Chapter 2 - Polynomials, RD Sharma Solutions - (Part-4) | RD Sharma Solutions for Class 10 Mathematics PDF Download

Page No 2.57

Ques.1. Apply division algorithm to find the quotient q(x) and remainder r(x) in dividing f(x) by g(x) in each of the following :
(i) f(x) = x³ − 6x² + 11x − 6, g(x) = x² + x + 1
(ii) f(x) = 10x⁴ + 17x³ − 62x² + 30x − 3, g(x) = 2x² + 7x + 1
(iii) f(x) = 4x³ + 8x + 8x² + 7, g(x) = 2x² − x + 1
(iv) f(x) = 15x³ − 20x² + 13x − 12, g(x) = 2 − 2x + x²
Ans. (i) We have
f(x) = x³ - 6x² + 11x - 6
g(x) = x² + x + 1
Here, degree [f(x)] = 3 and 
Degree (g(x)) = 2
Therefore, quotient q(x) is of degree 3 - 2 = 1 and the remainder r(x) is of degree less than 2
Let q(x) = ax + b and
r(x) = cx + d
Using division algorithm, we have


Equating the co-efficients of various powers of on both sides, we get
On equating the co-efficient of x³

On equating the co-efficient of x²

Substituting a = 1
-6 = 1 + b
-6 - 1 = b
-7 = b
On equating the co-efficient of

Substituting a = 1; and b = -7 we get, 
11 = 1 + (-7) + c
11 = -6 + c
11 + 6 = c
17 = c
On equating the constant terms
-6 = b + d
Substituting b = -7 we get,
-6 = -7 + d
-6 + 7 = d
1 = d
Therefore,
Quotient q(x) = ax + b
= (1x - 7)
And remainder r(x) = cx + d
= (17x + 1)
Hence, the quotient and remainder is given by,
. 
(ii) We have
f(x) = 10x⁴ + 17x³ - 62x² + 30x - 3
g(x) = 2x² + 7x + 1
Here, Degree (f(x)) = 4 and
Degree (g(x)) = 2
Therefore, quotient q(x) is of degree 4 - 2 = 2 and remainder r(x) is of degree less than 2
(=degree(g(x)))
Let g(x) = ax² + bx + c and
r(x) = px + q
Using division algorithm, we have


Equating the co-efficients of various powers x on both sides, we get 
On equating the co-efficient of x⁴
2a = 10
a = 10/2
a = 5
On equating the co-efficient of x³
7a + 2b = 17
Substituting a = 5 we get
7 x 5 + 2b = 17
35 + 2b = 17
2b = 17 - 35
2b = -18
b = (-18)/2
b = -9
On equating the co-efficient of x²
a + 7b + 2c = -62
Substituting a = 5 and b = -9, we get
5 + 7 x -9 + 2c  = -62
5 - 63 + 2c = -62
2c = -62 + 63 - 5
2c = -4
c = (-4)/2
c = -2
On equating the co-efficient of x
b + 7c + p = 30
Substituting b = -9 and c = -2, we get


p = 30 + 23
p = 53
On equating constant term, we get
c + q = -3
Substituting c = -2, we get
-2 + q = -3
q = -3 + 2
q = -1
Therefore, quotient q(x) = ax² + bx + c
= 5x² - 9x - 2
Remainder r(x) = px + q
= 53x - 1
Hence, the quotient and remainder are q(x) = 5x² - 9x - 2 and r(x) = 53x - 1.
(iii) we have
f(x) = 4x³ + 8x + 8x² + 7
g(x) = 2x² - x + 1
Here, Degree (f(x)) = 3 and
Degree (g(x)) = 2
Therefore, quotient q(x) is of degree 3 - 2 = 1 and
Remainder r(x) is of degree less than 2
Let q(x) = ax + b and
r(x) = cx + d
Using division algorithm, we have


Equating the co-efficient of various Powers of on both sides, we get
On equating the co-efficient of x³
2a = 4
a = 4/2
a = 2
On equating the co-efficient of x²
8 = -a + 2b
Substituting a = 2 we get

5 = b
On equating the co-efficient of x
a - b + c = 8
Substituting a = 2 and b = 5 we get
2 - 5 + c = 8
-3 + c = 8
c = 8 + 3
c = 11
On equating the constant term, we get
b + d = 7
Substituting b = 5, we get 
5 + d = 7
d = 7 - 5
d = 2
Therefore, quotient q(x) = ax + b
= 2x + 5
Remainder r(x) = cx + d
= 11x + 2
Hence, the quotient and remainder are q(x) = 2x + 5 and r(x) = 11x + 2.
(iv) Given,
f(x) = 15x³ - 20x² + 13x - 12
g(x) = 2 - 2x + x²
Here, Degree (f(x)) = 3 and
Degree (g(x)) = 2
Therefore, quotient q(x) is of degree 3 - 2 = 1 and
Remainder r(x) is of degree less than 2
Let q(x) = ax + b and
r(x) = cx + d
Using division algorithm, we have


Equating the co-efficients of various powers of x on both sides, we get
On equating the co-efficient of x³

On equating the co-efficient of x²
2a - b = 20
Substituting a = 15, we get
2 x 15 - b = 20
30 - b = 20
-b = 20 - 30

On equating the co-efficient of x
2a - 2b + c = 13
Substituting a = 15 and b = 10, we get

10 + c = 13
c = 13 - 10
c = 3
On equating constant term
2b + d = -12
Substituting b = 10, we get

Therefore, quotient q(x) = ax + b
= 15x + 10
Remainder r(x) = 3x - 32
= 3x - 32
Hence, the quotient and remainder are q(x) = 15x + 10 and r(x) = 3x - 32.

Ques.2. Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm :
(i) g(t)=t²−3, f(t)=2t⁴+3t³−2t²−9t−12
(ii) g(x)=x³−3x+1, f(x)=x⁵−4x³+x²+3x+1
(iii) g(x)=2x²−x+3, f(x)=6x⁵−x⁴+4x³−5x²−x−15
Ans. (i) Given g(t) = t² - 3
f(t) = 2t⁴ + 3t³ - 2t² - 9t - 12
Here, degree (f(t)) = 4 and
Degree (g(t)) = 2
Therefore, quotient q(t) is of degree 4 - 2 = 2
Remainder r(t) is of degree 1 or less
Let q(t) = at² + bt + c and
r(t) = pt + q
Using division algorithm, we have


Equating co-efficient of various powers of t, we get
On equating the co-efficient of t⁴

On equating the co-efficient of t³

On equating the co-efficient of t²
2 = 3a - c
Substituting a = 2, we get

c = 4
On equating the co-efficient of t
9 = 3b - p
Substituting b = 3, we get

p = 0
On equating constant term
-12 = -3c + q
Substituting c = 4, we get 
-12 = 3 x 4 + q
-12 = -12 + q
-12 + 12 = +q
0 = q
Quotient q(t) = at² + bt + c
= 2t² + 3t + 4
Remainder  r(t) = pt + q
= 0t  + 0
= 0
Clearly, r(t) = 0
Hence, g(t) is a factor of f(t).
(ii) Given
f(x) = x⁵ - 4x³ + x² + 3x + 1
g(x) = x³ - 3x + 1
Here, Degree (f(x)) = 5 and
Degree (g(x)) = 3
Therefore, quotient q(x) is of degree 5 - 3 = 2
Remainder r(x) is of degree 1
Let q(x) = ax² + bx + c and
r(x) = px + q
Using division algorithm, we have


Equating the co-efficient of various powers of x on both sides, we get
On equating the co-efficient of x⁵

On equating the co-efficient of x⁴

On equating the co-efficient of x³
3a - c = 4
Substituting a = 1 we get
3 x 1 - c = 4
3 - x = 4
-c = 4 - 3
-c = 1
On equating the co-efficient of x
b - 3c + p = 3
Substituting b = 0 and c = -1, we get

On equating constant term, we get
c + q = 1
Substituting c = -1, we get
-1 + q = 1
q = 1 + 1
q = 2
Therefore, quotient q(x) = ax² + bx + c
= 1x² + 0x - 1
= x² - 1
Remainder r(x) = px + q

Clearly, r(x) = 2
Hence, g(x) is not a factor of f(x).
(iii) Given,
f(x) = 6x⁵ - x⁴ + 4x³ - 5x² - x - 15
g(x) = 2x² - x + 3
Here, Degree (f(x)) = 5 and
Degree (g(x)) = 2
Therefore, quotient q(x) is of degree 5 - 2 = 3 and
Remainder r(x) is of degree less than 1
Let q(x) = ax³ + bx + cx + d and
r(x) = px + q
Using division algorithm, we have







Equating the co-efficient of various powers of x on both sides, we get 
On equating the co-efficient of  x⁵

On equating the co-efficient of x⁴
a - 2b = 1
Substituting, a = 3 we get

b = 1
On equating the co-efficient of x³
3a - b + 2c = 4
Substituting a = 3 and b = 1, we get

2c = -4
c = (-4)/2
c = -2
On equating the co-efficient of x²
c - 3b - 2d = 5
Substituting c = -2, b = 1 we get
-2 -3 x 1 - 2d = 5
-2 - 3 - 2d = 5
-5 - 2d = 5
-2d = 5 + 5
-2d = 10
d = 10/-2
d = -5
On equating the co-efficient of x
-3c + d - p = 1
Substituting c = -2 and d = -5, we get

-p = 0
0 = p
On equating constant term
3d + q = -15
Substituting d = -5, we get

Therefore, Quotient q(x) = ax³ + bx² + cx + d
= 3x³ + 1x² - 2x - 5
Remainder r(x) = px + q
= 0x + 0
= 0
Clearly, r(x) = 0
Hence,g(x) is a factor of f(x).

Ques.3. Obtain all zeros of the polynomial f(x) = 2x⁴ + x³ − 14x² − 19x − 6, if two of its zeros are −2 and −1.
Ans. We know that, if x = α is a zero of a polynomial, and then x - α is a factor of f(x).
Since -2 and -1 are zeros of f(x).
Therefore 
(x + 2)(x + 1) = x² + 2x + x + 2
= x² + 3x + 2
x² + 3x + 2 is a factor of f(x).Now, We divide 2x⁴ + x³ - 14x² - 19x - 6 by g(x) = x² + 3x + 2 to find the other zeros of f(x).

By using division algorithm we have,

Hence, the zeros of the given polynomials are. 

Ques.4. Obtain all zeros of f(x) = x³ + 13x² + 32x + 20, if one of its zeros is −2.
Ans. Since −2 is one zero of f(x).
Therefore, we know that, if x = α is a zero of a polynomial, then x - α is a factor of f(x) = x + 2 is a factor of f(x).
Now, we divide f(x) = x³ + 13x² + 32x + 20 by g(x) = (x + 2) to find the others zeros of f(x).

By using that division algorithm we have, 

Hence, the zeros of the given polynomials are -2, -1, and -10.