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NCERT New Syllabus Mathematics
Class: 10
Exercise 2.2
Topic: Polynomials
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.
In general, if α and β are
the zeroes of the quadratic polynomial
p(x) = ax2+bx+c, where a ≠ 0, then (x−α)
and (x−β) are factors of p(x).
Therefore, we can
express the quadratic polynomial as:
ax2+bx+c = k(x − α) (x − β) where k is a
constant.
ax2+bx+c = k(x − α) (x − β)
ax2+bx+c = k[x2 − (α + β) x+α⋅β]
ax2+bx+c = kx2 − k(α + β) x+
k.α⋅β
By comparing the
coefficients of x2, 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.
Note:
1) For the quadratic polynomial ax2+bx+c,
sum of zeros = − (coefficient of x)/(coefficient of x2)
product of zeros = (constant)/(coefficient of x2)
2) In general, it can be proved that if α, β, and γ are the zeroes of the cubic
polynomial ax3 + bx2 + cx + d, then
α + β + γ = − b/a = − (coefficient of x2)/(coefficient of x3)
αβ + βγ + γα = c/a = (coefficient of x)/(coefficient of x3)
α β γ = − d/a = − (constant)/(coefficient of x3)
EXERCISE 2.2
Q1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x2 – 2x – 8 (ii) 4s2 – 4s + 1 (iii) 6x2 – 3 – 7x
(iv) 4u2 + 8u (v) t2 – 15 (vi) 3x2 – x – 4
Explanation:
1) Consider the quadratic polynomial ax2 + bx + c, which we will denote as p(x).
2) That is, p(x) = ax2 + bx + c.
3) The zeroes of p(x) are the values of x that satisfy p(x) = 0, or ax2 + bx + c = 0.
4) To find the zeroes of p(x), we need to solve the quadratic equation
ax2 + bx + c = 0.
5) It is known that if α and β are the zeroes of the quadratic equation
ax2 + bx + c = 0, then:
sum of the zeroes (𝝰 + 𝛃) = − (b/a) = − (Coefficient of x)/Coefficient of x2)
product of the zeroes (𝝰 x 𝛃) = (c/a) = (Constant term)/Coefficient of x2)
Solution:
(i) x2 – 2x – 8
1) Quadratic polynomial = x2 – 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).
= x2 – 4x + 2x – 8= (x2 – 4x) + (2x – 8)= x(x – 4) + 2(x – 4)= (x – 4)(x + 2)
2) So x2 – 2x – 8 = (x – 4)(x + 2)
3) The value of p(x) is 0 when (x – 4) = 0 or (x + 2) = 0.
4) So, p(x) is 0 when x = 4 or x = – 2.
5) Therefore zeroes of x2 – 2x – 8 are 4 and – 2. i.e 𝝰 = 4 and 𝛃 = – 2,
a = 1, b = – 2 and c = – 8
6) Verification of Relationships Between Zeroes and Coefficients:
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.
7) Therefore, p(x) is 0 when x = – 2 or x = 4, and the relationships between
zeroes and coefficients are verified.
(ii) 4s2 – 4s + 1
1) Quadratic polynomial = 4s2 – 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).
= 4s2 – 2s – 2s + 1= (4s2 – 2s) – (2s – 1)= 2s(2s – 1) – (2s – 1)= (2s – 1)(2s – 1)
2) So 4s2 – 4s + 1 = (2s – 1)(2s – 1)
3) The value of p(s) is 0 when (2s – 1) = 0 or (2s – 1) = 0.
4) So, p(s) is 0 when s = 1/2 or s = 1/2.
5) Therefore zeroes of 4s2 – 4s + 1 are 1/2 and 1/2. i.e 𝝰 = 1/2 and 𝛃 = 1/2,
a = 4, b = – 4 and c = 1
6) Verification of Relationships Between Zeroes and Coefficients:
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.
7) Therefore, p(s) is 0 when s = 1/2 or s = 1/2, and the relationships between
zeroes and coefficients are verified.
(iii) 6x2 – 3 – 7x
1) Quadratic polynomial = 6x2 – 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).
= 6x2 + 2x – 9x – 3= (6x2 + 2x) – (9x + 3)= 2x(3x + 1) – 3(3x + 1)= (3x + 1)(2x – 3)
2) So 6x2 – 7x – 3 = (3x + 1)(2x – 3)
3) The value of p(x) is 0 when (3x + 1) = 0 or (2x – 3) = 0.
4) So, p(x) is 0 when x = –1/3 or x = 3/2.
5) Therefore zeroes of 6x2 – 7x – 3 are –1/3 and 3/2. i.e 𝝰 = –1/3 and 𝛃 = 3/2,
a = 6, b = – 7 and c = – 3
6) Verification of Relationships Between Zeroes and Coefficients:
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.
7) Therefore, p(x) is 0 when x = – 1/3 or x = 3/2, and the relationships between
zeroes and coefficients are verified.
(iv) 4u2 + 8u
1) Quadratic polynomial = 4u2 + 8u
here we can take 4u common directly.
= 4u(u + 2)
2) So 4u2 + 8u = 4u(u + 2)
3) The value of p(u) is 0 when u = 0 or (u + 2) = 0.
4) So, p(u) is 0 when u = 0 or u = – 2.
5) Therefore zeroes of 4u2 + 8u are 0 and – 2. i.e 𝝰 = 0 and 𝛃 = – 2,
a = 4, b = 8 and c = 0
6) Verification of Relationships Between Zeroes and Coefficients:
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.
7) Therefore, p(u) is 0 when u = – 2 or u = 0, and the relationships between
zeroes and coefficients are verified.
(v) t2 – 15
1) Quadratic polynomial = t2 – 15
here, use the formula
(a2 – b2) = (a – b) (a + b) directly.
= (t + √15) (t – √15)
2) So t2 – 15 = (t + √15) (t – √15)
3) The value of p(t) is 0 when (t – √15) = 0 or (t + √15) = 0.
4) So, p(t) is 0 when t = √15 or t = –√15.
5) Therefore zeroes of t2 – 15 are √15 and –√15. i.e 𝝰 = √15 and 𝛃 = –√15,
a = 1, b = 0 and c = – 15
6) Verification of Relationships Between Zeroes and Coefficients:
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.
7) Therefore, p(t) is 0 when t = – √15 or t = √15, and the relationships between
zeroes and coefficients are verified.
(vi) 3x2 – x – 4
1) Quadratic polynomial = 3x2 – 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).
= 3x2 + 3x – 4x – 4= (3x2 + 3x) – (4x + 4)= 3x(x + 1) – 4(x + 1)= (3x – 4)(x + 1)
2) So 3x2 – x – 4 = (3x – 4)(x + 1)
3) The value of p(x) is 0 when (3x – 4) = 0 or (x + 1) = 0.
4) So, p(x) is 0 when x = 4/3 or x = – 1.
5) Therefore zeroes of 3x2 – x – 4 are 4/3 and – 1. i.e 𝝰 = 4/3 and 𝛃 = – 1,
a = 3, b = – 1 and c = – 4
6) Verification of Relationships Between Zeroes and Coefficients:
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.
7) Therefore, p(x) is 0 when x = – 1 or x = 4/3, and the relationships between
zeroes and coefficients are verified.
zeroes and coefficients are verified.
Q2. Find a quadratic polynomial each with the given numbers as the sum and
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:
1) Let 𝝰 and 𝛃 be the zeroes of the polynomial p(x).
2) So p(x) = (x – 𝝰)(x – 𝛃)
p(x) = x2 – (𝝰 + 𝛃) x + (𝝰 x 𝛃)
Solution:
(i) 1/4, – 1
1) Let 𝝰 and 𝛃 be the zeroes of the polynomial p(x)
2) Given that,
the sum of zeroes is (𝝰 + 𝛃) = 1/4, and the product of zeroes is (𝝰 x 𝛃) = –1.
3) Using the standard form of a quadratic polynomial, we have:
p(x) = x2 – (𝝰 + 𝛃) x + (𝝰 x 𝛃)
p(x) = x2 – (1/4) x + (–1)
p(x) = 4x2 – x + (– 4)
p(x) = 4x2 – x – 4
4) Therefore, the quadratic polynomial is 4x2 – x – 4.
(ii) √2, 1/3
1) Let 𝝰 and 𝛃 be the zeroes of the polynomial p(x)
2) Given that,
the sum of zeroes is (𝝰 + 𝛃) = √2, and the product of zeroes is (𝝰 x 𝛃) = 1/3.
3) Using the standard form of a quadratic polynomial, we have:
p(x) = x2 – (𝝰 + 𝛃) x + (𝝰 x 𝛃)
p(x) = x2 – (√2) x + (1/3)
p(x) = x2 – √2x + (1/3)
p(x) = 3x2 – 3√2x + 1
4) Therefore, the quadratic polynomial is 3x2 – 3√2x + 1.
(iii) 0, √5
1) Let 𝝰 and 𝛃 be the zeroes of the polynomial p(x)
2) Given that,
the sum of zeroes is (𝝰 + 𝛃) = 0, and the product of zeroes is (𝝰 x 𝛃) = √5.
3) Using the standard form of a quadratic polynomial, we have:
p(x) = x2 – (𝝰 + 𝛃) x + (𝝰 x 𝛃)
p(x) = x2 – (0) x + (√5)
p(x) = x2 – 0x + √5
p(x) = x2 + √5
4) Therefore, the quadratic polynomial is x2 + √5.
(iv) 1, 1
1) Let 𝝰 and 𝛃 be the zeroes of the polynomial p(x)
2) Given that,
the sum of zeroes is (𝝰 + 𝛃) = 1, and the product of zeroes is (𝝰 x 𝛃) = 1.
3) Using the standard form of a quadratic polynomial, we have:
p(x) = x2 – (𝝰 + 𝛃) x + (𝝰 x 𝛃)
p(x) = x2 – (1) x + (1)
p(x) = x2 – x + 1
4) Therefore, the quadratic polynomial is x2 – x + 1.
(v) – 1/4, 1/4
1) Let 𝝰 and 𝛃 be the zeroes of the polynomial p(x)
2) Given that,
the sum of zeroes is (𝝰 + 𝛃) = – 1/4, and the product of zeroes is (𝝰 x 𝛃) = 1/4.
3) Using the standard form of a quadratic polynomial, we have:
p(x) = x2 – (𝝰 + 𝛃) x + (𝝰 x 𝛃)
p(x) = x2 – (– 1/4) x + (1/4)
p(x) = 4x2 + x + 1
4) Therefore, the quadratic polynomial is 4x2 + x + 1.
(vi) 4, 1
1) Let 𝝰 and 𝛃 be the zeroes of the polynomial p(x)
2) Given that, the sum of zeroes is (𝝰 + 𝛃) = 4, and the product of zeroes is (𝝰 x 𝛃) = 1.
3) Using the standard form of a quadratic polynomial, we have:
p(x) = x2 – (𝝰 + 𝛃) x + (𝝰 x 𝛃)
p(x) = x2 – (4) x + (1)
p(x) = x2 – 4x + 1
4) Therefore, the quadratic polynomial is x2 – 4x + 1.
Conclusion: The Power and Elegance of Polynomials
Polynomials are not just expressions with variables and coefficients – they are the keys to unlocking complex mathematical ideas. From their applications in algebra to their relevance in real-world problems, mastering polynomials equips you with a versatile tool in mathematics. As you progress through this fascinating journey, remember that every equation you solve and every concept you understand builds your foundation for more advanced topics. Embrace the elegance of polynomials and keep solving with curiosity!
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