**NCERT New Syllabus Class 10 - Polynomials Exercise 2.1**

**Understanding Polynomials: A Key Chapter in Class 10 Mathematics**

Polynomials form a fundamental building block in algebra and are crucial in understanding various higher-level concepts in mathematics. In this chapter of the 10th-grade NCERT syllabus, students will explore how to express algebraic expressions in polynomials and analyze their behavior through various methods. From identifying degrees to solving for roots, this chapter will empower you to solve equations efficiently, laying a strong foundation for future mathematical concepts like calculus and advanced algebra.

^{2}+bx+c, where a ≠ 0, then (x−α) and (x−β) are factors of p(x).

^{2}+bx+c = k(x − α) (x − β) where k is a constant.

^{2}+bx+c = k(x − α) (x − β)

^{2}+bx+c = k[x

^{2 }− (α + β) x+α⋅β]

^{2}+bx+c = kx

^{2 }− k(α + β) x+ k.α⋅β

^{2}, x, and the constant term on both sides, we get:

a = k, b = − k (α + β), c = k⋅α⋅β

From this, we can conclude that:

α + β = − b/a

αβ = c/a.

^{2}+bx+c,

sum of zeros = − (coefficient of x)/(coefficient of x^{2})

product of zeros = (constant)/(coefficient of x^{2})

polynomial ax^{3}+ bx^{2}+ cx + d, then

α + β + γ =−b/a =−(coefficient of x^{2})/(coefficient of x^{3})

αβ +βγ +γα= c/a = (coefficient of x)/(coefficient of x^{3})

α β γ = − d/a = − (constant)/(coefficient of x^{3})

### EXERCISE 2.2

(i) x– 2x – 8 (ii) 4s^{2}– 4s + 1 (iii) 6x^{2}– 3 – 7x^{2}

(iv) 4u+ 8u (v) t^{2}– 15 (vi) 3x^{2}– x – 4^{2}

### Explanation:

^{2}+ bx + c, which we will denote as p(x).

^{2}+ bx + c.

^{2}+ bx + c = 0.

ax^{2}+ bx + c = 0.

ax^{2}+ bx + c = 0, then:

^{2})

^{2})

### Solution:

**– 2x – 8**

^{2}^{2}– 2x – 8.

The last term is 8 and its sign is negative. We need to factor 8 x 1 in such a way that the difference between the factors is – 2. The factors of 8 x 1 = (– 4) x (2).

(– 4 + 2 = – 2).

= x^{2}– 4x + 2x – 8= (x^{2}– 4x) + (2x – 8)= x(x – 4) + 2(x – 4)= (x – 4)(x + 2)

^{2}– 2x – 8 = (x – 4)(x + 2)

^{2}– 2x – 8 are 4 and – 2. i.e 𝝰 = 4 and 𝛃 = – 2,

a = 1, b = – 2 and c = – 8

The sum of Zeroes:

(𝝰 + 𝛃) = – (b/a)

LHS = (𝝰 + 𝛃)

LHS = (4 + (– 2))

LHS = (2) ------------------ equation 1

RHS = – (b/a)

RHS = – (– 2)/1

RHS = (2) ------------------ equation 2

From equation 1 and equation 2, we have LHS = RHS.

Product of Zeroes:

(𝝰 x 𝛃) = (c/a)

LHS = (𝝰 x 𝛃)

LHS = (4 x (– 2))

LHS = (– 8) ------------------ equation 3

RHS = (c/a)

RHS = (– 8)/1

RHS = (– 8) ------------------ equation 4

From equation 3 and equation 4, we have LHS = RHS.

**p(x) is 0 when x =**

**– 2**

**or x = 4, and**

**the relationships between**

zeroes and coefficients are verified.

**– 4s + 1**

^{2}^{2}– 4s + 1

The last term is 1 and its sign is positive. The first term is 4. So, we need to factor 4 x 1 in such a way that their sum equals – 4. The factors of

4 x 1 = (– 2) x (– 2)

(– 2 – 2 = – 4).

= 4s^{2}– 2s – 2s + 1= (4s^{2}– 2s) – (2s – 1)= 2s(2s – 1) – (2s – 1)= (2s – 1)(2s – 1)

^{2}– 4s + 1 = (2s – 1)(2s – 1)

^{2}– 4s + 1 are 1/2 and 1/2. i.e 𝝰 = 1/2 and 𝛃 = 1/2,

a = 4, b = – 4 and c = 1

The sum of Zeroes:

(𝝰 + 𝛃) = – (b/a)

LHS = (𝝰 + 𝛃)

LHS = ((1/2) + (1/2))

LHS = (1) ------------------ equation 1

RHS = – (b/a)

RHS = – (– 4)/4

RHS = (1) ------------------ equation 2

From equation 1 and equation 2, we have LHS = RHS.

Product of Zeroes:

(𝝰 x 𝛃) = (c/a)

LHS = (𝝰 x 𝛃)

LHS = ((1/2) x (1/2))

LHS = (1/4) ------------------ equation 3

RHS = (c/a)

RHS = 1/4 ------------------ equation 4

From equation 3 and equation 4, we have LHS = RHS.

**p(s) is 0 when s =**

**1/2**

**or s = 1/2, and**

**the relationships between**

zeroes and coefficients are verified.

**– 3 – 7x**

^{2}^{2}– 7x – 3.

The last term is 3 and its sign is negative. The first term is 6. So we need to factor 6 x 3 in such a way that the difference between the factors is – 7.

The factors of 6 x (– 3) = (– 3 x 3) x (2).

(– 9 + 2 = – 7).

= 6x^{2}+ 2x – 9x – 3= (6x^{2}+ 2x) – (9x + 3)= 2x(3x + 1) – 3(3x + 1)= (3x + 1)(2x – 3)

^{2}– 7x – 3 = (3x + 1)(2x – 3)

^{2}– 7x – 3 are –1/3 and 3/2. i.e 𝝰 = –1/3 and 𝛃 = 3/2,

a = 6, b = – 7 and c = – 3

The sum of Zeroes:

(𝝰 + 𝛃) = – (b/a)

LHS = (𝝰 + 𝛃)

LHS = ((–1/3) + (3/2))

LHS = (– 2 + 9)/6

LHS = 7/6 ------------------ equation 1

RHS = – (b/a)

RHS = – (– 7)/6

RHS = (7/6) ------------------ equation 2

From equation 1 and equation 2, we have LHS = RHS.

Product of Zeroes:

(𝝰 x 𝛃) = (c/a)

LHS = (𝝰 x 𝛃)

LHS = ((–1/3) x (3/2))

LHS = (–1/2) ------------------ equation 3

RHS = (c/a)

RHS = – 3/6

RHS = – 1/2 ------------------ equation 4

From equation 3 and equation 4, we have LHS = RHS.

**p(x) is 0 when x =**

**– 1/3**

**or x = 3/2, and**

**the relationships between**

zeroes and coefficients are verified.

**+ 8u**

^{2}^{2}+ 8u

here we can take 4u common directly.

= 4u(u + 2)

^{2}+ 8u = 4u(u + 2)

^{2}+ 8u are 0 and – 2. i.e 𝝰 = 0 and 𝛃 = – 2,

a = 4, b = 8 and c = 0

The sum of Zeroes:

(𝝰 + 𝛃) = – (b/a)

LHS = (𝝰 + 𝛃)

LHS = ((0) + (– 2))

LHS = (– 2) ------------------ equation 1

RHS = – (b/a)

RHS = – (8)/4

RHS = (– 2) ------------------ equation 2

From equation 1 and equation 2, we have LHS = RHS.

Product of Zeroes:

(𝝰 x 𝛃) = (c/a)

LHS = (𝝰 x 𝛃)

LHS = ((0) x (– 2))

LHS = (0) ------------------ equation 3

RHS = (c/a)

RHS = 0/4

RHS = 0 ------------------ equation 4

From equation 3 and equation 4, we have LHS = RHS.

**p(u) is 0 when u =**

**– 2**

**or u = 0, and**

**the relationships between**

zeroes and coefficients are verified.

**– 15**

^{2}^{2}– 15

here, use the formula

(a^{2}– b^{2}) = (a – b) (a + b) directly.

= (t + √15) (t – √15)

^{2}– 15 = (t + √15) (t – √15)

^{2}– 15 are √15 and –√15. i.e 𝝰 = √15 and 𝛃 = –√15,

a = 1, b = 0 and c = – 15

The sum of Zeroes:

(𝝰 + 𝛃) = – (b/a)

LHS = (𝝰 + 𝛃)

LHS = ((√15) + (–√15))

LHS = (0) ------------------ equation 1

RHS = – (b/a)

RHS = – (0)/1

RHS = (0) ------------------ equation 2

From equation 1 and equation 2, we have LHS = RHS.

Product of Zeroes:

(𝝰 x 𝛃) = (c/a)

LHS = (𝝰 x 𝛃)

LHS = ((√15) x (–√15))

LHS = (– 15) ------------------ equation 3

RHS = (c/a)

RHS = – 15/1

RHS = – 15 ------------------ equation 4

From equation 3 and equation 4, we have LHS = RHS.

**p(t) is 0 when t =**

**–**

**√15 or t = √15**

**, and**

**the relationships between**

zeroes and coefficients are verified.

**– x – 4**

^{2}^{2}– x – 4

The last term is 4 and its sign is negative. The first term is 3. So we need to factor 3 x 4 in such a way that the difference between the factors is – 1.

The factors of 3 x (– 4) = (– 4 x 3).

(– 4 + 3 = – 1).

= 3x^{2}+ 3x – 4x – 4= (3x^{2}+ 3x) – (4x + 4)= 3x(x + 1) – 4(x + 1)= (3x – 4)(x + 1)

^{2}– x – 4 = (3x – 4)(x + 1)

^{2}– x – 4 are 4/3 and – 1. i.e 𝝰 = 4/3 and 𝛃 = – 1,

a = 3, b = – 1 and c = – 4

The sum of Zeroes:

(𝝰 + 𝛃) = – (b/a)

LHS = (𝝰 + 𝛃)

LHS = (4/3 + (– 1))

LHS = (1/3) ------------------ equation 1

RHS = – (b/a)

RHS = – (– 1)/3

RHS = (1/3) ------------------ equation 2

From equation 1 and equation 2, we have LHS = RHS.

Product of Zeroes:

(𝝰 x 𝛃) = (c/a)

LHS = (𝝰 x 𝛃)

LHS = (4/3 x (– 1))

LHS = (– 4/3) ------------------ equation 3

RHS = (c/a)

RHS = – 4/3 ------------------ equation 4

From equation 3 and equation 4, we have LHS = RHS.

**p(x) is 0 when x =**

**– 1**

**or x = 4/3**

**, and**

**the relationships between**

**zeroes and coefficients are verified**

**.**

product of its zeroes respectively.

(i) 1/4, – 1 (ii) √2, 1/3 (iii) 0, √5

(iv) 1, 1 (v) – 1/4, 1/4 (vi) 4, 1

### Explanation:

p(x) = x^{2}– (𝝰 + 𝛃) x + (𝝰 x 𝛃)

### Solution:

**(i) 1/4,**

**–**

**1**

the sum of zeroes is (𝝰 + 𝛃) = 1/4, and the product of zeroes is (𝝰 x 𝛃) = –1.

p(x) = x^{2}– (𝝰 + 𝛃) x + (𝝰 x 𝛃)

p(x) = x^{2}– (1/4) x + (–1)

p(x) = 4x^{2}– x + (– 4)

p(x) = 4x^{2}– x – 4

**the quadratic polynomial is 4x**

^{2}– x –**4**.

**(ii)**

**√**

**2, 1/3**

the sum of zeroes is (𝝰 + 𝛃) = √2, and the product of zeroes is (𝝰 x 𝛃) = 1/3.

p(x) = x^{2}– (𝝰 + 𝛃) x + (𝝰 x 𝛃)

p(x) = x^{2}– (√2) x + (1/3)

p(x) = x^{2}– √2x + (1/3)

p(x) = 3x^{2}– 3√2x + 1

**the quadratic polynomial is 3x**.

^{2}– 3√2x + 1**(iii) 0,**√5

the sum of zeroes is (𝝰 + 𝛃) = 0, and the product of zeroes is (𝝰 x 𝛃) = √5.

p(x) = x^{2}– (𝝰 + 𝛃) x + (𝝰 x 𝛃)

p(x) = x^{2}– (0) x + (√5)

p(x) = x^{2}– 0x + √5

p(x) = x^{2}+ √5

**the quadratic polynomial is x**.

^{2}+ √5the sum of zeroes is (𝝰 + 𝛃) = 1, and the product of zeroes is (𝝰 x 𝛃) = 1.

p(x) = x^{2}– (𝝰 + 𝛃) x + (𝝰 x 𝛃)

p(x) = x^{2}– (1) x + (1)

p(x) = x^{2}– x + 1

**the quadratic polynomial is x**.

^{2}– x + 1**(v) – 1/4, 1/4**

the sum of zeroes is (𝝰 + 𝛃) = – 1/4, and the product of zeroes is (𝝰 x 𝛃) = 1/4.

p(x) = x^{2}– (𝝰 + 𝛃) x + (𝝰 x 𝛃)

p(x) = x^{2}– (– 1/4) x + (1/4)

p(x) = 4x^{2}+ x + 1

**the quadratic polynomial is 4x**.

^{2}+ x + 1the sum of zeroes is (𝝰 + 𝛃) = 4, and the product of zeroes is (𝝰 x 𝛃) = 1.

p(x) = x^{2}– (𝝰 + 𝛃) x + (𝝰 x 𝛃)

p(x) = x^{2}– (4) x + (1)

p(x) = x^{2}– 4x + 1

**the quadratic polynomial is x**.

^{2}– 4x + 1**Conclusion: The Power and Elegance of Polynomials**

## No comments:

## Post a Comment