NCERT Solutions for Class 10th Mathematics: Chapter 2 Polynomials

Class 10: Mathematics Chapter 2 solutions. Complete Class 10 Mathematics Chapter 2 Notes.

NCERT Solutions for Class 10th Mathematics: Chapter 2 Polynomials

NCERT 10th Mathematics Chapter 2, class 10 Mathematics Chapter 2 solutions

Page No: 28

Exercises 2.1

1. The graphs of y = p(x) are given in following figure, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

Answer

(i) The number of zeroes is 0 as the graph does not cut the x-axis at any point.
(ii) The number of zeroes is 1 as the graph intersects the x-axis at only 1 point.
(iii) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.
(iv) The number of zeroes is 2 as the graph intersects the x-axis at 2 points.
(v) The number of zeroes is 4 as the graph intersects the x-axis at 4 points.
(vi) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.

1. 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

Answer

(i) x² – 2x – 8
= (x - 4) (x + 2)
The value of x² – 2x – 8 is zero when x - 4 = 0 or x + 2 = 0, i.e., when x = 4 or x = -2
Therefore, the zeroes of x² – 2x – 8 are 4 and -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
= (2s-1)²
The value of 4s² - 4s + 1 is zero when 2s - 1 = 0, i.e., s = 1/2

Therefore, the zeroes of 4s² - 4s + 1 are 1/2 and 1/2.

Sum of zeroes = 1/2 + 1/2 = 1 = -(-4)/4 = -(Coefficient of s)/Coefficient of s²
Product of zeroes = 1/2 × 1/2 = 1/4 = Constant term/Coefficient of s².

(iii) 6x² – 3 – 7x
= 6x² – 7x – 3
= (3x + 1) (2x - 3)
The value of 6x² - 3 - 7x is zero when 3x + 1 = 0 or 2x - 3 = 0, i.e., x = -1/3 or x = 3/2

Therefore, the zeroes of 6x² - 3 - 7x are -1/3 and 3/2.

Sum of zeroes = -1/3 + 3/2 = 7/6 = -(-7)/6 = -(Coefficient of x)/Coefficient of x²
Product of zeroes = -1/3 × 3/2 = -1/2 = -3/6 = Constant term/Coefficient of x².

(iv) 4u² + 8u
= 4u² + 8u + 0
= 4u(u + 2)
The value of 4u² + 8u is zero when 4u = 0 or u + 2 = 0, i.e., u = 0 or u = - 2
Therefore, the zeroes of 4u² + 8u are 0 and - 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
= t² - 0.t - 15
= (t - √15) (t + √15)
The value of t² - 15 is zero when t - √15 = 0 or t + √15 = 0, i.e., when t = √15 or t = -√15
Therefore, the zeroes of t² - 15 are √15 and -√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 u².

(vi) 3x² – x – 4
= (3x - 4) (x + 1)
The value of 3x² – x – 4 is zero when 3x - 4 = 0 and x + 1 = 0,i.e., when x = 4/3 or x = -1
Therefore, the zeroes of 3x² – x – 4 are 4/3 and -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 10th Mathematics Chapter 2, class 10 Mathematics Chapter 2 solutions

2. 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

Answer

(i) 1/4 , -1
Let the polynomial be ax² + bx + c, and its zeroes be α and ß
α + ß = 1/4 = -b/a
αß = -1 = -4/4 = c/a
If a = 4, then b = -1, c = -4
Therefore, the quadratic polynomial is 4x² - x -4.

(ii) √2 , 1/3
Let the polynomial be ax² + bx + c, and its zeroes be α and ß
α + ß = √2 = 3√2/3 = -b/a
αß = 1/3 = c/a
If a = 3, then b = -3√2, c = 1
Therefore, the quadratic polynomial is 3x² -3√2x +1.

(iii) 0, √5
Let the polynomial be ax² + bx + c, and its zeroes be α and ß
α + ß = 0 = 0/1 = -b/a
αß = √5 = √5/1 = c/a
If a = 1, then b = 0, c = √5
Therefore, the quadratic polynomial is x² + √5.

(iv) 1, 1
Let the polynomial be ax² + bx + c, and its zeroes be α and ß
α + ß = 1 = 1/1 = -b/a
αß = 1 = 1/1 = c/a
If a = 1, then b = -1, c = 1
Therefore, the quadratic polynomial is x² - x +1.

(v) -1/4 ,1/4
Let the polynomial be ax² + bx + c, and its zeroes be α and ß
α + ß = -1/4 = -b/a
αß = 1/4 = c/a
If a = 4, then b = 1, c = 1
Therefore, the quadratic polynomial is 4x² + x+1

.(vi) 4,1
Let the polynomial be ax² + bx + c, and its zeroes be α and ß
α + ß = 4 = 4/1 = -b/a
αß = 1 = 1/1 = c/a
If a = 1, then b = -4, c = 1
Therefore, the quadratic polynomial is x² - 4x+1

Exercise 2.3

NCERT 10th Mathematics Chapter 2, class 10 Mathematics Chapter 2 solutions

1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

Answer

(i) p(x) = x³ – 3x² + 5x – 3, g(x) = x² – 2

Quotient = x-3 and remainder 7x - 9

(ii) p(x) = x⁴ – 3x² + 4x + 5, g(x) = x² + 1 – x

Quotient = x² + x - 3 and remainder 8

(iii) p(x) = x⁴ – 5x + 6, g(x) = 2 – x²

Quotient = -x² -2 and remainder -5x +10

2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

Answer

(i) t² – 3,  2t⁴ + 3t³ – 2t² – 9t – 12

t² – 3 exactly divides  2t⁴ + 3t³ – 2t² – 9t – 12 leaving no remainder. Hence, it is a factor of  2t⁴ + 3t³ – 2t² – 9t – 12.

(ii) x² + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2

x² + 3x + 1 exactly divides 3x⁴ + 5x³ – 7x² + 2x + 2 leaving no remainder. Hence, it is factor of 3x⁴ + 5x³ – 7x² + 2x + 2.

(iii) x³ – 3x + 1, x⁵ – 4x³ + x² + 3x + 1

x³ – 3x + 1 didn't divides exactly x⁵ – 4x³ + x² + 3x + 1 and leaves 2 as remainder. Hence, it not a factor of x⁵ – 4x³ + x² + 3x + 1.

3. Obtain all other zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeroes are √(5/3) and - √(5/3).

Answer

p(x) = 3x⁴ + 6x³ – 2x² – 10x – 5
Since the two zeroes are √(5/3) and - √(5/3).

We factorize x² + 2x + 1
= (x + 1)²
Therefore, its zero is given by x + 1 = 0
x = -1
As it has the term (x + 1)² , therefore, there will be 2 zeroes at x = - 1.
Hence, the zeroes of the given polynomial are √(5/3) and - √(5/3), - 1 and - 1.

NCERT 10th Mathematics Chapter 2, class 10 Mathematics Chapter 2 solutions

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).

Answer

Here in the given question,
Dividend = x³ - 3x² + x + 2
Quotient = x - 2
Remainder = -2x + 4
Divisor = g(x)
We know that,
Dividend = Quotient × Divisor + Remainder
⇒ x³ - 3x² + x + 2 = (x - 2) × g(x) + (-2x + 4)⇒ x³ - 3x² + x + 2 - (-2x + 4) = (x - 2) × g(x)
⇒ x³ - 3x² + 3x - 2 = (x - 2) × g(x)
⇒ g(x) =  (x³ - 3x² + 3x - 2)/(x - 2)

∴ g(x) = (x² - x + 1)

5.Give examples of polynomial p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0

Answer

(i) Let us assume the division of 6x² + 2x + 2 by 2
Here, p(x) = 6x² + 2x + 2
g(x) = 2
q(x) = 3x² + x + 1
r(x) = 0
Degree of p(x) and q(x) is same i.e. 2.
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
Or, 6x² + 2x + 2 = 2x (3x² + x + 1)
Hence, division algorithm is satisfied.

(ii) Let us assume the division of x³+ x by x²,
Here, p(x) = x³ + x
g(x) = x²
q(x) = x and r(x) = x
Clearly, the degree of q(x) and r(x) is the same i.e., 1.
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
x³ + x = (x² ) × x + x
x³ + x = x³ + x
Thus, the division algorithm is satisfied.

(iii) Let us assume the division of x³+ 1 by x².
Here, p(x) = x³ + 1
g(x) = x²
q(x) = x and r(x) = 1
Clearly, the degree of r(x) is 0.
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
x³ + 1 = (x² ) × x + 1
x³ + 1 = x³ + 1
Thus, the division algorithm is satisfied.

Exercise 2.4 (Optional)

1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:
(i) 2x³ + x² - 5x + 2; 1/2, 1, -2
(ii) x³ - 4x² + 5x - 2; 2, 1, 1

Answer

(i) p(x) = 2x³ + x² - 5x + 2
Now for zeroes, putting the given value in x.

p(1/2) = 2(1/2)³ + (1/2)² - 5(1/2)+ 2
= (2×1/8) + 1/4 - 5/2 + 2
= 1/4 + 1/4 - 5/2 + 2
= 1/2 - 5/2 + 2 = 0

p(1) = 2(1)³ + (1)² - 5(1)+ 2
= (2×1) + 1 - 5 + 2
= 2 + 1 - 5 + 2 = 0

p(-2) = 2(-2)³ + (-2)² - 5(-2)+ 2
= (2 × -8) + 4 + 10 + 2
= -16 + 16 = 0

Thus, 1/2, 1 and -2 are the zeroes of the given polynomial.

Comparing the given polynomial with ax³ + bx² + cx + d, we get a=2, b=1, c=-5, d=2
Also, α=1/2, β=1 and γ=-2
Now,
-b/a = α+β+γ
⇒ 1/2 = 1/2 + 1 - 2
⇒ 1/2 = 1/2

c/a = 

αβ+βγ+γα
⇒ -5/2 = (1/2 × 1) + (1 × -2) + (-2 × 1/2)
⇒ -5/2 = 1/2 - 2 - 1
⇒ -5/2 = -5/2

-d/a = αβγ
⇒ -2/2 = (1/2 × 1 × -2)
⇒ -1 = 1

Thus, the relationship between zeroes and the coefficients are verified.

(ii)  p(x) = x³ - 4x² + 5x - 2
Now for zeroes, putting the given value in x.

p(2) = 2³ - 4(2)² + 5(2)- 2
= 8 - 16 + 10 - 2
= 0

p(1) = 1³ - 4(1)² + 5(1)- 2
= 1 - 4 + 5 - 2
= 0

p(1) = 1³ - 4(1)² + 5(1)- 2
= 1 - 4 + 5 - 2
= 0

Thus, 2, 1 and 1 are the zeroes of the given polynomial.

Comparing the given polynomial with ax³ + bx² + cx + d, we get a=1, b=-4, c=5, d=-2
Also, α=2, β=1 and γ=1
Now,
-b/a = α+β+γ
⇒ 4/1 = 2 + 1 + 1
⇒ 4 = 4

c/a = 

αβ+βγ+γα
⇒ 5/1 = (2 × 1) + (1 × 1) + (1 × 2)
⇒ 5 = 2 + 1 + 2
⇒ 5 = 5

-d/a = αβγ
⇒ 2/1 = (2 × 1 × 1)
⇒ 2 = 2

Thus, the relationship between zeroes and the coefficients are verified.

NCERT 10th Mathematics Chapter 2, class 10 Mathematics Chapter 2 solutions

2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively.

Answer

Let the polynomial be ax³ + bx² + cx + d and the zeroes be α, β and γ
Then, α + β + γ = -(-2)/1 = 2 = -b/a
αβ + βγ + γα = -7 = -7/1 = c/a
αβγ = -14 = -14/1 = -d/a

∴ a = 1, b = -2, c = -7 and d = 14
So, one cubic polynomial which satisfy the given conditions will be x³ - 2x²  - 7x + 14

3. If the zeroes of the polynomial x³ – 3x² + x + 1 are a–b, a, a+b, find a and b.

Answer

Since, (a - b), a, (a + b) are the zeroes of the polynomial x³ – 3x² + x + 1.
Therefore, sum of the zeroes = (a - b) + a + (a + b) = -(-3)/1 = 3

⇒ 3a = 3 ⇒ a =1

∴ Sum of the products of is zeroes taken two at a time = a(a - b) + a(a + b) + (a + b) (a - b) =1/1 = 1
a² - ab + a² + ab + a² - b² = 1
⇒ 3a² - b² =1

Putting the value of a,

⇒ 3(1)² - b² = 1
⇒ 3 - b² = 1
⇒ b² = 2
⇒ b = ±√2
Hence, a = 1 and b = ±√2

4. If two zeroes of the polynomial x⁴ – 6x³ – 26x² + 138x – 35 are  2±√3,  find other zeroes.

Answer

2+√3 and 2-√3 are two zeroes of the polynomial p(x) = x⁴ – 6x³ – 26x² + 138x – 35.
Let x = 2±√3
So, x-2 = ±√3
On squaring, we get x² - 4x + 4 = 3,
⇒ x² - 4x + 1= 0

Now, dividing p(x) by x² - 4x + 1

∴ p(x) = x⁴ - 6x³ - 26x² + 138x - 35
= (x² - 4x + 1) (x² - 2x - 35)
= (x² - 4x + 1) (x² - 7x + 5x - 35)
= (x² - 4x + 1) [x(x - 7) + 5 (x - 7)]
= (x² - 4x + 1) (x + 5) (x - 7)

∴ (x + 5) and (x - 7) are other factors of p(x).
∴ - 5 and 7 are other zeroes of the given polynomial.

5. If the polynomial x⁴ – 6x³ + 16x² – 25x + 10 is divided by another polynomial x² – 2x + k, the remainder comes out to be x + a, find k and a.

Answer

On dividing x⁴ – 6x³ + 16x² – 25x + 10 by x² – 2x + k

∴ Remainder = (2k - 9)x - (8 - k)k + 10 

But the remainder is given as x+ a. 

On comparing their coefficients,

2k - 9 = 1
⇒ k = 10
⇒ k = 5 and,
-(8-k)k + 10 = a
⇒ a = -(8 - 5)5 + 10 =- 15 + 10 = -5
Hence, k = 5 and a = -5

NCERT 10th Mathematics Chapter 2, class 10 Mathematics Chapter 2 solutions

NCERT Solutions for Class 10th Mathematics: Chapter 2: Download PDF

NCERT Solutions for Class 10th Mathematics: Chapter 2 Polynomials

Download PDF: NCERT Solutions for Class 10th Mathematics: Chapter 2 Polynomials PDF

Chapterwise NCERT Solutions for Class 10 Maths:

• Chapter 1 Real Numbers
• Chapter 2 Polynomials
• Chapter 3 Pair of Linear Equations in Two Variables
• Chapter 4 Quadratic Equations
• Chapter 5 Arithmetic Progressions
• Chapter 6 Triangles
• Chapter 7 Coordinate Geometry
• Chapter 8 Introduction to Trigonometry
• Chapter 9 Applications of Trigonometry
• Chapter 10 Circle
• Chapter 11 Constructions
• Chapter 12 Areas related to Circles
• Chapter 13 Surface Areas and Volumes
• Chapter 14 Statistics
• Chapter 15 Probability

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