Class 10 Chapter 2 Maths Polynomials

by
Polynomials

Table of Contents

Chapter 2: Polynomials of Class 10 Mathematics. This chapter covers important concepts such as the degree of a polynomial, types of polynomials, factorization of polynomials, and polynomial identities.

Introduction

In Chapter 2, we study Polynomials, which are algebraic expressions that involve sums and differences of variables raised to various powers. A polynomial is a mathematical expression of the form:P(x)=anxn+an−1xn−1+⋯+a1x+a0P(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0P(x)=an​xn+an−1​xn−1+⋯+a1​x+a0​

Where an,an−1,…,a0a_n, a_{n-1}, \dots, a_0an​,an−1​,…,a0​ are constants, and nnn is a non-negative integer. This chapter includes:

  • The degree of polynomials,
  • The types of polynomials based on degree and number of terms,
  • The zeroes of polynomials,
  • The factorization of polynomials, and
  • Polynomial identities.

Exercise 2.1: Types of Polynomials and Basic Operations

Q1. Identify the degree and the leading coefficient of the following polynomials:

  1. 5×3−4×2+2x−15x^3 – 4x^2 + 2x – 15×3−4×2+2x−1
  2. 2×4−3×2+x−72x^4 – 3x^2 + x – 72×4−3×2+x−7
  3. x5+3×3−2×2+4x+5x^5 + 3x^3 – 2x^2 + 4x + 5×5+3×3−2×2+4x+5

Solution:

  1. Polynomial: 5×3−4×2+2x−15x^3 – 4x^2 + 2x – 15×3−4×2+2x−1
    • Degree: 3 (Highest power of xxx)
    • Leading coefficient: 5 (coefficient of x3x^3×3)
  2. Polynomial: 2×4−3×2+x−72x^4 – 3x^2 + x – 72×4−3×2+x−7
    • Degree: 4
    • Leading coefficient: 2 (coefficient of x4x^4×4)
  3. Polynomial: x5+3×3−2×2+4x+5x^5 + 3x^3 – 2x^2 + 4x + 5×5+3×3−2×2+4x+5
    • Degree: 5
    • Leading coefficient: 1 (coefficient of x5x^5×5)

Q2. Write the following polynomials in standard form:

  1. 4x−7+2x24x – 7 + 2x^24x−7+2×2
  2. x3−3×2+5x+2x^3 – 3x^2 + 5x + 2×3−3×2+5x+2
  3. 9x+2−4x39x + 2 – 4x^39x+2−4×3

Solution:

To write a polynomial in standard form, arrange the terms in descending order of degree:

  1. Polynomial: 4x−7+2x24x – 7 + 2x^24x−7+2×2
    • Standard form: 2×2+4x−72x^2 + 4x – 72×2+4x−7
  2. Polynomial: x3−3×2+5x+2x^3 – 3x^2 + 5x + 2×3−3×2+5x+2
    • Already in standard form: x3−3×2+5x+2x^3 – 3x^2 + 5x + 2×3−3×2+5x+2
  3. Polynomial: 9x+2−4x39x + 2 – 4x^39x+2−4×3
    • Standard form: −4×3+9x+2-4x^3 + 9x + 2−4×3+9x+2

Q3. Find the value of the following polynomials at x=2x = 2x=2:

  1. 2×2+3x−52x^2 + 3x – 52×2+3x−5
  2. x3−4x+6x^3 – 4x + 6×3−4x+6
  3. 5×4−3×2+x+85x^4 – 3x^2 + x + 85×4−3×2+x+8

Solution:

  1. Polynomial: 2×2+3x−52x^2 + 3x – 52×2+3x−5Substitute x=2x = 2x=2:2(2)2+3(2)−5=2(4)+6−5=8+6−5=92(2)^2 + 3(2) – 5 = 2(4) + 6 – 5 = 8 + 6 – 5 = 92(2)2+3(2)−5=2(4)+6−5=8+6−5=9Value at x=2x = 2x=2 is 9.
  2. Polynomial: x3−4x+6x^3 – 4x + 6×3−4x+6Substitute x=2x = 2x=2:(2)3−4(2)+6=8−8+6=6(2)^3 – 4(2) + 6 = 8 – 8 + 6 = 6(2)3−4(2)+6=8−8+6=6Value at x=2x = 2x=2 is 6.
  3. Polynomial: 5×4−3×2+x+85x^4 – 3x^2 + x + 85×4−3×2+x+8Substitute x=2x = 2x=2:5(2)4−3(2)2+2+8=5(16)−3(4)+2+8=80−12+2+8=785(2)^4 – 3(2)^2 + 2 + 8 = 5(16) – 3(4) + 2 + 8 = 80 – 12 + 2 + 8 = 785(2)4−3(2)2+2+8=5(16)−3(4)+2+8=80−12+2+8=78Value at x=2x = 2x=2 is 78.

Exercise 2.2: Zeroes of Polynomials

Q1. Find the zeroes of the polynomial x2−5x+6x^2 – 5x + 6×2−5x+6.

Solution:

We need to solve for xxx in the equation:x2−5x+6=0x^2 – 5x + 6 = 0x2−5x+6=0

We can factorize the polynomial:x2−5x+6=(x−2)(x−3)x^2 – 5x + 6 = (x – 2)(x – 3)x2−5x+6=(x−2)(x−3)

Now, set each factor equal to zero:x−2=0⇒x=2x – 2 = 0 \quad \Rightarrow \quad x = 2x−2=0⇒x=2 x−3=0⇒x=3x – 3 = 0 \quad \Rightarrow \quad x = 3x−3=0⇒x=3

The zeroes of the polynomial are 2 and 3.

Q2. Find the zeroes of the polynomial x2−4xx^2 – 4xx2−4x.

Solution:

We need to solve for xxx in the equation:x2−4x=0x^2 – 4x = 0x2−4x=0

Factor the polynomial:x(x−4)=0x(x – 4) = 0x(x−4)=0

Now, set each factor equal to zero:x=0orx−4=0⇒x=4x = 0 \quad \text{or} \quad x – 4 = 0 \quad \Rightarrow \quad x = 4x=0orx−4=0⇒x=4

The zeroes of the polynomial are 0 and 4.

Q3. Find the zeroes of the polynomial 2×2−3x−52x^2 – 3x – 52×2−3x−5.

Solution:

We need to solve for xxx in the equation:2×2−3x−5=02x^2 – 3x – 5 = 02×2−3x−5=0

Use the quadratic formula:x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}x=2a−b±b2−4ac​​

Where a=2a = 2a=2, b=−3b = -3b=−3, and c=−5c = -5c=−5.

Substitute these values into the quadratic formula:x=−(−3)±(−3)2−4(2)(−5)2(2)x = \frac{-(-3) \pm \sqrt{(-3)^2 – 4(2)(-5)}}{2(2)}x=2(2)−(−3)±(−3)2−4(2)(−5)​​ x=3±9+404x = \frac{3 \pm \sqrt{9 + 40}}{4}x=43±9+40​​ x=3±494x = \frac{3 \pm \sqrt{49}}{4}x=43±49​​ x=3±74x = \frac{3 \pm 7}{4}x=43±7​

Now, solve for the two roots:

  1. x=3+74=104=2.5x = \frac{3 + 7}{4} = \frac{10}{4} = 2.5x=43+7​=410​=2.5
  2. x=3−74=−44=−1x = \frac{3 – 7}{4} = \frac{-4}{4} = -1x=43−7​=4−4​=−1

The zeroes of the polynomial are 2.5 and -1.

Exercise 2.3: Factorization of Polynomials

Q1. Factorize the polynomial x2+7x+10x^2 + 7x + 10×2+7x+10.

Solution:

We need to factor the polynomial x2+7x+10x^2 + 7x + 10×2+7x+10. We look for two numbers whose product is 10 and whose sum is 7. These numbers are 2 and 5.

So, the factorization is:x2+7x+10=(x+2)(x+5)x^2 + 7x + 10 = (x + 2)(x + 5)x2+7x+10=(x+2)(x+5)

Q2. Factorize the polynomial x2−8x+16x^2 – 8x + 16×2−8x+16.

Solution:

We need to factor the polynomial x2−8x+16x^2 – 8x + 16×2−8x+16. We look for two numbers whose product is 16 and whose sum is -8. These numbers are -4 and -4.

So, the factorization is:x2−8x+16=(x−4)(x−4)=(x−4)2x^2 – 8x + 16 = (x – 4)(x – 4) = (x – 4)^2×2−8x+16=(x−4)(x−4)=(x−4)2

Q3. Factorize the polynomial x2−4x^2 – 4×2−4.

Solution:

We need to factor the polynomial x2−4x^2 – 4×2−4. This is a difference of squares:x2−4=(x−2)(x+2)x^2 – 4 = (x – 2)(x + 2)x2−4=(x−2)(x+2)

Exercise 2.4: Polynomial Identities

Q1. Verify the identity (x+y)2=x2+2xy+y2(x + y)^2 = x^2 + 2xy + y^2(x+y)2=x2+2xy+y2 by expansion.

Solution:

We will expand (x+y)2(x + y)^2(x+y)2 using the distributive property:(x+y)2=(x+y)(x+y)(x + y)^2 = (x + y)(x + y)(x+y)2=(x+y)(x+y)

Using the distributive property:(x+y)(x+y)=x(x+y)+y(x+y)(x + y)(x + y) = x(x + y) + y(x + y)(x+y)(x+y)=x(x+y)+y(x+y) =x2+xy+xy+y2= x^2 + xy + xy + y^2=x2+xy+xy+y2 =x2+2xy+y2= x^2 + 2xy + y^2=x2+2xy+y2

Thus, the identity is verified.

Q2. Verify the identity (a+b)(a−b)=a2−b2(a + b)(a – b) = a^2 – b^2(a+b)(a−b)=a2−b2.

Solution:

We will expand (a+b)(a−b)(a + b)(a – b)(a+b)(a−b) using the distributive property:(a+b)(a−b)=a(a−b)+b(a−b)(a + b)(a – b) = a(a – b) + b(a – b)(a+b)(a−b)=a(a−b)+b(a−b) =a2−ab+ab−b2= a^2 – ab + ab – b^2=a2−ab+ab−b2 =a2−b2= a^2 – b^2=a2−b2

Thus, the identity is verified.

Additional Important Questions for Chapter 2: Polynomials

1. Find the zeroes of the polynomial x2−3x−10x^2 – 3x – 10×2−3x−10.

Solution:

We need to solve the equation x2−3x−10=0x^2 – 3x – 10 = 0x2−3x−10=0.

To factorize the polynomial, we look for two numbers whose product is -10 and whose sum is -3. These numbers are -5 and 2.

So, the factorization is:x2−3x−10=(x−5)(x+2)x^2 – 3x – 10 = (x – 5)(x + 2)x2−3x−10=(x−5)(x+2)

Now, set each factor equal to zero:x−5=0⇒x=5x – 5 = 0 \quad \Rightarrow \quad x = 5x−5=0⇒x=5x+2=0⇒x=−2x + 2 = 0 \quad \Rightarrow \quad x = -2x+2=0⇒x=−2

The zeroes of the polynomial are 5 and -2.

2. Factorize the polynomial x2+3x−10x^2 + 3x – 10×2+3x−10.

Solution:

We need to factorize the polynomial x2+3x−10x^2 + 3x – 10×2+3x−10.

We look for two numbers whose product is -10 and whose sum is 3. These numbers are 5 and -2.

So, the factorization is:x2+3x−10=(x+5)(x−2)x^2 + 3x – 10 = (x + 5)(x – 2)x2+3x−10=(x+5)(x−2)

3. Verify the identity (x+y)2−(x−y)2=4xy(x + y)^2 – (x – y)^2 = 4xy(x+y)2−(x−y)2=4xy.

Solution:

We need to verify the identity (x+y)2−(x−y)2=4xy(x + y)^2 – (x – y)^2 = 4xy(x+y)2−(x−y)2=4xy.

Start by expanding both sides.

The left-hand side is:(x+y)2−(x−y)2(x + y)^2 – (x – y)^2(x+y)2−(x−y)2

Now, expand both squares:(x+y)2=x2+2xy+y2(x + y)^2 = x^2 + 2xy + y^2(x+y)2=x2+2xy+y2(x−y)2=x2−2xy+y2(x – y)^2 = x^2 – 2xy + y^2(x−y)2=x2−2xy+y2

Subtract the second expansion from the first:(x2+2xy+y2)−(x2−2xy+y2)(x^2 + 2xy + y^2) – (x^2 – 2xy + y^2)(x2+2xy+y2)−(x2−2xy+y2)

Simplify:x2+2xy+y2−x2+2xy−y2x^2 + 2xy + y^2 – x^2 + 2xy – y^2×2+2xy+y2−x2+2xy−y2=4xy= 4xy=4xy

Thus, the identity is verified.

4. If x+1x + 1x+1 is a factor of the polynomial x2+px+6x^2 + px + 6×2+px+6, find the value of ppp.

Solution:

If x+1x + 1x+1 is a factor of the polynomial x2+px+6x^2 + px + 6×2+px+6, then by the factor theorem, substituting x=−1x = -1x=−1 in the polynomial should give us 0.

Substitute x=−1x = -1x=−1 into the polynomial:(−1)2+p(−1)+6=0(-1)^2 + p(-1) + 6 = 0(−1)2+p(−1)+6=0

Simplify:1−p+6=01 – p + 6 = 01−p+6=07−p=07 – p = 07−p=0p=7p = 7p=7

Thus, the value of ppp is 7.

5. Find the factorization of the polynomial x2+5x+6x^2 + 5x + 6×2+5x+6.

Solution:

We need to factor the polynomial x2+5x+6x^2 + 5x + 6×2+5x+6.

We look for two numbers whose product is 6 and whose sum is 5. These numbers are 2 and 3.

So, the factorization is:x2+5x+6=(x+2)(x+3)x^2 + 5x + 6 = (x + 2)(x + 3)x2+5x+6=(x+2)(x+3)

6. Find the value of kkk if x−2x – 2x−2 is a factor of the polynomial x2+kx−6x^2 + kx – 6×2+kx−6.

Solution:

If x−2x – 2x−2 is a factor of the polynomial x2+kx−6x^2 + kx – 6×2+kx−6, then by the factor theorem, substituting x=2x = 2x=2 in the polynomial should give us 0.

Substitute x=2x = 2x=2 into the polynomial:(2)2+k(2)−6=0(2)^2 + k(2) – 6 = 0(2)2+k(2)−6=04+2k−6=04 + 2k – 6 = 04+2k−6=02k−2=02k – 2 = 02k−2=02k=22k = 22k=2k=1k = 1k=1

Thus, the value of kkk is 1.

7. Factorize the polynomial x2−9x+20x^2 – 9x + 20×2−9x+20.

Solution:

We need to factor the polynomial x2−9x+20x^2 – 9x + 20×2−9x+20.

We look for two numbers whose product is 20 and whose sum is -9. These numbers are -4 and -5.

So, the factorization is:x2−9x+20=(x−4)(x−5)x^2 – 9x + 20 = (x – 4)(x – 5)x2−9x+20=(x−4)(x−5)

8. Find the zeroes of the polynomial x2−2x−15x^2 – 2x – 15×2−2x−15.

Solution:

We need to solve for xxx in the equation x2−2x−15=0x^2 – 2x – 15 = 0x2−2x−15=0.

To factorize the polynomial, we look for two numbers whose product is -15 and whose sum is -2. These numbers are -5 and 3.

So, the factorization is:x2−2x−15=(x−5)(x+3)x^2 – 2x – 15 = (x – 5)(x + 3)x2−2x−15=(x−5)(x+3)

Now, set each factor equal to zero:x−5=0⇒x=5x – 5 = 0 \quad \Rightarrow \quad x = 5x−5=0⇒x=5x+3=0⇒x=−3x + 3 = 0 \quad \Rightarrow \quad x = -3x+3=0⇒x=−3

The zeroes of the polynomial are 5 and -3.

9. Verify the identity (x−y)2=x2−2xy+y2(x – y)^2 = x^2 – 2xy + y^2(x−y)2=x2−2xy+y2.

Solution:

We need to verify the identity (x−y)2=x2−2xy+y2(x – y)^2 = x^2 – 2xy + y^2(x−y)2=x2−2xy+y2.

Start by expanding the left-hand side:(x−y)2=(x−y)(x−y)(x – y)^2 = (x – y)(x – y)(x−y)2=(x−y)(x−y)

Using the distributive property:(x−y)(x−y)=x(x−y)−y(x−y)(x – y)(x – y) = x(x – y) – y(x – y)(x−y)(x−y)=x(x−y)−y(x−y)=x2−xy−xy+y2= x^2 – xy – xy + y^2=x2−xy−xy+y2=x2−2xy+y2= x^2 – 2xy + y^2=x2−2xy+y2

Thus, the identity is verified.

10. Factorize the polynomial x2−6x+9x^2 – 6x + 9×2−6x+9.

Solution:

We need to factor the polynomial x2−6x+9x^2 – 6x + 9×2−6x+9.

We look for two numbers whose product is 9 and whose sum is -6. These numbers are -3 and -3.

So, the factorization is:x2−6x+9=(x−3)(x−3)=(x−3)2x^2 – 6x + 9 = (x – 3)(x – 3) = (x – 3)^2×2−6x+9=(x−3)(x−3)=(x−3)2

Conclusion

These additional questions cover a variety of key concepts in polynomials, including zeroes of polynomials, factorization, and polynomial identities. By practicing these questions, you can strengthen your understanding and problem-solving skills, ensuring you’re well-prepared for exams. Keep practicing more such questions, and you’ll be able to handle even the most challenging problems in this chapter.

To deepen your understanding of polynomials, practice solving a variety of problems involving different methods of factorization, such as grouping, splitting the middle term, and using identities. Additionally, explore real-life applications of polynomials, such as in physics, engineering, and economics, to appreciate their practical significance beyond the classroom.


Class 10 Chapter 1 Maths Real Numbers

For more Updates Click Here.

Related Posts:

Leave a Comment