Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems

Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems

Remainder and Factor Theorems Exercise 8A – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 1
Solution:
By remainder theorem we know that when a polynomial f (x) is divided by x – a, then the remainder is f(a).
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 2

Question 2.
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 3
Solution:
(x – a) is a factor of a polynomial f(x) if the remainder, when f(x) is divided by (x – a), is 0, i.e., if f(a) = 0.
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 4

Question 3.
Use the Remainder Theorem to find which of the following is a factor of 2x3 + 3x2 – 5x – 6.
(i) x + 1
(ii) 2x – 1
(iii) x + 2
Solution:
By remainder theorem we know that when a polynomial f (x) is divided by x – a, then the remainder is f(a).
Let f(x) = 2x3 + 3x2 – 5x – 6
(i) f (-1) = 2(-1)3 + 3(-1)2 – 5(-1) – 6 = -2 + 3 + 5 – 6 = 0
Thus, (x + 1) is a factor of the polynomial f(x).
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 6
Thus, (2x – 1) is not a factor of the polynomial f(x).
(iii) f (-2) = 2(-2)3 + 3(-2)2 – 5(-2) – 6 = -16 + 12 + 10 – 6 = 0
Thus, (x + 2) is a factor of the polynomial f(x).

Question 4.
(i) If 2x + 1 is a factor of 2x2 + ax – 3, find the value of a.
(ii) Find the value of k, if 3x – 4 is a factor of expression 3x2 + 2x – k.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 7

Question 5.
Find the values of constants a and b when x – 2 and x + 3 both are the factors of expression x3 + ax2 + bx – 12.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 8

Question 6.
find the value of k, if 2x + 1 is a factor of (3k + 2)x3 + (k – 1).
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 9

Question 7.
Find the value of a, if x – 2 is a factor of 2x5 – 6x4 – 2ax3 + 6ax2 + 4ax + 8.
Solution:
f(x) = 2x5 – 6x4 – 2ax3 + 6ax2 + 4ax + 8
x – 2 = 0 ⇒  x = 2
Since, x – 2 is a factor of f(x), remainder = 0.
2(2)5 – 6(2)4 – 2a(2)3 + 6a(2)2 + 4a(2) + 8 = 0
64 – 96 – 16a + 24a + 8a + 8 = 0
-24 + 16a = 0
16a = 24
a = 1.5

Question 8.
Find the values of m and n so that x – 1 and x + 2 both are factors of x3 + (3m + 1) x2 + nx – 18.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 10

Question 9.
When x3 + 2x2 – kx + 4 is divided by x – 2, the remainder is k. Find the value of constant k.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 11

Question 10.
Find the value of a, if the division of ax3 + 9x2 + 4x – 10 by x + 3 leaves a remainder 5.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 12

Question 11.
If x3 + ax2 + bx + 6 has x – 2 as a factor and leaves a remainder 3 when divided by x – 3, find the values of a and b.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 13

Question 12.
The expression 2x3 + ax2 + bx – 2 leaves remainder 7 and 0 when divided by 2x – 3 and x + 2 respectively. Calculate the values of a and b.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 14

Question 13.
What number should be added to 3x3 – 5x2 + 6x so that when resulting polynomial is divided by x – 3, the remainder is 8?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 15

Question 14.
What number should be subtracted from x3 + 3x2 – 8x + 14 so that on dividing it with x – 2, the remainder is 10.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 16

Question 15.
The polynomials 2x3 – 7x2 + ax – 6 and x3 – 8x2 + (2a + 1)x – 16 leaves the same remainder when divided by x – 2. Find the value of ‘a’.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 17

Question 16.
If (x – 2) is a factor of the expression 2x3 + ax2 + bx – 14 and when the expression is divided by (x – 3), it leaves a remainder 52, find the values of a and b
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 18

Question 17.
Find ‘a‘ if the two polynomials ax3 + 3x2 – 9 and 2x3 + 4x + a, leave the same remainder when divided by x + 3.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 19

Remainder and Factor Theorems Exercise 8B – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Using the Factor Theorem, show that:
(i) (x – 2) is a factor of x3 – 2x2 – 9x + 18. Hence, factorise the expression x3 – 2x2 – 9x + 18 completely.
(ii) (x + 5) is a factor of 2x3 + 5x2 – 28x – 15. Hence, factorise the expression 2x3 + 5x2 – 28x – 15 completely.
(iii) (3x + 2) is a factor of 3x3 + 2x2 – 3x – 2. Hence, factorise the expression 3x3 + 2x2 – 3x – 2 completely.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 20
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 21

Question 2.
Using the Remainder Theorem, factorise each of the following completely.
(i) 3x+ 2x2 − 19x + 6
(ii) 2x3 + x2 – 13x + 6
(iii) 3x3 + 2x2 – 23x – 30
(iv) 4x3 + 7x2 – 36x – 63
(v) x3 + x2 – 4x – 4
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 22
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 23
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 24

Question 3.
Using the Remainder Theorem, factorise the expression 3x3 + 10x2 + x – 6. Hence, solve the equation 3x3 + 10x2 + x – 6 = 0.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 25.

Question 4.
Factorise the expression f (x) = 2x3 – 7x2 – 3x + 18. Hence, find all possible values of x for which f(x) = 0.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 26

Question 5.
Given that x – 2 and x + 1 are factors of f(x) = x3 + 3x2 + ax + b; calculate the values of a and b. Hence, find all the factors of f(x).
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 27

Question 6.
The expression 4x3 – bx2 + x – c leaves remainders 0 and 30 when divided by x + 1 and 2x – 3 respectively. Calculate the values of b and c. Hence, factorise the expression completely.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 28
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 29

Question 7.
If x + a is a common factor of expressions f(x) = x2 + px + q and g(x) = x2 + mx + n;
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 30
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 31

Question 8.
The polynomials ax3 + 3x2 – 3 and 2x3 – 5x + a, when divided by x – 4, leave the same remainder in each case. Find the value of a.
Solution:
Let f(x) = ax3 + 3x2 – 3
When f(x) is divided by (x – 4), remainder = f(4)
f(4) = a(4)3 + 3(4)2 – 3 = 64a + 45
Let g(x) = 2x3 – 5x + a
When g(x) is divided by (x – 4), remainder = g(4)
g(4) = 2(4)3 – 5(4) + a = a + 108
It is given that f(4) = g(4)
64a + 45 = a + 108
63a = 63
a = 1

Question 9.
Find the value of ‘a’, if (x – a) is a factor of x3 – ax2 + x + 2.
Solution:
Let f(x) = x3 – ax2 + x + 2
It is given that (x – a) is a factor of f(x).
Remainder = f(a) = 0
a3 – a3 + a + 2 = 0
a + 2 = 0
a = -2

Question 10.
Find the number that must be subtracted from the polynomial 3y3 + y2 – 22y + 15, so that the resulting polynomial is completely divisible by y + 3.
Solution:
Let the number to be subtracted from the given polynomial be k.
Let f(y) = 3y3 + y2 – 22y + 15 – k
It is given that f(y) is divisible by (y + 3).
Remainder = f(-3) = 0
3(-3)3 + (-3)2 – 22(-3) + 15 – k = 0
-81 + 9 + 66 + 15 – k = 0
9 – k = 0
k = 9

Remainder and Factor Theorems Exercise 8C – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Show that (x – 1) is a factor of x3 – 7x2 + 14x – 8. Hence, completely factorise the given expression.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 32

Question 2.
Using Remainder Theorem, factorise:
x3 + 10x2 – 37x + 26 completely.
Solution:

Question 3.
When x3 + 3x2 – mx + 4 is divided by x – 2, the remainder is m + 3. Find the value of m.
Solution:
Let f(x) = x3 + 3x2 – mx + 4
According to the given information,
f(2) = m + 3
(2)3 + 3(2)2 – m(2) + 4 = m + 3
8 + 12 – 2m + 4 = m + 3
24 – 3 = m + 2m
3m = 21
m = 7

Question 4.
What should be subtracted from 3x3 – 8x2 + 4x – 3, so that the resulting expression has x + 2 as a factor?
Solution:
Let the required number be k.
Let f(x) = 3x3 – 8x2 + 4x – 3 – k
According to the given information,
f (-2) = 0
3(-2)3 – 8(-2)2 + 4(-2) – 3 – k = 0
-24 – 32 – 8 – 3 – k = 0
-67 – k = 0
k = -67
Thus, the required number is -67.

Question 5.
If (x + 1) and (x – 2) are factors of x3 + (a + 1)x2 – (b – 2)x – 6, find the values of a and b. And then, factorise the given expression completely.
Solution:
Let f(x) = x3 + (a + 1)x2 – (b – 2)x – 6
Since, (x + 1) is a factor of f(x).
Remainder = f(-1) = 0
(-1)3 + (a + 1)(-1)2 – (b – 2) (-1) – 6 = 0
-1 + (a + 1) + (b – 2) – 6 = 0
a + b – 8 = 0 …(i)
Since, (x – 2) is a factor of f(x).
Remainder = f(2) = 0
(2)3 + (a + 1) (2)2 – (b – 2) (2) – 6 = 0
8 + 4a + 4 – 2b + 4 – 6 = 0
4a – 2b + 10 = 0
2a – b + 5 = 0 …(ii)
Adding (i) and (ii), we get,
3a – 3 = 0
a = 1
Substituting the value of a in (i), we get,
1 + b – 8 = 0
b = 7
f(x) = x3 + 2x2 – 5x – 6
Now, (x + 1) and (x – 2) are factors of f(x). Hence, (x + 1) (x – 2) = x2 – x – 2 is a factor of f(x).
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 34
f(x) = x3 + 2x2 – 5x – 6 = (x + 1) (x – 2) (x + 3)

Question 6.
If x – 2 is a factor of x2 + ax + b and a + b = 1, find the values of a and b.
Solution:
Let f(x) = x2 + ax + b
Since, (x – 2) is a factor of f(x).
Remainder = f(2) = 0
(2)2 + a(2) + b = 0
4 + 2a + b = 0
2a + b = -4 …(i)
It is given that:
a + b = 1 …(ii)
Subtracting (ii) from (i), we get,
a = -5
Substituting the value of a in (ii), we get,
b = 1 – (-5) = 6

Question 7.
Factorise x3 + 6x2 + 11x + 6 completely using factor theorem.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 35

Question 8.
Find the value of ‘m’, if mx3 + 2x2 – 3 and x2 – mx + 4 leave the same remainder when each is divided by x – 2.
Solution:
Let f(x) = mx3 + 2x2 – 3
g(x) = x2 – mx + 4
It is given that f(x) and g(x) leave the same remainder when divided by (x – 2). Therefore, we have:
f (2) = g (2)
m(2)3 + 2(2)2 – 3 = (2)2 – m(2) + 4
8m + 8 – 3 = 4 – 2m + 4
10m = 3
m = 3/10

Question 9.
The polynomial px3 + 4x2 – 3x + q is completely divisible by x2 – 1; find the values of p and q. Also, for these values of p and q factorize the given polynomial completely.
Solution:
Let f(x) = px3 + 4x2 – 3x + q
It is given that f(x) is completely divisible by (x2 – 1) = (x + 1)(x – 1).
Therefore, f(1) = 0 and f(-1) = 0
f(1) = p(1)3 + 4(1)2 – 3(1) + q = 0
p + q + 1 = 0 …(i)
f(-1) = p(-1)3 + 4(-1)2 – 3(-1) + q = 0
-p + q + 7 = 0 …(ii)
Adding (i) and (ii), we get,
2q + 8 = 0
q = -4
Substituting the value of q in (i), we get,
p = -q – 1 = 4 – 1 = 3
f(x) = 3x3 + 4x2 – 3x – 4
Given that f(x) is completely divisible by (x2 – 1).
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 36

Question 10.
Find the number which should be added to x2 + x + 3 so that the resulting polynomial is completely divisible by (x + 3).
Solution:
Let the required number be k.
Let f(x) = x2 + x + 3 + k
It is given that f(x) is divisible by (x + 3).
Remainder = 0
f (-3) = 0
(-3)2 + (-3) + 3 + k = 0
9 – 3 + 3 + k = 0
9 + k = 0
k = -9
Thus, the required number is -9.

Question 11.
When the polynomial x3 + 2x2 – 5ax – 7 is divided by (x – 1), the remainder is A and when the polynomial x3 + ax2 – 12x + 16 is divided by (x + 2), the remainder is B. Find the value of ‘a’ if 2A + B = 0.
Solution:
It is given that when the polynomial x3 + 2x2 – 5ax – 7 is divided by (x – 1), the remainder is A.
(1)3 + 2(1)2 – 5a(1) – 7 = A
1 + 2 – 5a – 7 = A
– 5a – 4 = A …(i)
It is also given that when the polynomial x3 + ax2 – 12x + 16 is divided by (x + 2), the remainder is B.
x3 + ax2 – 12x + 16 = B
(-2)3 + a(-2)2 – 12(-2) + 16 = B
-8 + 4a + 24 + 16 = B
4a + 32 = B …(ii)
It is also given that 2A + B = 0
Using (i) and (ii), we get,
2(-5a – 4) + 4a + 32 = 0
-10a – 8 + 4a + 32 = 0
-6a + 24 = 0
6a = 24
a = 4

Question 12.
(3x + 5) is a factor of the polynomial (a – 1)x3 + (a + 1)x2 – (2a + 1)x – 15. Find the value of ‘a’, factorise the given polynomial completely.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 37

Question 13.
When divided by x – 3 the polynomials x3 – px2 + x + 6 and 2x3 – x2 – (p + 3) x – 6 leave the same remainder. Find the value of ‘p’.
Solution:
If (x – 3) divides f(x) = x3 – px2 + x + 6, then,
Remainder = f(3) = 33 – p(3)2 + 3 + 6 = 36 – 9p
If (x – 3) divides g(x) = 2x3 – x2 – (p + 3) x – 6, then
Remainder = g(3) = 2(3)3 – (3)2 – (p + 3) (3) – 6 = 30 – 3p
Now, f(3) = g(3)
⇒ 36 – 9p = 30 – 3p
⇒ -6p = -6
⇒  p = 1

Question 14.
Use the Remainder Theorem to factorise the following expression:
2x3 + x2 – 13x + 6
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 38

Question 15.
Using remainder theorem, find the value of k if on dividing 2x3 + 3x2 – kx + 5 by x – 2, leaves a remainder 7.
Solution:
Let f(x) = 2x3 + 3x2 – kx + 5
Using Remainder Theorem, we have
f(2) = 7
∴ 2(2)3 + 3(2)2 – k(2) + 5 = 7
∴ 16 + 12 – 2k + 5 = 7
∴ 33 – 2k = 7
∴ 2k = 26
∴ k = 13

Question 16.
What must be subtracted from 16x3 – 8x2 + 4x + 7 so that the resulting expression has 2x + 1 as a factor?
Solution:
Here, f(x) = 16x3 – 8x2 + 4x + 7
Let the number subtracted be k from the given polynomial f(x).
Given that 2x + 1 is a factor of f(x).
Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems - 39
Therefore 1 must be subtracted from 16x3 – 8x2 + 4x + 7 so that the resulting expression has 2x + 1 as a factor.

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