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NCERT Solutions for Class 10 Maths Chapter 2 - Polynomials (Exercise 2.1)

 Exercise 2.1

Q1: The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.
NCERT Solutions for Class 10 Maths Chapter 2 - Polynomials (Exercise 2.1)

Sol:
Graphical method to find zeroes:
Total number of zeroes in any polynomial equation = total number of times the curve intersects x-axis.
(i) In the given graph, the number of zeroes of p(x) is 0 because the graph is parallel to x-axis does not cut it at any point.
(ii) In the given graph, the number of zeroes of p(x) is 1 because the graph intersects the x-axis at only one point.
(iii) In the given graph, the number of zeroes of p(x) is 3 because the graph intersects the x-axis at any three points.
(iv) In the given graph, the number of zeroes of p(x) is 2 because the graph intersects the x-axis at two points.
(v) In the given graph, the number of zeroes of p(x) is 4 because the graph intersects the x-axis at four points.
(vi) In the given graph, the number of zeroes of p(x) is 3 because the graph intersects the x-axis at three points.


(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
Sol: ⇒x2– 4x+2x–8 = x(x–4)+2(x–4) = (x-4)(x+2)
Therefore, zeroes of polynomial equation x2–2x–8 are (4, -2)
Sum of zeroes = 4–2 = 2 = -(-2)/1 = -(Coefficient of x)/(Coefficient of x2)
Product of zeroes = 4×(-2) = -8 =-(8)/1 = (Constant term)/(Coefficient of x2)

(ii) 4s2–4s+1
Sol: ⇒4s2–2s–2s+1 = 2s(2s–1)–1(2s-1) = (2s–1)(2s–1)
Therefore, zeroes of polynomial equation 4s2–4s+1 are (1/2, 1/2)
Sum of zeroes = (½)+(1/2) = 1 = -(-4)/4 = -(Coefficient of s)/(Coefficient of s2)
Product of zeros = (1/2)×(1/2) = 1/4 = (Constant term)/(Coefficient of s)

(iii) 6x2–3–7x
Sol: ⇒6x2–7x–3 = 6x– 9x + 2x – 3 = 3x(2x – 3) +1(2x – 3) = (3x+1)(2x-3)
Therefore, zeroes of polynomial equation 6x2–3–7x are (-1/3, 3/2)
Sum of zeroes = -(1/3)+(3/2) = (7/6) = -(Coefficient of x)/(Coefficient of x2)
Product of zeroes = -(1/3)×(3/2) = -(1/2) = (Constant term) /(Coefficient of x)

(iv) 4u2+8u
Sol: ⇒ 4u(u+2)
Therefore, zeroes of polynomial equation 4u2 + 8u are (0, -2).
Sum of zeroes = 0+(-2) = -2 = -(8/4) = = -(Coefficient of u)/(Coefficient of u2)
Product of zeroes = 0×-2 = 0 = 0/4 = (Constant term)/(Coefficient of u)

(v) t2–15
⇒ t2 = 15 or t = ± √15
Therefore, zeroes of polynomial equation t2 –15 are (√15, -√15)
Sum of zeroes =√15 + (-√15) = 0 = -(0/1)= -(Coefficient of t) / (Coefficient of t2)
Product of zeroes = √15 × (-√15) = -15 = -15/1 = (Constant term) / (Coefficient of t)

(vi) 3x2–x–4
⇒ 3x2–4x+3x–4 = x(3x-4)+1(3x-4) = (3x – 4)(x + 1)
Therefore, zeroes of polynomial equation3x2 – x – 4 are (4/3, -1)
Sum of zeroes = (4/3)+(-1) = (1/3)= -(-1/3) = -(Coefficient of x) / (Coefficient of x2)
Product of zeroes=(4/3)×(-1) = (-4/3) = (Constant term) /(Coefficient of x)


Q2: Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes, respectively.
(i) 1/4 , -1
Sol: From the formulas of sum and product of zeroes, we know,
Sum of zeroes = α+β
Product of zeroes = α β
Sum of zeroes = α+β = 1/4
Product of zeroes = α β = -1
∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-
x2–(α+β)x +αβ = 0
x2–(1/4)x +(-1) = 0
4x2–x-4 = 0
Thus4x2–x–4 is the quadratic polynomial.

(ii) √2, 1/3
Sol: Sum of zeroes = α + β =√2
Product of zeroes = α β = 1/3
∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-
x2–(α+β)x +αβ = 0
x2 –(√2)x + (1/3) = 0
3x2-3√2x+1 = 0
Thus, 3x2-3√2x+1 is the quadratic polynomial.

(iii) 0, √5
Sol: Given,
Sum of zeroes = α+β = 0
Product of zeroes = α β = √5
∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-
x2–(α+β)x +αβ = 0
x2–(0)x +√5= 0
Thus, x2+√5 is the quadratic polynomial.

(iv) 1, 1
Sol: Given,
Sum of zeroes = α+β = 1
Product of zeroes = α β = 1
∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-
x2–(α+β)x +αβ = 0
x2–x+1 = 0
Thus, x2–x+1 is the quadratic polynomial.

(v) -1/4, 1/4
Sol: Given,
Sum of zeroes = α+β = -1/4
Product of zeroes = α β = 1/4
∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-
x2–(α+β)x +αβ = 0
x2–(-1/4)x +(1/4) = 0
4x2+x+1 = 0
Thus, 4x2+x+1 is the quadratic polynomial.

(vi) 4, 1
Sol: Given,
Sum of zeroes = α+β =4
Product of zeroes = αβ = 1
∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-
x2–(α+β)x+αβ = 0
x2–4x+1 = 0
Thus, x2–4x+1 is the quadratic polynomial.

The document NCERT Solutions for Class 10 Maths Chapter 2 - Polynomials (Exercise 2.1) is a part of the Class 10 Course Mathematics (Maths) Class 10.
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FAQs on NCERT Solutions for Class 10 Maths Chapter 2 - Polynomials (Exercise 2.1)

1. What are polynomials and how are they classified?
Ans. Polynomials are algebraic expressions that consist of variables raised to non-negative integer powers and coefficients. They are classified based on their degree: a polynomial of degree 0 is a constant, degree 1 is linear, degree 2 is quadratic, degree 3 is cubic, and so on. Additionally, they can be classified based on the number of terms as monomials (one term), binomials (two terms), and trinomials (three terms).
2. How do you add and subtract polynomials?
Ans. To add or subtract polynomials, you combine like terms. Like terms are terms that have the same variable raised to the same power. For addition, you sum the coefficients of the like terms, and for subtraction, you subtract the coefficients. Ensure that you maintain the correct signs throughout the operation.
3. What is the significance of the Remainder Theorem in polynomial division?
Ans. The Remainder Theorem states that when a polynomial f(x) is divided by a linear divisor of the form (x - c), the remainder of this division is equal to f(c). This theorem is significant because it provides a quick way to evaluate polynomials at specific points without performing long division.
4. How can you factor polynomials, and why is it important?
Ans. Factoring polynomials involves expressing them as a product of simpler polynomials or monomials. Techniques for factoring include taking out the greatest common factor (GCF), using quadratic factoring methods, and applying special product formulas (like difference of squares). Factoring is important because it simplifies expressions, solves equations, and helps in understanding the roots of polynomials.
5. What are the types of zeros of a polynomial, and how can they be determined?
Ans. The zeros of a polynomial are the values of x for which the polynomial equals zero. They can be real or complex, and their nature can be determined using methods like factoring, synthetic division, or applying the quadratic formula for quadratic polynomials. The Fundamental Theorem of Algebra states that a polynomial of degree n has exactly n roots (counting multiplicities), which helps in understanding the behavior of the polynomial graphically.
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