| Ncert Solution Class 10th (Chapter – 2) Exercise – 2.1 Question No. 1 1. 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.
Solution: 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 | Ncert Solution Class 10th (Chapter – 2) Exercise – 2.2 Question No. 1 1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. (i) x2– 2x – 8 Solution: x2 – 2x – 8
⇒ 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 Solution: 4s2– 4s + 1
⇒ 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 s2 ) (iii) 6x2– 3 – 7x Solution: 6x2–3–7x
⇒ 6x2 – 7x –3
⇒ 6x2 – 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) = -(3/6) = (Constant term) /(Coefficient of x2 ) (iv) 4u2+ 8u Solution: 4u2 + 8u
⇒ 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 u2 ) (v) t2 – 15 Solution: 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 t2 ) (vi) 3x2 – x – 4 Solution: 3x2– x – 4
⇒ 3x2 – 4x + 3x – 4
⇒ x(3x – 4) + 1(3x – 4)
⇒ (3x – 4)(x + 1)
Therefore, zeroes of polynomial equation 3x2 – 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 x2 ) |
| Ncert Solution Class 10th (Chapter – 2) Exercise – 2.2 Question No. 2 2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. (i) 1/4, -1 Solution: From the formulas of sum and product of zeroes, we know,
Sum of zeroes = α + β
Product of zeroes = αβ
Sum of zeroes = α + β = 1/4Product 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– (1/4)x + (-1) = 0
4x2 – x – 4 = 0
Thus,4x2– x – 4 is the quadratic polynomial. (ii) √2, 1/3 Solution: 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
Solution: Given,
Sum of zeroes = α + β = 0
Product of zeroes = αβ = √5
∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly
x2– (α + β)x + αβ = 0
x2– (0)x + √5= 0
Thus, x2 + √5 is the quadratic polynomial. (iv) 1, 1 Solution: Given,
Sum of zeroes = α + β = 1
Product of zeroes = αβ = 1
∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly
x2– (α + β)x + αβ = 0
x2 – x + 1 = 0
Thus ,x2 – x + 1 is the quadratic polynomial. (v) -1/4, 1/4
Solution: Given,
Sum of zeroes = α + β = -1/4
Product of zeroes = αβ = 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
Solution: 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.
Exercise – 2.3 | Ncert Solution Class 10th (Chapter – 2) Exercise – 2.3 Question No. 1 1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following: (i) p(x) = x3– 3x2 + 5x – 3 , g(x) = x2– 2 Solution: Given,
Dividend = p(x) = x3– 3x2 + 5x – 3
Divisor = g(x) = x2 – 2
Therefore, upon division we get,
Quotient = x – 3
Remainder = 7x – 9
(ii) p(x) = x4– 3x2 + 4x + 5 , g(x) = x2 + 1 – x Solution: Given,
Dividend = p(x) = x4 – 3x2 + 4x +5
Divisor = g(x) = x2 + 1 – x
Therefore, upon division we get,
Quotient = x2 + x – 3
Remainder = 8 (iii) p(x) = x4– 5x + 6, g(x) = 2 – x2 Solution: Given,
Dividend = p(x) = x4 – 5x + 6 = x4 + 0x2 – 5x + 6
Divisor = g(x) = 2 – x2 = – x2 + 2
Therefore, upon division we get,
Quotient = – x2– 2
Remainder = – 5x + 10 |
| Ncert Solution Class 10th (Chapter – 2) Exercise – 2.3 Question No. 2 2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial: (i) t2– 3, 2t4 + 3t3– 2t2– 9t – 12 Solution: Given,
First polynomial = t2– 3
Second polynomial = 2t4 + 3t3– 2t2 – 9t -12
As we can see, the remainder is left as 0. Therefore, we say that, t2– 3 is a factor of 2t4 + 3t3– 2t2 – 9t -12. (ii) x2 + 3x + 1 , 3x4 + 5x3 – 7x2 + 2x + 2 Solution: Given,
First polynomial = x2 + 3x + 1
Second polynomial = 3x4 + 5x3– 7x2 + 2x + 2
we can see,the remainder is left as 0.Therefore, say that, x2 + 3x + 1 is a factor of 3x4 + 5x3 – 7x2 + 2x + 2. (iii) x3– 3x + 1, x5– 4x3 + x2 + 3x + 1 Solution: Given,
First polynomial = x3– 3x + 1
Second polynomial = x5– 4x3 + x2 + 3x + 1
As we can see, the remainder is not equal to 0. Therefore, we say that, x3– 3x + 1 is not a factor of x5– 4x3+ x2 + 3x + 1 |
| Ncert Solutions Class 10th (Chapter – 2) Exercise – 2.3 Question No. 3 3. O btain all other zeroes of 3x4 + 6x3– 2x2 – 10x – 5, if two of its zeroes are √(5/3) and – √(5/3). Solution: Since this is a polynomial equation of degree 4, hence there will be total 4 roots.
√(5/3) and – √(5/3) are zeroes of polynomial f(x).
∴ (x – √(5/3)) (x + √(5/3) = x2– (5/3) = 0
(3x2 − 5) = 0, is a factor of given polynomial f(x).
Now, when we will divide f(x) by (3x2 − 5) the quotient obtained will also be a factor of f(x) and the remainder will be 0.
Therefore, 3x4 + 6x3 − 2x2 − 10x – 5 = (3x2 – 5)(x2 + 2x + 1)
Now, on further factorizing (x2 + 2x + 1) we get,
x2 + 2x + 1 = x2 + x + x + 1 = 0
x(x + 1) + 1(x + 1) = 0
(x + 1)(x + 1) = 0
So, its zeroes are given by: x = −1 and x = −1.
Therefore, all four zeroes of given polynomial equation are:
√(5/3),- √(5/3) , −1 and −1.
Hence, is the answer. |
| Ncert Solution Class 10th (Chapter – 2) Exercise – 2.3 Question No. 4 4. On dividing x3– 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and – 2x + 4, respectively. Find g(x). Solution: Given,
Dividend, p(x) = x3– 3x2 + x + 2
Quotient = x – 2
Remainder = – 2x + 4
We have to find the value of Divisor, g(x) =?
As we know,
Dividend = Divisor × Quotient + Remainder
∴ x3– 3x2 + x + 2 = g(x) × (x – 2) + (- 2x + 4)
x3– 3x2 + x + 2 -(- 2x + 4) = g(x) × (x – 2)
Therefore, g(x) × (x – 2) = x3– 3 x2 + 3x – 2
Now, for finding g(x) we will divide x3– 3x2 + 3 x – 2 with (x – 2)
Therefore, g(x) = (x2 – x + 1) |
| Ncert Solution Class 10th (Chapter – 2) Exercise – 2.3 Question No. 5 5. Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
(i) Deg p(x) = deg q(x) Solution: According to the division algorithm, dividend p(x) and divisor g(x) are two polynomials, where g(x) ≠ 0. Then we can find the value of quotient q(x) and remainder r(x), with the help of below given formula;
Dividend = Divisor × Quotient + Remainder
∴ p(x) = g(x) × q(x) + r(x)
Where r(x) = 0 or degree of r(x) < degree of g(x).
Now let us proof the three given cases as per division algorithm by taking examples for each.(i) deg p(x) = deg q(x)
Degree of dividend is equal to degree of quotient, only when the divisor is a constant term.
Let us take an example, p(x) = 3x2 + 3x + 3 is a polynomial to be divided by g(x) = 3.
So, (3x2 + 3x + 3)/3 = x2 + x + 1 = q(x)
Thus, you can see, the degree of quotient q(x) = 2, which also equal to the degree of dividend p(x).
Hence, division algorithm is satisfied here. (ii) Deg q(x) = deg r(x) Solution: Deg q(x) = deg r(x)
Let us take an example, p(x) = x2 + 3 is a polynomial to be divided by g(x) = x – 1.
So, x2 + 3 = (x – 1)×(x) + (x + 3)
Hence, quotient q(x) = x
Also, remainder r(x) = x + 3
Thus, you can see, the degree of quotient q(x) = 1, which is also equal to the degree of remainder r(x).
Hence, division algorithm is satisfied here. (iii) Deg r(x) = 0 Solution: Deg r(x) = 0
The degree of remainder is 0 only when the remainder left after division algorithm is constant.
Let us take an example, p(x) = x2 + 1 is a polynomial to be divided by g(x) = x.
So, x2 + 1 = (x) × (x) + 1
Hence, quotient q(x) = x
And, remainder r(x) = 1
Clearly, the degree of remainder here is 0.
Hence, division algorithm is satisfied here. |
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