NCERT Solutions Class 10 Mathematics Chapter 2 Polynomials

Exercise 2.1

Q.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.
Sol.1) (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.

Exercise 2.2

Q.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
Sol.1) (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.

Q.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
Sol.2) (i) 1/4, −1
Let the polynomial be 𝑎𝑥2 + 𝑏𝑥 + 𝑐, and its zeroes be 𝛼 and ß

Exercise 2.3

Q.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
(ii) 𝑝(𝑥) = 𝑥⁴ – 3𝑥² + 4𝑥 + 5, 𝑔(𝑥) = 𝑥² + 1 – 𝑥
(iii) 𝑝(𝑥) = 𝑥⁴ – 5𝑥 + 6, 𝑔(𝑥) = 2 – 𝑥²
Sol.1) (i) 𝑝(𝑥) = 𝑥³ – 3𝑥² + 5𝑥 – 3, 𝑔(𝑥) = 𝑥² – 2

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

Q.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
(ii) 𝑥² + 3𝑥 + 1, 3𝑥⁴ + 5𝑥³ – 7𝑥² + 2𝑥 + 2
(iii) 𝑥³ – 3𝑥 + 1, 𝑥⁵ – 4𝑥³ + 𝑥² + 3𝑥 + 1
Sol.2) (i) 𝑡²– 3, 2𝑡⁴ + 3𝑡³– 2𝑡² – 9𝑡 – 12

Q.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).
Sol.4) 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) = (𝑥² − 𝑥 + 1)

Q.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
Sol.5) (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 𝑥² + 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

Q.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
Sol.1) (i) 𝑝(𝑥) = 2𝑥³ + 𝑥² − 5𝑥 + 2
Now for zeroes, putting the given value in 𝑥.

−(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) = 1 3 − 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.

Q.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.
Sol.2) 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

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

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

Q.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.
Sol.5) 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