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NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

NCERT Solutions Exercise 2.1 Class 10 Polynomials

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

Solution:
The number of zeroes of p(x) in each graph given; are

NCERT Solutions Exercise 2.2 Class 10 Polynomials

Question 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
     

Solution:

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

For all different real values of k, we can have different quadratic polynomials of the form 3×2 - 3√2x +1 having sum of zeroes = √2 and product of zeroes = 1/3

NCERT Solutions Exercise 2.3 Class 10 Polynomials

Question 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
     

Solution:
(i) Here p(x) = x3 -3x2 + 5x – 3 and g(x) = x2 -2
dividing p(x) by g(x) ⇒

Quotient = x - 3, Remainder = 7x - 9
(ii) Here p(x) = x4- 3x2 + 4x + 5 and g(x) = x2 + 1 -x
dividing p(x) by g(x) ⇒

Quotient = x2 + x - 3, Remainder = 8
(iii) Herep(x) = x4- 5x + 6 and g(x) = 2-x2
Rearranging g(x) = -x2 + 2
dividing p(x) by g(x) ⇒
Quotient = -x2 - 2
Remainder = -5x + 10.

Question 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 + t3 – 2t2 – 9t – 12
  • (ii) x2 + 3x + 1,3x4+5x3-7x2+2x + 2
  • (iii) x3 -3x + 1, x5 – 4x3 + x2 + 3x + l
     

Solution:
(i) First polynomial = t2 – 3,
Second polynomial = 2t4 + 3t3 – 2t2 – 9t – 12
dividing second polynomial
by first polynomial ⇒
∵ Remainder is zero.
∴First polynomial is a factor of second polynomial.

(ii) First polynomial = x2 + 3x + 1
Second polynomial = 3x4 + 5x3 – 7x2 + 2x + 2
dividing second polynomial
by first polynomial ⇒
∵ Remainder is zero.
∴ First polynomial is a factor of second polynomial.

(iii) First polynomial = x3 - 3x + 1
Second polynomial = x5 - 4x3 + x2 + 3x + 1.
∵ Remainder ≠ 0.
∴ First polynomial is not a factor of second polynomial.

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

Solution:

Question 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:

p(x) = x3 – 3 x 2 + x + 2 g(x) = ?
Quotient = x – 2; Remainder = -2x + 4
On dividing p(x) by g(x), we have
p(x) = g(x) x quotient + remainder
⇒ x3– 3x2 + x + 2 = g(x) (x – 2) + (-2x + 4)
⇒ x3 – 3x2 + x + 2 + 2 x- 4 = g(x) x (x-2)
⇒ x3 – 3x2 + 3x – 2 = g(x) (x – 2)

Question 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)
  • (ii) deg q(x) = deg r(x)
  • (iii) deg r(x) = 0
     

Solution:
(i) p(x),g(x),q(x),r(x)
deg p(x) = deg q(x)
∴ both g(x) and r(x) are constant terms.
p(x) = 2x2– 2x + 14; g(x) = 2
q(x) = x2 – x + 7; r(x) = 0

(ii) deg q(x) = deg r(x)
∴ this is possible when
deg of both q(x) and r(x) should be less than p(x) and g(x).
p(x) = x3+ x2 + x + 1; g(x) = x2 – 1
q(x) = x + 1, r(x) = 2c + 2

(iii) deg r(x) is 0.
This is possible when product of q(x) and g(c) form a polynomial whose degree is equal to degree of p(x) and constant term.

NCERT Solutions Exercise 2.4 Class 10 Polynomials

Question 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;  1/4, 1, -2
(ii) x3 – 4x2 + 5x – 2; 2, 1, 1
 

Solution:
(i) Comparing the given polynomial with ax3 + bx2 + cx + d, we get:
a = 2, b – 1, c = -5 and d = 2.
∴  p(x) = 2x3 + x2 – 5x + 2

(ii) Compearing the given polynomial with ax3 + bx2 + cx + d, we get:
a = 1, b = -4, c = 5 and d = – 2.
∴ p (x) = x3 – 4x2 + 5x – 2
⇒ p(2) = (2)3 – 4(2)2 + 5 x 2 – 2
= 8 – 16+ 10 – 2 = 0
p(1) = (1)3 – 4(1)2 + 5 x 1-2
= 1 – 4 + 1 – 2
= 6-6 = 0
Hence, 2, 1 and 1 are the zeroes of x3 – 4x2 + 5x – 2.
Hence verified.
Now we take α = 2, β = 1 and γ = 1.
α + β + γ = 2 + 1 + 1 = 41 = −ba
αβ + βγ + γα = 2 + 1 + 2 = 51 = ca
αβγ = 2 x 1 x 1  = 21 = −da.
Hence verified.

Question 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.
 

Solution:
Let α , β and  γ be the zeroes of the required polynomial.
Then α + β + γ = 2, αβ + βγ + γα = -7 and αβγ = -14.
∴ Cubic polynomial
= x3 – (α + β + γ)x2 + (αβ + βγ + γα)x – αβγ
= x3 – 2x2 – 1x + 14
Hence, the required cubic polynomial is x3 – 2x2 – 7x + 14.

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

Solution:
Let α , β and  γ be the zeroes of polynomial x3 – 3x2 + x + 1.
Then α =  a-b, β = a and γ = a + b.
∴ Sum of zeroes = α + β + γ
⇒ 3 = (a − b) + a + (a + b)
⇒ (a − b) + a + (a + b) = 3
⇒ a−b + a + a + b = 3
⇒ 3a = 3
⇒ a =3/3 = 1 …(i)
Product of zeroes = αβγ
⇒ −1 = (a − b) a (a + b)
⇒ (a − b) a (a + b) = −1
⇒ (a2− b2)a = −1
⇒a3− ab2;= −1… (ii)
Putting the value of a from equation (i) in equation (ii), we get:
(1)3−(1)b2= −1
⇒ 1 – b2 = −1
⇒ – b2 = −1 − 1
⇒−b2 = 2
⇒ b = ± √2
Hence, a = 1 and b = ± √2.

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

Solution:
Since two zeroes are 2 + √3 and 2 - √3,
∴  [x-(2 + √3)] [x- (2 - √3)]
= (x-2- √3)(x-2 + √3)
= (x-2)2- (√3)2
x2 - 4x + 1 is a factor of the given polynomial.
Now, we divide the given polynomial by x2 - 4x + 1.

So, 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 – 7) (x + 5)
Hence, the other zeroes of the given polynomial are 7 and -5.

Question 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.
 

Solution:
We have
p(x) = x4 – 6x3 + 16x2 – 25x + 10
Remainder = x + a   … (i)
Now, we divide the given polynomial 6x3 + 16x2 – 25x + 10 by x2 – 2x + k.

Using equation (i), we get:
(-9 + 2k)x + 10-8 k + k2 = x + a
On comparing the like coefficients, we have:
-9 + 2k = 1
⇒ 2k = 10
⇒ k = 10/2 = 5 ..(ii)
and 10 -8k + k2- a ..(iii)
Substituting the value of k = 5, we get:
10 - 8(5) + (5)2 = a
⇒ 10 - 40 + 25 = a
⇒ 35 - 40 = a
⇒ a = -5
Hence, k = 5 and a = -5.

Important Question

NCERT CBSE for Class 10 Maths Chapter 2 Polynomials Important Questions

Polynomials Class 10 Important Questions Very Short Answer (1 Mark)

Question 1.
If the sum of zeroes of the quadratic polynomial 3x2 - kx + 6 is 3, then find the value of k.
Year of Question:(2012)

Solution:
Here a = 3, b = -k, c = 6
Sum of the zeroes, (α + β) = −b/a = 3 ...(given)
⇒ −(−k)/3 = 3
⇒ k = 9

Question 2.
If α and β are the zeroes of the polynomial ax2 + bx + c, find the value of α2 + β2.
Year of Question:(2013)

Solution:

Question 3.
If the sum of the zeroes of the polynomial p(x) = (k2 - 14) x2 - 2x - 12 is 1, then find the value of k.
Year of Question:(2017 D)

Solution:
p(x) = (k2 - 14) x2 - 2x - 12
Here a = k2 - 14, b = -2, c = -12
Sum of the zeroes, (α + β) = 1 .[Given]
⇒ −b/a = 1
⇒ −(−2)/k2−14 = 1
⇒ k2 - 14 = 2
⇒ k2= 16
⇒ k = ±4

Question 4.
If α and β are the zeroes of a polynomial such that α + β = -6 and αβ = 5, then find the polynomial.
Year of Question:(2016 D)

Solution:
Quadratic polynomial is x2 - Sx + P = 0
⇒ x2 - (-6)x + 5 = 0
⇒ x2 + 6x + 5 = 0

Question 5.
A quadratic polynomial, whose zeroes are -4 and -5, is ..
Year of Question:(2016 D)

Solution:
x2 + 9x + 20 is the required polynomial.

Polynomials Class 10 Important Questions Short Answer-I (2 Marks)

Question 6.
Find the condition that zeroes of polynomial p(x) = ax2 + bx + c are reciprocal of each other.
Year of Question:(2017 OD)

Solution:
Let α and 1/α be the zeroes of P(x).
P(a) = ax2 + bx + c .(given)
Product of zeroes = ca
⇒ α × 1/α = c/a
⇒ 1 = c/a
⇒ a = c (Required condition)
Coefficient of x2 = Constant term

Question 7.
Form a quadratic polynomial whose zeroes are 3 + √2 and 3 - √2.
Year of Question:(2012)

Solution:
Sum of zeroes,
S = (3 + √2) + (3 - √2) = 6
Product of zeroes,
P = (3 + √2) x (3 - √2) = (3)2 - (√2)2 = 9 - 2 = 7
Quadratic polynomial = x2 - Sx + P = x2 - 6x + 7

Question 8.
Find a quadratic polynomial, the stun and product of whose zeroes are √3 and 1/√3 respectively.
Year of Question:(2014)

Solution:
Sum of zeroes, (S) = √3
Product of zeroes, (P) = 1√3
Quadratic polynomial = x2 - Sx + P

Question 9.
Find a quadratic polynomial, the sum and product of whose zeroes are 0 and -√2 respectively.
Year of Question:(2015)

Solution:
Quadratic polynomial is
x2 - (Sum of zeroes) x + (Product of zeroes)
= x2 - (0)x + (-√2)
= x2 - √2

Question 10.
Find the zeroes of the quadratic polynomial √3 x2 - 8x + 4√3.
Year of Question:(2013)

Solution:

Question 11.
If the zeroes of the polynomial x2 + px + q are double in value to the zeroes of 2x2 - 5x - 3, find the value of p and q.
Year of Question:(2012)

Solution:
We have, 2x2 - 5x - 3 = 0
= 2x2 - 6x + x - 3
= 2x(x - 3) + 1(x - 3)
= (x - 3) (2x + 1)
Zeroes are:
x - 3 = 0 or 2x + 1 = 0
⇒ x = 3 or x = −1/2
Since the zeroes of required polynomial is double of given polynomial.
Zeroes of the required polynomial are:
3 × 2, (−1/2 × 2), i.e., 6, -1
Sum of zeroes, S = 6 + (-1) = 5
Product of zeroes, P = 6 × (-1) = -6
Quadratic polynomial is x2 - Sx + P
⇒ x2 - 5x - 6 .(i)
Comparing (i) with x2 + px + q
p = -5, q = -6

Question 12.
Can (x - 2) be the remainder on division of a polynomial p(x) by (2x + 3)? Justify your answer.
Year of Question:(2016 OD)

Solution:
In case of division of a polynomial by another polynomial, the degree of the remainder (polynomial) is always less than that of the divisor. (x - 2) can not be the remainder when p(x) is divided by (2x + 3) as the degree is the same.

Question 13.
Find a quadratic polynomial whose zeroes are 3+√5/5 and 3−√5/5.
Year of Question:(2013)

Solution:

Question 14.
Find the quadratic polynomial whose zeroes are -2 and -5. Verify the relationship between zeroes and coefficients of the polynomial.
Year of Question:(2013)

Solution:
Sum of zeroes, S = (-2) + (-5) = -7
Product of zeroes, P = (-2)(-5) = 10
Quadratic polynomial is x2 - Sx + P = 0
= x2 - (-7)x + 10
= x2 + 7x + 10
Verification:
Here a = 1, b = 7, c = 10
Sum of zeroes = (-2) + (-5) = 7

Question 15.
Find the zeroes of the quadratic polynomial 3x2 - 75 and verify the relationship between the zeroes and the coefficients.
Year of Question:(2014)

Solution:
We have, 3x2 - 75
= 3(x2 - 25)
= 3(x2 - 52)
= 3(x - 5)(x + 5)
Zeroes are:
x - 5 = 0 or x + 5 = 0
x = 5 or x = -5
Verification:
Here a = 3, b = 0, c = -75
Sum of the zeroes = 5 + (-5) = 0

Question 16.
Find the zeroes of p(x) = 2x2 - x - 6 and verify the relationship of zeroes with these co-efficients.
Year of Question:(2017 OD)

Solution:
p(x) = 2x2 - x - 6 .[Given]
= 2x2 - 4x + 3x - 6
= 2x (x - 2) + 3 (x - 2)
= (x - 2) (2x + 3)
Zeroes are:
x - 2 = 0 or 2x + 3 = 0
x = 2 or x = −3/2
Verification:
Here a = 2, b = -1, c = -6

Question 17.
What must be subtracted from the polynomial f(x) = x4 + 2x3 - 13x2 - 12x + 21 so that the resulting polynomial is exactly divisible by x2- 4x + 3?
Year of Question:(2012, 2017 D)

Solution:

(2x - 3) should be subtracted from x4 + 2x3 - 13x2 - 12x + 21.

Polynomials Class 10 Important Questions Short Answer-II (3 Marks)

Question 18.
Verify whether 2, 3 and 1/2 are the zeroes of the polynomial p(x) = 2x3 - 11x2 + 17x - 6.
Year of Question:(2012, 2017 D)

Solution:
p(x) = 2x3 - 11x2 + 17x - 6
When x = 2,
p(2) = 2(2)3 - 11(2)2 + 17(2) - 6 = 16 - 44 + 34 - 6 = 0
When x = 3, p(3) = 2(3)3- 11(3)2 + 17(3) - 6 = 54 - 99 + 51 - 6 = 0

Yes, x = 2, 3 and 1/2 all are the zeroes of the given polynomial.

Question 19.
Show that 12 and −3/2 are the zeroes of the polynomial 4x2 + 4x - 3 and verify the relationship between zeroes and co-efficients of polynomial.
Year of Question:(2013)

Solution:
Let P(x) = 4x2 + 4x - 3

Question 20.
Find a quadratic polynomial, the sum and product of whose zeroes are -8 and 12 respectively. Hence find the zeroes.
Year of Question:(2014)

Solution:
Let Sum of zeroes (α + β) = S = -8 .[Given]
Product of zeroes (αβ) = P = 12 .[Given]
Quadratic polynomial is x2 - Sx + P
= x2 - (-8)x + 12
= x2 + 8x + 12
= x2 + 6x + 2x + 12
= x(x + 6) + 2(x + 6)
= (x + 2)(x + 6)
Zeroes are:
x + 2 = 0 or x + 6 = 0
x = -2 or x = -6

Question 21.
Find a quadratic polynomial, the sum and product of whose zeroes are 0 and −3/5 respectively. Hence find the zeroes.
Year of Question:(2015)

Solution:
Quadratic polynomial = x2 - (Sum)x + Product

Question 22.
Find the zeroes of the quadratic polynomial 6x2 - 3 - 7x and verify the relationship between the zeroes and the coefficients of the polynomial.
Year of Question:(2015, 2016 OD)

Solution:
We have, 6x2 - 3 - 7x
= 6x2 - 7x - 3
= 6x2 - 9x + 2x - 3
= 3x(2x - 3) + 1(2x - 3)
= (2x - 3) (3x + 1)

Question 23.
Find the zeroes of the quadratic polynomial f(x) = x2 - 3x - 28 and verify the relationship between the zeroes and the co-efficients of the polynomial.
Year of Question:(2012, 2017 D)

Solution:
p(x) = x2 - 3x - 28
= x2 - 7x + 4x - 28
= x(x - 7) + 4(x - 7)
= (x - 7) (x + 4)
Zeroes are:
x - 7 = 0 or x + 4 = 0
x = 7 or x = -4

Question 24.
If α and β are the zeroes of the polynomial 6y2- 7y + 2, find a quadratic polynomial whose zeroes are 1/α and 1/α.
Year of Question:(2012)

Solution:
Given: 6y2 - 7y + 2
Here a = 6, b = -7, c = 2

Question 25.
Divide 3x2 + 5x - 1 by x + 2 and verify the division algorithm.
Year of Question:(2013 OD)

Solution:

Quotient = 3x - 1
Remainder = 1
Verification:
Divisor × Quotient + Remainder
= (x + 2) × (3x - 1) + 1
= 3x2- x + 6x - 2 + 1
= 3x2 + 5x - 1
= Dividend

Question 26.
On dividing 3x3 + 4x2 + 5x - 13 by a polynomial g(x) the quotient and remainder were 3x +10 and 16x - 43 respectively. Find the polynomial g(x).
Year of Question:(2017 OD)

Solution:
Let 3x3 + 4x2 + 5x - 13 = P(x)
q(x) = 3x + 10, r(x) = 16x - 43 .[Given]
As we know, P(x) = g(x) . q(x) + r(x)
3x3 + 4x2 + 5x - 13 = g(x) . (3x + 10) + (16x - 43)
3x3 + 4x2 + 5x - 13 - 16x + 43 = g(x) . (3x + 10)

Question 27.
Check whether polynomial x - 1 is a factor of the polynomial x3 - 8x2 + 19x - 12. Verify by division algorithm.
Year of Question:(2014)

Solution:
Let P(x) = x3 - 8x2 + 19x - 12
Put x = 1
P(1) = (1)3 - 8(1)2+ 19(1) - 12
= 1 - 8 + 19 - 12
= 20 - 20
= 0
Remainder = 0
(x - 1) is a facter of P(x).
Verification:

Since remainder = 0
(x - 1) is a factor of P(x).

Polynomials Class 10 Important Questions Long Answer (4 Marks)

Question 28.
Divide 4x3 + 2x2 + 5x - 6 by 2x2 + 1 + 3x and verify the division algorithm.
Year of Question:(2013)

Solution:

Quotient = 2x - 2
Remainder = 9x - 4
Verification:
Divisor × Quotient + Remainder
= (2x2 + 3x + 1) × (2x - 2) + 9x - 4
= 4x3 - 4x2 + 6x2 - 6x + 2x - 2 + 9x - 4
= 4x3 + 2x2 + 5x - 6
= Dividend

Question 29.
Given that x - √5 is a factor of the polynomial x3- 3√5 x2 - 5x + 15√5, find all the zeroes of the polynomial.
Year of Question:(2012, 2016)

Solution:
Let P(x) = x3 - 3√5 x2 - 5x + 15√5
x - √5 is a factor of the given polynomial.
Put x = -√5,

Other zero:
x - 3√5 = 0 ⇒ x = 3√5
All the zeroes of P(x) are -√5, √5 and 3√5.

Question 30.
If a polynomial x4+ 5x3 + 4x2 - 10x - 12 has two zeroes as -2 and -3, then find the other zeroes.
Year of Question:(2014)

Solution:
Since two zeroes are -2 and -3.
(x + 2)(x + 3) = x2 + 3x + 2x + 6 = x2 + 5x + 6
Dividing the given equation with x2 + 5x + 6, we get

x4 + 5x3 + 4x2 - 10x - 12
= (x2 + 5x + 6)(x2- 2)
= (x + 2)(x + 3)(x - √2 )(x + √2 )
Other zeroes are:
x - √2 = 0 or x + √2 = 0
x = √2 or x = -√2

Question 31.
Find all the zeroes of the polynomial 8x4 + 8x3 - 18x2 - 20x - 5, if it is given that two of its zeroes are √5/2 and −√5/2.
Year of Question: (2014, 2016 D)

Solution:

Question 32.
If p(x) = x3 - 2x2 + kx + 5 is divided by (x - 2), the remainder is 11. Find k. Hence find all the zeroes of x3+ kx2 + 3x + 1.
Year of Question:(2012)

Solution:
p(x) = x3 - 2x2 + kx + 5,
When x - 2,
p(2) = (2)3 - 2(2)2 + k(2) + 5
⇒ 11 = 8 - 8 + 2k + 5
⇒ 11 - 5 = 2k
⇒ 6 = 2k
⇒ k = 3
Let q(x) = x3 + kx2 + 3x + 1
= x3 + 3x2 + 3x + 1
= x3 + 1 + 3x2 + 3x
= (x)3 + (1)3 + 3x(x + 1)
= (x + 1)3
= (x + 1) (x + 1) (x + 1) .[∵ a3 + b3 + 3ab (a + b) = (a + b)3]
All zeroes are:
x + 1 = 0 ⇒ x = -1
x + 1 = 0 ⇒ x = -1
x + 1 = 0 ⇒ x = -1
Hence zeroes are -1, -1 and -1.

Question 33.
If α and β are zeroes of p(x) = kx2 + 4x + 4, such that α2 + β2= 24, find k.
Year of Question:(2013)

Solution:
We have, p(x) = kx2 + 4x + 4
Here a = k, b = 4, c = 4

⇒ 24k2 = 16 - 8k
⇒ 24k2 + 8k - 16 = 0
⇒ 3k2 + k - 2 = 0 .[Dividing both sides by 8]
⇒ 3k2 + 3k - 2k - 2 = 0
⇒ 3k(k + 1) - 2(k + 1) = 0
⇒ (k + 1)(3k - 2) = 0
⇒ k + 1 = 0 or 3k - 2 = 0
⇒ k = -1 or k = 2/3

Question 34.
If α and β are the zeroes of the polynomial p(x) = 2x2 + 5x + k, satisfying the relation, α2 + β2 + αβ = 21/4 then find the value of k.
Year of Question:(2017 OD)

Solution:
Given polynomial is p(x) = 2x2 + 5x + k
Here a = 2, b = 5, c = k

Question 35.
What must be subtracted from p(x) = 8x4 + 14x3 - 2x2 + 8x - 12 so that 4x2 + 3x - 2 is factor of p(x)? This question was given to group of students for working together.
Year of Question:(2015)

Solution:

Polynomial to be subtracted by (15x - 14).

Question 36.
Find the values of a and b so that x4 + x3 + 8x2 +ax - b is divisible by x2 + 1.
Year of Question:(2015)

Solution:

If x4 + x3 + 8x2 +ax - b is divisible by x2 + 1
Remainder = 0
(a - 1)x - b - 7 = 0
(a - 1)x + (-b - 7) = 0 . x + 0
a - 1 = 0, -b - 7 = 0
a = 1, b = -7
a = 1, b = -7

Question 37.
If a polynomial 3x4 - 4x3 - 16x2 + 15x + 14 is divided by another polynomial x2 - 4, the remainder comes out to be px + q. Find the value of p and q.
Year of Question:(2014)

Solution:

Question 38.
If the polynomial (x4 + 2x3 + 8x2 + 12x + 18) is divided by another polynomial (x2 + 5), the remainder comes out to be (px + q), find the values of p and q.
Year of Question:(2014)

Solution:

Remainder = 2x + 3
px + q = 2x + 3
p = 2 and q = 3.

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