a polynomial p(x) of degree ≥ 1 is divided by a linear polynomial ax + b, a ≠ 0, then remainder is equal to p\(\left(\frac{-b}{a}\right)\).

Corollary II. If a polynomial p(x) vanishes when x = a and also x = b, then p(x) is divisible by (x – a) (x – b).

2. Factor Theorem: If p(x) is a polynomial of degree ≥ 1 and a is any real number, then:

(i) (x – a) is a factor of p(x), if p(a) = 0 and

(ii) p(a) = 0, if (x – a) is a factor of p(x).

Proof:

When a polynomial p(x) is divided by (x – a), then

p(x) = (x – a).q(x) + r(x)…….(i)

where q(x) is the quotient and r(x) is the remainder.

By Remainder theorem, we have

remainder = p(a).

Putting r(x) = p(a) in (i), we get

p(x) = (x – a).q(x) + p(a).

(i) If p(a) = 0, then p(x) = (x – a).q(x)

⇒ (x – a) is a factor of p(x).

(ii) If (x – a) is a factor of p(x) (given), then p(x) = (x – a).g(x), for some polynomial g(x)

In this case, when p(x) is divided by (x – a), the remainder is

p(a) = (a – a).g(a)

⇒ p(a) = 0 × g(a)

⇒ p(a) = 0. Hence proved.

Factorisation of Polynomials:

(a) Factorisation of a quadratic polynomial by splitting the middle term : Quadratic polynomial ax^{2} + bx + c, where a, b, c ≠ 0 divided into two classes:

(i) When a = 1 i.e., x^{2} + bx + c.

(ii) When a ≠ 1 i.e., ax^{2} + bx + c.

Resolution of x^{2} + bx + c into two factors: Examine the following example:

(x + 3)(x + 4) = x^{2} + (3 + 4)x + 3 × 4 = x^{2} + 7x + 12

Here, we obtain product 12 by multiplying together 3 × 4 and 7 is obtain by adding together (+3) and (+4).

Hence, in order to find out factors of x^{2} + 7x + 12, we must find two numbers whose product is 12 and sum is 7.

Since, (x + p) (x + q) = x^{2} + (p + q)x + pq

∴ x^{2} + (p + q)x + pq = (x + p) (x + q)

We observe that, the product pq is obtained by multiplying together p and q, and (p + q) is the sum of the coefficient of the two second terms. Conversely, to find the factors, we have to find two numbers whose sum is (p + q) and product is pq.

We have the following procedure for factorising the quadratic polynomial x^{2} + bx +c:

Step 1: Rewrite the quadratic polynomial in standard form (x^{2} + bx + c), if not already in this form.

Step 2: Factorise constant term e into two factors p and q such that p × q = c and p + q = b.

Step 3: Split the middle term i.e., the term x into two terms with the coefficients p and q.

Step 4: By using grouping factorisation method, factorise the obtained four terms.

(b) Factorisation of a quadratic polynomial by splitting the middle term (continued): For factors of the quadratic polynomial of the form ax^{2} + bx + c, where a ≠ 1 and a, b, c ≠ 0. We have the following procedure for factorising the quadratic polynomial ax^{2} + bx + c.

Step 1: Rewrite the quadratic polynomial in standard form (ax^{2} + bx + c), if not already in this form.

Step 2: Find the product of constant term and the coefficient of x^{2} i.e., a × c = ac.

Step 3: (i) Observe that, if ac is positive (+ve), then find two factors p and q of ac such that p + q = b.

(ii) If ac is negative (-ve), then find two factors p and q of ac such that p – q = b.

Step 4: Split the middle term into two terms with coefficients p and q.

Step 5: Factorize the four terms by grouping.

(c) Factorisation of cubic polynomials by factor theorem: In this section, we have studied how to factorize cubic polynomials. The idea is to find out a linear factor (x – a) of the given polynomial p(x) and write it as p(x) = (x – a).q(x).

The degree of q(x) is one less than that of p(x). Therefore, q(x) is a quadratic polynomial which can be factorised by the method of splitting the middle term or we may again try to find a linear factor (x – b) of q(x). Thus, we have p(x) = (x – a).(x – b) g(x).

Here, the degree of g(x) is one less than q(x) i.e., the degree of g(x) is 1. Thus p(x) = (x – a).(x – b) g(x), which is product of linear factors.

For example: Factorise x^{3} – 2x^{2} – 5x + 6 by factor theorem.

Solution:

First Method: Let

p(x) = x^{3} – 2x^{2} – 5x + 6.

Step 1. First check at which value, the given polynomial p(x) becomes zero. These values can be any one of the factors of constant term.

The factors of constant term (6) are ±1, ±2, ±3, ±6.

p(1) = p(1)^{3} – 2 × (1)^{2} – 5 × 1 + 6

= 1 – 2 – 5 + 6

= 7 – 7 = 0.

∴ By factor theorem (x – 1) is a factor of p(x).

Step 2. Now, divide x^{3} – 2x^{2} – 5x + 6 by x – 1

The other factor of p(x) is x^{2} – x – 6.

Step 3. Now, p(x) = (x – 1) (x^{2} – x – 6)

= (x – 1)[x^{2} – (3 – 2)x – 6)

[∵ -6 = -(3 × 2), 3 – 2 = 1]

= (x – 1)[x^{2} – 3x + 2x – 6[

= (x – 1)[x(x – 3) + 2(x – 3)]

= (x – 1)(x – 3)(x + 2)

Hence, x^{3} – 2x^{2} – 5x + 6 = (x – 1)

(x – 3)(x + 2).

Second method: Let

p(x) = x^{3} – 2x^{2} – 5x + 6

First check at which value, the given p(x) becomes zero. These values can be any one of the factors of constant term. Factors of constant term 6 are ±1, ± 2, ± 3, ±6.

Put x = 1,

p(1) = (1)^{3} – 2 × (1)^{2} – 5 × 1 + 6

= 1 – 2 – 5 + 6

= 7 – 7 = 0

⇒ (x – 1) is a factor of x^{3} – 2x^{2} – 5x + 6.

Put x = -2,

p(-2) = (-2)^{3} – 2 × (-2)^{2} – 5 × (-2) + 6

= – 8 – 8 + 10 + 6

= -16 + 16 = 0

⇒ (x + 2) is a factor of p(x).

Put x = 3,

p(3) = (3)^{3} – 2 × (3)^{2} – 5 × 3 + 6

= 27 – 18 – 15 + 6

= 33 – 33 = 0

⇒ (x – 3) is a factor of p(x).

Since, the given expression is a cubic polynomial.

So, it will have three factors of first degree.

∴ x^{3} – 2x^{2} – 5x + 6 = k(x – 1)(x + 2)(x – 3) ……(i)

Here, k is the numerical coefficient yet to be found. Now putting x = 0 (value other than 1, 2, 3) on both sides of equation (i),

we get

6 = k(0 – 1) (0 + 2) (0 – 3)

⇒ 6 = 6k ⇒ k = \(\frac{6}{6}\) ⇒ k = 1

Putting the value of k in (i), we get

x^{3} – 2x^{2} – 5x + 6 = (x – 1)(x + 2)(x – 3).

Algebraic Identities: In previous classes, we have learnt about some algebraic identities.

An algebraic identity is an algebraic equation that is true for all values of the variables present in the equation.

Identity I: (x + y)^{2} = x^{2} + 2xy + y^{2}.

Identity II: (x – y)^{2} = x^{2} – 2xy + y^{2}.

Identity III: x^{2} – y^{2} = (x + y) (x – y).

Identity IV: (x + a)(x + b) = x^{2} + (a + b)x + ab.

(a) Algebraic Identities for Cubes of Binomials

Identity V: (x + y)^{3} = x^{3} + y^{3} + 3xy(x + y)

= x^{3} + y^{3} + 3x^{2}y + 3xy^{2}.

Identity VI: (x – y)^{3} = x^{3} – y^{3} – 3xy(x – y)

= x^{3} – y^{3} – 3x^{2}y + 3xy^{2}.

(b) Algebraic Identities Relating to Trinomials

Identity VII: (x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2xy + 2yz + 2zx.

Identity VIII: x^{3} + y^{3} + z^{3} – 3xyz

= (x + y + z) (x^{2} + y^{2} + z^{2} – xy – yz – zx)

Note: If x + y + z = 0, then

x^{3} + y^{3} + z^{3} – 3xyz = 0(x^{2} + y^{2} + z^{2} – xy – yz – zx)

x^{3} + y^{3} + z^{3} – 3xyz = 0

x^{3} + y^{3} + z^{3} = 3xyz.

(c) Sum and Difference of Cubes

We will use the following identity for factorisation of sum and difference of two cubes.

Identity IX: x^{3} + y^{3} = (x + y)(x^{2} – xy + y^{2}).

Identity X: x^{3} – y^{3} = (x – y)(x^{2} + xy + y^{2}).