Chapter 2

Polynomials

NCERT maths class 9 ch 2

Polynomials are expressions with at least one non-zero coefficient term. There can be several terms in a polynomial. Every articulation in the polynomial is referred to as a term. If x2 + 5x + 2 is a polynomial, then x2, 5x, and 2 are the polynomial's terms. The polynomial contains a coefficient for each term. For example, if the polynomial is 2x + 1, then the coefficient of x is 2.

Polynomials can also be used to represent real numbers. For example, numbers with almost no factors, such as 3, 6, 7 are also polynomials. 

The degree of the polynomial in x3 + y3 + 3xy(x + y), for example, is 3. The degree of a non-zero polynomial is zero. Aside from these, there are several types of polynomials in maths class 9th chapter 2, such as -

  • A direct polynomial is of degree one 
  • Quadratic polynomial is of degree two 
  • Cubic polynomials are of degree three

Types of polynomial in class 9 maths ch 2

  • Monomial - Algebraic articulations with just one term are called monomials.
  • Binomial - Algebraic articulations with two terms are called binomials.
  • Trinomial - Algebraic articulations with three terms are called trinomials.

Basically, logarithmic articulations having multiple terms are all known as polynomials.

Significant points regarding polynomials in chapter 2 class 9 maths

  • A polynomial can have many terms, but not endless terms.
  • The power of a variable polynomial cannot be negative. This implies a variable with power - 2, - 3, - 4, and so forth is not permitted. In the event that the force of a variable in a logarithmic articulation is negative, that cannot be viewed as a polynomial.
  • The level of a non-zero steady polynomial is zero.

Q1. Which of the accompanying articulations are polynomials in a single variable and which are not? State the basis behind your responses.

 (i) x2-5x+6

(ii) y2

(iii) 9 +t

(iv) y

(v) 5x10+ 10y3+ 15t50

Solution: 

(i) x2-5x+6 \( \rightarrow \) Polynomial in single variable x

(ii) y2 \( \rightarrow \) Polynomial in single variable y

(iii) 9 +t\( \rightarrow \) Not a polynomial since the power of the variable in the first term is not a whole number.

(iv) y \( \rightarrow \) Not a polynomial since the power of the variable in the second term is – 1, which is not a whole number.

(v) 5x10+ 10y3+ 15t50 \( \rightarrow \) Not a polynomial in one variable since there are three variables: x, y, t.

Q2. Compose the coefficients of x2 in every one of the accompanying

(i) 2 + 3x2 + x

(ii) 4 –2x2 +2 x3

(iii) 8x2 +8x

(iv) \( \sqrt{11} \) x – 5

Solution:

  1. Coefficient of x2 = 3
  2. Coefficient of x2 = −2
  3. Coefficient of x2 = 8
  4. Coefficient of x2 = 0 , since there is no term of x2


Q3. Give one example of a binomial of the degree 35 and a monomial of the degree 100.

Solution:

(i) Binomial of degree 35 can be 5x35 - 7.

(ii) A monomial of degree 100 can be \( \sqrt{5} \) y100.

Q4. Write the degree of each of the following polynomials.

(i) 5x3+2x2 + 2x

(ii) 4 – 3y2

(iii) 4t – \( \sqrt{7} \)

(iv) 5

Solution:

(i) The given polynomial is 5x3+2x2 + 2x. The highest power of the variable x is 3.

So, the degree of the polynomial is 3.

(ii) The given polynomial is 4 - 3y2. The highest power of the variable y is 2.

So, the degree of the polynomial is 2.

(iii) The given polynomial is 4t – \( \sqrt{7} \). The highest power of variable t is 1. 

So, the degree of the polynomial is 1.

(iv) Since, 5 = 5x° [\( \because \) x°=1]

So, the degree of the polynomial is 0.

Q5. Find the zero of the polynomial in each of the following cases.

(i) p(x)=x+5

(ii) p(x) = x – 5

(iii) p(x) = 2x + 5

Solution:

(i) We have, p(x) = x + 5. Since, p(x) = 0

\( \Rightarrow \) x + 5 = 0

\( \Rightarrow \) x = -5.

Thus, zero of x + 5 is -5.

(ii) We have, p(x) = x – 5.

Since, p(x) = 0 \( \Rightarrow \) x – 5 = 0 \( \Rightarrow \) x = -5

Thus, zero of x – 5 is 5.

(iii) We have, p(x) = 2x + 5. Since, p(x) = 0

\( \Rightarrow \) 2x + 5 =0

\( \Rightarrow \) 2x = -5

\( \Rightarrow \) x = −5/2

Thus, zero of 2x + 5 is −5/2

Q6. Determine which of the following polynomials has (x +1) a factor.

(i) x3+x2+x +1

(ii) x4 + x3 + x2 + x + 1

(iii) x4 + 3x3 + 3x2 + x + 1

(iv) x3 – x2 – (2 +\( \sqrt{2} \) )x + \( \sqrt{2} \)

Solution:

The zero of x + 1 is -1.

(i) Let p (x) =x3+x2+x +1

\( \therefore \) p (-1) = (-1)3 + (-1)2 + (-1) + 1.

= -1 + 1 – 1 + 1

\( \Rightarrow \) p (- 1) = 0

So, (x+ 1) is a factor of x3 + x2 + x + 1.

(ii) Let p (x) = x4 + x3 + x2 + x + 1

\( \therefore \) P(-1) = (-1)4 + (-1)3 + (-1)2 + (-1)+1

= 1 – 1 + 1 – 1 + 1

\( \Rightarrow \) P (-1) \( \neq \) 1

So, (x + 1) is not a factor of x4 + x3 + x2 + x + 1.

(iii) Let p (x) = x4 + 3x3 + 3x2 + x + 1.

∴ p (-1)= (-1)4 + 3 (-1)3 + 3 (-1)2 + (- 1) + 1

= 1 – 3 + 3 – 1 + 1 = 1

\( \Rightarrow \)p (-1) \( \neq \) 0

So, (x + 1) is not a factor of x4 + 3x3 + 3x2 + x + 1.

(iv) Let p (x) = x3 – x2 – (2 + \( \sqrt{2} \)) x + \( \sqrt{2} \)

\( \therefore \) p (- 1) =(- 1)3- (-1)2 – (2 + \( \sqrt{2} \))(-1) + \( \sqrt{2} \)

= -1 – 1 + 2 + \( \sqrt{2} \) + \( \sqrt{2} \)

= 2\( \sqrt{2} \)

\( \Rightarrow \) p (-1) \( \neq \) 0

So, (x + 1) is not a factor of x3 – x2 – (2 + \( \sqrt{2} \)) x + √2.

 

Q7. Factorise the following

(i) 16x2 – 10x +1

(ii) 2x2 + 9x + 4

(iii) 6x2 -9x – 6

(iv) 5x2 – x – 6

Solution:

(i) We have,

16x2 – 10x + 1 = 16x2 – 8x- 2x + 1

= 8x (2x – 1 ) -1 (2x – 1)

= (8x -1) (2x -1)

Thus, 16x2 – 10x + 1 = (8x -1) (2x -1)

(ii) We have, 2x2 + 9x + 4 = 2x2 + x + 8x + 4

= x(2x + 1) + 4(2x + 1)

= (2x + 1)(x + 4)

Thus, 2x2 + 9x + 4 = (2x + 1)(x + 4)

(iii) We have, 6x2 -9x – 6 = 6x2 -12x +3x – 6

= 6x(x - 2) – 3(x -2)

= (6x-3)(x -2)

Thus, 6x2 -9x – 6 = (6x-3)(x -2)

(iv) We have, 5x2 – x – 6 = 5x2 +5x -6x – 6

= 5x(x +1) -6(x +1) = (5x – 6)(x + 1)

Thus, 5x2 – x – 6 = (5x – 6)(x + 1)

 

Q8. Evaluate the following products without multiplying directly

(i) 103 \( \times \)107

(ii) 95  \( \times \) 96

(iii) 104  \( \times \) 96

Solution:

(i) Using (x + a)(x + b) = x2 + (a + b)x + ab, we have 

103 x 107 = (100 + 3) (100 + 7)

= ( 100)2 + (3 + 7) (100)+ (3 \( \times \) 7)

= 10000 + (10) \( \times \) 100 + 21

= 10000 + 1000 + 21=11021

 

(ii) We have, 95 \( \times \) 96 = (100 – 5) (100 – 4)

= ( 100)2 + [(- 5) + (- 4)] 100 + (- 5 \( \times \) – 4)

[Using (x + a)(x + b) = x2 + (a + b)x + ab]

= 10000 + (-9) + 20 = 9120

= 10000 + (-900) + 20 = 9120 

 

(iii) We have 104 \( \times \) 96 = (100 + 4) (100 – 4)

= (100)2- 42

[Using (a + b)(a -b) = a2– b2]

= 10000 – 16 = 9984

 

Q9. Factorise

(i) x3 – 2x2 – x + 2

(ii) x3 – 3x2 – 9x – 5

(iii) x3 + 13x2 + 32x + 20

(iv) 2y3 + y2 – 2y – 1

Solution:

(i) We have, x3 – 2x2 – x + 2

Rearranging the terms, we have x3 – x – 2x2 + 2

= x(x2 – 1) – 2(x2 -1) = (x2 – 1)(x – 2)

= [(x)2 – (1)2](x – 2)

= (x – 1)(x + 1)(x – 2)

[\( \because \) (a2 – b2) = (a + b)(a-b)]

Thus, x3 – 2x2 – x + 2 = (x – 1)(x + 1)(x – 2)

 

(ii) We have, x3 – 3x2 – 9x – 5

= x3 + x2 – 4x2 – 4x – 5x – 5 ,

= x2 (x + 1) – 4x(x + 1) – 5(x + 1)

= (x + 1)(x2 – 4x – 5)

= (x + 1)(x2 – 5x + x – 5)

= (x + 1)[x(x – 5) + 1(x – 5)]

= (x + 1)(x – 5)(x + 1)

Thus, x3 – 3x2 – 9x – 5 = (x + 1)(x – 5)(x +1)

 

(iii) We have, x3 + 13x2 + 32x + 20

= x3 + x2 + 12x2 + 12x + 20x + 20

= x2(x + 1) + 12x(x +1) + 20(x + 1)

= (x + 1)(x2 + 12x + 20)

= (x + 1)(x2 + 2x + 10x + 20)

= (x + 1)[x(x + 2) + 10(x + 2)]

= (x + 1)(x + 2)(x + 10)

Thus, x3 + 13x2 + 32x + 20

= (x + 1)(x + 2)(x + 10)

 

(iv) We have, 2y3 + y2 – 2y – 1

= 2y3 – 2y2 + 3y2 – 3y + y – 1

= 2y2(y – 1) + 3y(y – 1) + 1(y – 1)

= (y – 1)(2y2 + 3y + 1)

= (y – 1)(2y2 + 2y + y + 1)

= (y – 1)[2y(y + 1) + 1(y + 1)]

= (y – 1)(y + 1)(2y + 1)

Thus, 2y3 + y2 – 2y – 1

= (y – 1)(y + 1)(2y +1)

 

Q10. Evaluate the following using suitable identities

(i) (97)3

(ii) (105)3

(iii) (992)3

 

Solution:

(i) We have, 97 = (100 - 3)

\( \therefore \) 973 = (100 – 3)3

= (100)3 – (3)3 – 3(100)(3)(100 -3)

[Using (a – b)3 = a3 – b3 – 3ab (a – b)]

= 1000000 – 27 – 900(100 – 3)

= 1000000 - 27 – 90000 + 270

= 1000270 – 90027 = 910243

 

(ii) We have, 105 =100 + 5

\( \therefore \) 1053 = (100 + 5)3

= (100)3 + (5)3 + 3(100)(5)(100 + 5)

[Using (a + b)3 = a3 + b3 + 3ab (a + b)]

= 1000000 + 125 + 1500(100 + 5)

= 1000000 + 125 + 150000 + 7500 = 1157620

 

(iii) We have, 992 = 1000 – 8

\( \therefore \)(998)3 = (1000-8)3

= (1000)3– (8)3 – 3(1000)(8)(1000 – 8)

[Using (a – b)3 = a3 – b3 – 3ab (a – b)]

= 1000000000 – 512 – 24000(1000 – 8)

= 1000000000 – 512 – 24000000 +192000

= 976191488

This chapter helps you to get a comprehension of polynomials and how to calculate them to effectively finish the test. The study of polynomials is one of the main subjects in arithmetics, so you must learn polynomials well. Additionally, polynomials are extraordinary ways of fostering a specific reasoning ability. 

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