cbse ncert solutions for class 10 mathematics chapter 2 · 2019-08-20 · class- x-cbse-mathematics...

Class- X-CBSE-Mathematics Polynomials

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CBSE NCERT Solutions for Class 10 Mathematics Chapter 2 Back of Chapter Questions

1. The graphs of 𝑦𝑦 = 𝑝𝑝(𝑥𝑥) are given in following figure, for some polynomials𝑝𝑝(𝑥𝑥). Find the number of zeroes of 𝑝𝑝(𝑥𝑥), in each case.

(i)

(ii)

(iii)

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(iv)

(v)

(vi)

Solution:

(i) Since the graph of 𝒑𝒑(𝑥𝑥) does not cut the X-axis at all. Therefore, thenumber of zeroes is 𝟎𝟎.

(ii) As the graph of 𝒑𝒑(𝑥𝑥) intersects the X-axis at only 𝟏𝟏 point. Therefore, thenumber of zeroes is 𝟏𝟏.





(iii) Since the graph of 𝒑𝒑(𝑥𝑥) intersects the X-axis at 𝟑𝟑 points. Hence, thenumber of zeroes is 3.

(iv) As the graph of 𝒑𝒑(𝑥𝑥) intersects the X-axis at 𝟐𝟐 points. So, the number ofzeroes is 𝟐𝟐.

(v) Since the graph of 𝒑𝒑(𝑥𝑥) intersects the X-axis at 𝟒𝟒 points. Therefore, thenumber of zeroes is 𝟒𝟒.

(vi) As the graph of 𝒑𝒑(𝑥𝑥) intersects the X-axis at 𝟑𝟑 points. So, the number ofzeroes is 𝟑𝟑.

♦ ♦ ♦

EXERCISE 2.2

1. Find the zeroes of the following quadratic polynomials and verify the relationshipbetween the zeroes and the coefficients.

(i) 𝑥𝑥2 − 2𝑥𝑥 − 8

(ii) 4𝑠𝑠2 − 4𝑠𝑠 + 1

(iii) 6𝑥𝑥2 − 3 − 7𝑥𝑥

(iv) 4𝑢𝑢2 + 8𝑢𝑢

(v) 𝑡𝑡2 − 15

(vi) 3𝑥𝑥2 − 𝑥𝑥 − 4

Solution:

(i) 𝑥𝑥2 − 2𝑥𝑥 − 8

= 𝑥𝑥2 − 4𝑥𝑥 + 2𝑥𝑥 − 8 [Factorisation by splitting the middle term]

= 𝑥𝑥(𝑥𝑥 − 4) + 2(𝑥𝑥 − 4)

= (𝑥𝑥 − 4)(𝑥𝑥 + 2)

We know that the zeroes of the quadratic polynomial 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 arethe same as the roots of the quadratic equation 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0.

Therefore, by equating the given polynomial to zero. We get,





𝑥𝑥2 − 2𝑥𝑥 − 8 = 0

⇒ (𝑥𝑥 − 4)(𝑥𝑥 + 2) = 0

⇒ 𝑥𝑥 − 4 = 0 or 𝑥𝑥 + 2 = 0

⇒ 𝑥𝑥 = 4 or 𝑥𝑥 = −2

Therefore, the zeroes of 𝑥𝑥² − 2𝑥𝑥 − 8 are 4 and −2.

Sum of zeroes = 4 − 2 = 2 = −(−2)1

= −(Coefficient of 𝑥𝑥)Coefficient of 𝑥𝑥2

Product of zeroes = 4 × (−2) = −8 = (−8)1

= Constant termCoefficient of 𝑥𝑥2

Hence, the relationship between the zeroes and the coefficients is verified.

(ii) 4𝑠𝑠2 − 4𝑠𝑠 + 1 = (2𝑠𝑠 − 1)2 [Since, 𝑎𝑎2 − 2𝑎𝑎𝑏𝑏 + 𝑏𝑏2 = (𝑎𝑎 − 𝑏𝑏)2]



4𝑠𝑠2 − 4𝑠𝑠 + 1 = 0

⇒ (2𝑠𝑠 − 1)2 = 0

Cancelling square on both the sides,

⇒ 2𝑠𝑠 − 1 = 0

⇒ 𝑠𝑠 =12

Therefore, the zeroes of 4𝑠𝑠2 − 4𝑠𝑠 + 1 are 12 and 1

2.

Sum of zeroes = 12

+ 12

= 1 = −(−4)4

= −(Coefficient of 𝑠𝑠)(Coefficient of 𝑠𝑠2)

Product of zeroes = 12

× 12

= 14

= Constant termCoefficient of 𝑠𝑠2


(iii) 6𝑥𝑥2 − 3 − 7𝑥𝑥 = 6𝑥𝑥2 − 7𝑥𝑥 − 3

= 6𝑥𝑥2 − 9𝑥𝑥 + 2𝑥𝑥 − 3 [Factorisation by splitting the middle term]





= 3𝑥𝑥(2𝑥𝑥 − 3) + (2𝑥𝑥 − 3)

= (3𝑥𝑥 + 1)(2𝑥𝑥 − 3)

We know that the zeroes of the quadratic polynomial 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 are the same as the roots of the quadratic equation 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0.


6𝑥𝑥2 − 3 − 7𝑥𝑥 = 0

⇒ 3𝑥𝑥 + 1 = 0 or 2𝑥𝑥 − 3 = 0

⇒ 𝑥𝑥 = −13

or 𝑥𝑥 = 32

Therefore, the zeroes of 6𝑥𝑥2 − 3 − 7𝑥𝑥 are −13

and 32

Sum of zeroes = −13

+ 32

= 76

= −(−7)6


Product of zeroes = −13

× 32

= −12

= −36



(iv) 4𝑢𝑢2 + 8𝑢𝑢 = 4𝑢𝑢2 + 8𝑢𝑢 + 0 = 4𝑢𝑢(𝑢𝑢 + 2)



4𝑢𝑢2 + 8𝑢𝑢 = 0

⇒ 4𝑢𝑢 = 0 or 𝑢𝑢 + 2 = 0

⇒ 𝑢𝑢 = 0 or 𝑢𝑢 = −2

So, the zeroes of 4𝑢𝑢2 + 8𝑢𝑢 are 0 and −2.

Sum of zeroes = 0 + (−2) = −2 = −84

= −(Coefficient of 𝑢𝑢)Coefficient of 𝑢𝑢2

Product of zeroes = 0 × (−2) = 0 = 04

= Constant termCoefficient of 𝑢𝑢2






(v) 𝑡𝑡2 − 15 = 𝑡𝑡2 − 0. 𝑡𝑡 − 15 = �𝑡𝑡 − √15��𝑡𝑡 + √15� [Since, 𝑎𝑎2 − 𝑏𝑏2 =(𝑎𝑎 + 𝑏𝑏)(𝑎𝑎 − 𝑏𝑏)]



𝑡𝑡2 − 15 = 0

⇒ 𝑡𝑡 − √15 = 0 or 𝑡𝑡 + √15 = 0

⇒ 𝑡𝑡 = √15 or 𝑡𝑡 = −√15

Therefore, the zeroes of 𝑡𝑡2 − 15 are √15 and −√15

Sum of zeroes = √15 + �−√15� = 0 = −01

= −(Coefficient of 𝑡𝑡)(Coefficient of 𝑡𝑡2)

Product of zeroes = �√15��−√15� = −15 = −151



(vi) 3𝑥𝑥2 − 𝑥𝑥 − 4

= 3𝑥𝑥2 − 4𝑥𝑥 + 3𝑥𝑥 − 4 [Factorisation by splitting the middle term]

= 𝑥𝑥(3𝑥𝑥 − 4) + (3𝑥𝑥 − 4)

= (3𝑥𝑥 − 4)(𝑥𝑥 + 1)

We know that the zeroes of the quadratic polynomial 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 are the same as the roots of the quadratic equation 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0.


3𝑥𝑥2 − 𝑥𝑥 − 4 = 0

⇒ 3𝑥𝑥 − 4 = 0 or 𝑥𝑥 + 1 = 0

⇒ 𝑥𝑥 =43

or 𝑥𝑥 = −1

Hence, the zeroes of 3𝑥𝑥2 − 𝑥𝑥 − 4 are 43 and −1.

Sum of zeroes = 43

+ (−1) = 13

= −(−1)3






Product of zeroes = 43

(−1) = −43



2. Find a quadratic polynomial each with the given numbers as the sum and productof its zeroes respectively.

(i) 𝟏𝟏𝟒𝟒

,−𝟏𝟏

(ii) √𝟐𝟐, 𝟏𝟏𝟑𝟑

(iii) 𝟎𝟎,√𝟓𝟓

(iv) 𝟏𝟏,𝟏𝟏

(v) −𝟏𝟏𝟒𝟒

, 𝟏𝟏𝟒𝟒

(vi) 𝟒𝟒,𝟏𝟏

Solution:

(i) We know that if 𝛼𝛼 𝑎𝑎𝑎𝑎𝑎𝑎 𝛽𝛽 are the zeroes of a quadratic polynomial𝑝𝑝(𝑥𝑥), 𝑡𝑡hen, the polynomial 𝑝𝑝(𝑥𝑥) can be written as 𝑝𝑝(𝑥𝑥) = 𝑎𝑎{𝑥𝑥2 −(𝛼𝛼 + 𝛽𝛽)𝑥𝑥 + 𝛼𝛼𝛽𝛽} or 𝑝𝑝(𝑥𝑥) = 𝑎𝑎{𝑥𝑥2 − (Sum of the zeroes)𝑥𝑥 +Product of the zeroes}, where 𝑎𝑎 is a non-zero real number.

Given: sum of the roots = 𝛼𝛼 + 𝛽𝛽 = 14 and product of the roots = 𝛼𝛼𝛽𝛽 = −1

Hence, the quadratic polynomial 𝑝𝑝(𝑥𝑥) can be written as:

𝑝𝑝(𝑥𝑥) = 𝑎𝑎{𝑥𝑥2 −14𝑥𝑥 − 1}

= 𝑎𝑎 �4𝑥𝑥2 − 𝑥𝑥 − 4

4�

By taking 𝑎𝑎 = 4, we get one of the quadratic polynomials which satisfy the given conditions.

Therefore, the quadratic polynomial is (4𝑥𝑥2 − 𝑥𝑥 − 4).

(ii) We know that if 𝛼𝛼 𝑎𝑎𝑎𝑎𝑎𝑎 𝛽𝛽 are the zeroes of a quadratic polynomial𝑝𝑝(𝑥𝑥), 𝑡𝑡hen, the polynomial 𝑝𝑝(𝑥𝑥) can be written as 𝑝𝑝(𝑥𝑥) = 𝑎𝑎{𝑥𝑥2 −





(𝛼𝛼 + 𝛽𝛽)𝑥𝑥 + 𝛼𝛼𝛽𝛽} or 𝑝𝑝(𝑥𝑥) = 𝑎𝑎{𝑥𝑥2 − (Sum of the zeroes)𝑥𝑥 +Product of the zeroes}, where 𝑎𝑎 is a non-zero real number.

Given: sum of the roots = 𝛼𝛼 + 𝛽𝛽 = √2 and product of the roots = 𝛼𝛼𝛽𝛽 = 13


𝑝𝑝(𝑥𝑥) = 𝑎𝑎{𝑥𝑥2 − √2𝑥𝑥 +13

}

= 𝑎𝑎 �3𝑥𝑥2 − 3√2𝑥𝑥 + 1

3�


Therefore, the quadratic polynomial is �3𝑥𝑥2 − 3√2𝑥𝑥 + 1�.

(iii) We know that if 𝛼𝛼 𝑎𝑎𝑎𝑎𝑎𝑎 𝛽𝛽 are the zeroes of a quadratic polynomial𝑝𝑝(𝑥𝑥), 𝑡𝑡hen, the polynomial 𝑝𝑝(𝑥𝑥) can be written as 𝑝𝑝(𝑥𝑥) = 𝑎𝑎{𝑥𝑥2 −(𝛼𝛼 + 𝛽𝛽)𝑥𝑥 + 𝛼𝛼𝛽𝛽} or 𝑝𝑝(𝑥𝑥) = 𝑎𝑎{𝑥𝑥2 − (Sum of the zeroes)𝑥𝑥 +Product of the zeroes}, where 𝑎𝑎 is a non-zero real number

Given: sum of the roots = 𝛼𝛼 + 𝛽𝛽 = 0 and product of the roots = 𝛼𝛼𝛽𝛽 = √5


𝑝𝑝(𝑥𝑥) = 𝑎𝑎{𝑥𝑥2 − 0. 𝑥𝑥 + √5}

= 𝑎𝑎{𝑥𝑥2 + √5}


Therefore, the quadratic polynomial is �𝑥𝑥2 + √5�.

(iv) We know that if 𝛼𝛼 𝑎𝑎𝑎𝑎𝑎𝑎 𝛽𝛽 are the zeroes of a quadratic polynomial𝑝𝑝(𝑥𝑥), 𝑡𝑡hen, the polynomial 𝑝𝑝(𝑥𝑥) can be written as 𝑝𝑝(𝑥𝑥) = 𝑎𝑎{𝑥𝑥2 −(𝛼𝛼 + 𝛽𝛽)𝑥𝑥 + 𝛼𝛼𝛽𝛽} or 𝑝𝑝(𝑥𝑥) = 𝑎𝑎{𝑥𝑥2 − (Sum of the zeroes)𝑥𝑥 +Product of the zeroes}, where 𝑎𝑎 is a non-zero real number.

Given: sum of the roots = 𝛼𝛼 + 𝛽𝛽 = 1 and product of the roots = 𝛼𝛼𝛽𝛽 = 1






𝑝𝑝(𝑥𝑥) = 𝑎𝑎{𝑥𝑥2 − 1. 𝑥𝑥 + 1}

= 𝑎𝑎{𝑥𝑥2 − 𝑥𝑥 + 1}


Therefore, the quadratic polynomial is (𝑥𝑥2 − 𝑥𝑥 + 1).

(v) We know that if 𝛼𝛼 𝑎𝑎𝑎𝑎𝑎𝑎 𝛽𝛽 are the zeroes of a quadratic polynomial𝑝𝑝(𝑥𝑥), then, the polynomial 𝑝𝑝(𝑥𝑥) can be written as 𝑝𝑝(𝑥𝑥) = 𝑎𝑎{𝑥𝑥2 −(𝛼𝛼 + 𝛽𝛽)𝑥𝑥 + 𝛼𝛼𝛽𝛽} or 𝑝𝑝(𝑥𝑥) = 𝑎𝑎{𝑥𝑥2 − (Sum of the zeroes)𝑥𝑥 +Product of the zeroes}, where 𝑎𝑎 is a non-zero real number.

Given: sum of the roots = 𝛼𝛼 + 𝛽𝛽 = −14 and product of the roots = 𝛼𝛼𝛽𝛽 = 1

4


𝑝𝑝(𝑥𝑥) = 𝑎𝑎{𝑥𝑥2 +14𝑥𝑥 +

14

}

= 𝑎𝑎 �4𝑥𝑥2 + 𝑥𝑥 + 1

4�


Therefore, the quadratic polynomial is (4𝑥𝑥2 + 𝑥𝑥 + 1).

(vi) We know that if 𝛼𝛼 𝑎𝑎𝑎𝑎𝑎𝑎 𝛽𝛽 are the zeroes of a quadratic polynomial𝑝𝑝(𝑥𝑥), 𝑡𝑡hen, the polynomial 𝑝𝑝(𝑥𝑥) can be written as 𝑝𝑝(𝑥𝑥) = 𝑎𝑎{𝑥𝑥2 −(𝛼𝛼 + 𝛽𝛽)𝑥𝑥 + 𝛼𝛼𝛽𝛽} or 𝑝𝑝(𝑥𝑥) = 𝑎𝑎{𝑥𝑥2 − (Sum of the zeroes)𝑥𝑥 +Product of the zeroes}, where 𝑎𝑎 is a non-zero real number.

Given: sum of the roots = 𝛼𝛼 + 𝛽𝛽 = 4 and product of the roots = 𝛼𝛼𝛽𝛽 = 1


𝑝𝑝(𝑥𝑥) = 𝑎𝑎{𝑥𝑥2 − 4𝑥𝑥 + 1}


Therefore, the quadratic polynomial is (𝑥𝑥2 − 4𝑥𝑥 + 1).

♦ ♦ ♦





EXERCISE 2.3

1. Divide the polynomial 𝑝𝑝(𝑥𝑥) by the polynomial 𝑔𝑔(𝑥𝑥) and find the quotient and remainder in each of the following:

(i) 𝑝𝑝(𝑥𝑥) = 𝑥𝑥3 − 3𝑥𝑥2 + 5𝑥𝑥 − 3, 𝑔𝑔(𝑥𝑥) = 𝑥𝑥2 − 2

(ii) 𝑝𝑝(𝑥𝑥) = 𝑥𝑥4 − 3𝑥𝑥2 + 4𝑥𝑥 + 5, 𝑔𝑔(𝑥𝑥) = 𝑥𝑥2 + 1 − 𝑥𝑥

(iii) 𝑝𝑝(𝑥𝑥) = 𝑥𝑥4 − 5𝑥𝑥 + 6, 𝑔𝑔(𝑥𝑥) = 2 − 𝑥𝑥2

Solution:

(i) 𝑝𝑝(𝑥𝑥) = 𝑥𝑥3 − 3𝑥𝑥2 + 5𝑥𝑥 − 3, 𝑔𝑔(𝑥𝑥) = 𝑥𝑥2 − 2

Here, both the polynomials are already arranged in the descending powers of variable.

The polynomial 𝑝𝑝(𝑥𝑥) can be divided by the polynomial 𝑔𝑔(𝑥𝑥) as follows:

Quotient = 𝑥𝑥 − 3

Remainder = 7𝑥𝑥 − 9

(ii) 𝑝𝑝(𝑥𝑥) = 𝑥𝑥4 − 3𝑥𝑥2 + 4𝑥𝑥 + 5 = 𝑥𝑥4 + 0. 𝑥𝑥3 − 3𝑥𝑥2 + 4𝑥𝑥 + 5,

Here, the polynomial 𝑝𝑝(𝑥𝑥) is already arranged in the descending powers of variable.

𝑔𝑔(𝑥𝑥) = 𝑥𝑥2 + 1 − 𝑥𝑥

Here, the polynomial 𝑔𝑔(𝑥𝑥) is not arranged in the descending powers of variable.

Now, 𝑔𝑔(𝑥𝑥) = 𝑥𝑥2 − 𝑥𝑥 + 1






Quotient = 𝑥𝑥2 + 𝑥𝑥 − 3

Remainder = 8

(iii) 𝑝𝑝(𝑥𝑥) = 𝑥𝑥4 − 5𝑥𝑥 + 6 = 𝑥𝑥4 + 0. 𝑥𝑥2 − 5𝑥𝑥 + 6

𝑔𝑔(𝑥𝑥) = 2 − 𝑥𝑥2

Here, the polynomial 𝑔𝑔(𝑥𝑥) is not arranged in the descending powers of variable.

Now, 𝑔𝑔(𝑥𝑥) = −𝑥𝑥2 + 2


Quotient = −𝑥𝑥2 − 2

Remainder = −5𝑥𝑥 + 10

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

(i) 𝑡𝑡2 − 3, 2𝑡𝑡4 + 3𝑡𝑡3 − 2𝑡𝑡2 − 9𝑡𝑡 − 12





(ii) 𝑥𝑥2 + 3𝑥𝑥 + 1, 3𝑥𝑥4 + 5𝑥𝑥3 − 7𝑥𝑥2 + 2𝑥𝑥 + 2

(iii) 𝑥𝑥3 − 3𝑥𝑥 + 1, 𝑥𝑥5 − 4𝑥𝑥3 + 𝑥𝑥2 + 3𝑥𝑥 + 1

Solution:

(i) The polynomial 2𝑡𝑡4 + 3𝑡𝑡3 − 2𝑡𝑡2 − 9𝑡𝑡 − 12 can be divided by the polynomial 𝑡𝑡2 − 3 = 𝑡𝑡2 + 0. 𝑡𝑡 − 3 as follows:

Since the remainder is 0, hence 𝑡𝑡2 − 3 is a factor of 2𝑡𝑡4 + 3𝑡𝑡3 − 2𝑡𝑡2 −9𝑡𝑡 − 12.

(ii) The polynomial 3𝑥𝑥4 + 5𝑥𝑥3 − 7𝑥𝑥2 + 2𝑥𝑥 + 2 can be divided by the polynomial 𝑥𝑥2 + 3𝑥𝑥 + 1 as follows:





Since the remainder is 0, hence 𝑥𝑥2 + 3𝑥𝑥 + 1 is a factor of 3𝑥𝑥4 + 5𝑥𝑥3 −7𝑥𝑥2 + 2𝑥𝑥 + 2

(iii) The polynomial 𝑥𝑥5 − 4𝑥𝑥3 + 𝑥𝑥2 + 3𝑥𝑥 + 1 can be divided by the polynomial 𝑥𝑥3 − 3𝑥𝑥 + 1 as follows:

Since the remainder is not equal to 0, hence 𝑥𝑥3 − 3𝑥𝑥 + 1 is not a factor of 𝑥𝑥5 − 4𝑥𝑥3 + 𝑥𝑥2 + 3𝑥𝑥 + 1.

3. Obtain all other zeroes of 3𝑥𝑥4 + 6𝑥𝑥3 − 2𝑥𝑥2 − 10𝑥𝑥 − 5, if two of its zeroes are

�53 and −�5

3

Solution:

Let 𝑝𝑝(𝑥𝑥) = 3𝑥𝑥4 + 6𝑥𝑥3 − 2𝑥𝑥2 − 10𝑥𝑥 − 5

It is given that the two zeroes of 𝑝𝑝(𝑥𝑥) are �53 and −�5

3

∴ �𝑥𝑥 − �53� �𝑥𝑥 + �5

3� = �𝑥𝑥2 − 5

3� is a factor of 𝑝𝑝(𝑥𝑥) {Since, (𝑎𝑎 − 𝑏𝑏)(𝑎𝑎 +

𝑏𝑏) = 𝑎𝑎2 − 𝑏𝑏2}

Therefore, on dividing the given polynomial by 𝑥𝑥2 − 53, we obtain remainder as 0.





Hence, 3𝑥𝑥4 + 6𝑥𝑥3 − 2𝑥𝑥2 − 10𝑥𝑥 − 5 = �𝑥𝑥2 − 53� (3𝑥𝑥2 + 6𝑥𝑥 + 3)

= 3 �𝑥𝑥2 −53� (𝑥𝑥2 + 2𝑥𝑥 + 1)

Now, 𝑥𝑥2 + 2𝑥𝑥 + 1 = (𝑥𝑥 + 1)2

Thus, the two zeroes of 𝑥𝑥2 + 2𝑥𝑥 + 1 are −1 and −1

Therefore, the zeroes of the given polynomial are �53

,−�53

,−1 and −1.

4. On dividing 𝑥𝑥3 − 3𝑥𝑥2 + 𝑥𝑥 + 2 by a polynomial 𝑔𝑔(𝑥𝑥), the quotient and remainder were 𝑥𝑥 − 2 and −2𝑥𝑥 + 4, respectively. Find 𝑔𝑔(𝑥𝑥).

Solution:

Dividend, 𝑝𝑝(𝑥𝑥) = 𝑥𝑥3 − 3𝑥𝑥2 + 𝑥𝑥 + 2

Quotient = (𝑥𝑥 − 2)

Remainder = (−2𝑥𝑥 + 4)

𝑔𝑔(𝑥𝑥) be the divisor.

According to the division algorithm,

Dividend = Divisor × Quotient + Remainder

𝑥𝑥3 − 3𝑥𝑥2 + 𝑥𝑥 + 2 = 𝑔𝑔(𝑥𝑥) × (𝑥𝑥 − 2) + (−2𝑥𝑥 + 4)

𝑥𝑥3 − 3𝑥𝑥2 + 𝑥𝑥 + 2 + 2𝑥𝑥 − 4 = 𝑔𝑔(𝑥𝑥)(𝑥𝑥 − 2)





𝑥𝑥3 − 3𝑥𝑥2 + 3𝑥𝑥 − 2 = 𝑔𝑔(𝑥𝑥)(𝑥𝑥 − 2)

Now, 𝑔𝑔(𝑥𝑥) is the quotient when 𝑥𝑥3 − 3𝑥𝑥2 + 3𝑥𝑥 − 2 is divided by 𝑥𝑥 − 2. (Since, Remainder = 0)

∴ 𝑔𝑔(𝑥𝑥) = 𝑥𝑥2 − 𝑥𝑥 + 1

6. Give examples of polynomials 𝑝𝑝(𝑥𝑥), 𝑔𝑔(𝑥𝑥), 𝑞𝑞(𝑥𝑥) and 𝑟𝑟(𝑥𝑥), which satisfy the division algorithm and

(i) deg𝑝𝑝(𝑥𝑥) = deg 𝑞𝑞(𝑥𝑥)

(ii) deg 𝑞𝑞(𝑥𝑥) = deg 𝑟𝑟(𝑥𝑥)

(iii) deg 𝑟𝑟(𝑥𝑥) = 0

Solution:

According to the division algorithm, if 𝑝𝑝(𝑥𝑥) and 𝑔𝑔(𝑥𝑥) are two polynomials with g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that

𝑝𝑝(𝑥𝑥) = 𝑔𝑔(𝑥𝑥) × 𝑞𝑞(𝑥𝑥) + 𝑟𝑟(𝑥𝑥), where 𝑟𝑟(𝑥𝑥) = 0 or degree of 𝑟𝑟(𝑥𝑥) < degree of 𝑔𝑔(𝑥𝑥).

(i) Degree of quotient will be equal to degree of dividend when divisor is constant.

Let us consider the division of 2𝑥𝑥2 + 2𝑥𝑥 − 16 by 2.

Here, 𝑝𝑝(𝑥𝑥) = 2𝑥𝑥2 + 2𝑥𝑥 − 16 and 𝑔𝑔(𝑥𝑥) = 2





𝑞𝑞(𝑥𝑥) = 𝑥𝑥2 + 𝑥𝑥 − 8 and 𝑟𝑟(𝑥𝑥) = 0

Clearly, the degree of 𝑝𝑝(𝑥𝑥) and 𝑞𝑞(𝑥𝑥) is the same which is 2.

Verification:

𝑝𝑝(𝑥𝑥) = 𝑔𝑔(𝑥𝑥) × 𝑞𝑞(𝑥𝑥) + 𝑟𝑟(𝑥𝑥)

2𝑥𝑥2 + 2𝑥𝑥 − 16 = 2(𝑥𝑥2 + 𝑥𝑥 − 8) + 0

= 2𝑥𝑥2 + 2𝑥𝑥 − 16

Thus, the division algorithm is satisfied.

(ii) Let us consider the division of 4𝑥𝑥 + 3 by 𝑥𝑥 + 2.

Here, 𝑝𝑝(𝑥𝑥) = 4𝑥𝑥 + 3 and 𝑔𝑔(𝑥𝑥) = 𝑥𝑥 + 2

𝑞𝑞(𝑥𝑥) = 4 and 𝑟𝑟(𝑥𝑥) = −5

Here, degree of 𝑞𝑞(𝑥𝑥) and 𝑟𝑟(𝑥𝑥) is the same which is 0.

Verification:


4𝑥𝑥 + 3 = (𝑥𝑥 + 2) × 4 + (−5)

4𝑥𝑥 + 3 = 4𝑥𝑥 + 3


(iii) Degree of remainder will be 0 when remainder obtained on division is a constant.

Let us consider the division of 4𝑥𝑥 + 3 by 𝑥𝑥 + 2.

Here, 𝑝𝑝(𝑥𝑥) = 4𝑥𝑥 + 3 and 𝑔𝑔(𝑥𝑥) = 𝑥𝑥 + 2

𝑞𝑞(𝑥𝑥) = 4 and 𝑟𝑟(𝑥𝑥) = −5

Here, we get remainder as a constant. Therefore, the degree of 𝑟𝑟(𝑥𝑥) is 0.

Verification:


4𝑥𝑥 + 3 = (𝑥𝑥 + 2) × 4 + (−5)





4𝑥𝑥 + 3 = 4𝑥𝑥 + 3


♦ ♦ ♦



cbse ncert solutions for class 10 mathematics chapter 2 · 2019-08-20 · class- x-cbse-mathematics...

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