Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems

Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems

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Selina ICSE Solutions for Class 10 Maths Chapter 8 Remainder and Factor Theorems

Exercise 8(A)

Question 1.

Solution:
By remainder theorem we know that when a polynomial f (x) is divided by x – a, then the remainder is f(a).

Question 2.

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.

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).

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:

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:

Question 6.
find the value of k, if 2x + 1 is a factor of (3k + 2)x³ + (k – 1).
Solution:

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:

Question 9.
When x³ + 2x² – kx + 4 is divided by x – 2, the remainder is k. Find the value of constant k.
Solution:

Question 10.
Find the value of a, if the division of ax³ + 9x² + 4x – 10 by x + 3 leaves a remainder 5.
Solution:

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:

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:

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:

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:

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:

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:

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:

Exercise 8(B)

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:

Question 2.
Using the Remainder Theorem, factorise each of the following completely.
(i) 3x³ + 2x² − 19x + 6
(ii) 2x³ + x² – 13x + 6
(iii) 3x³ + 2x² – 23x – 30
(iv) 4x³ + 7x² – 36x – 63
(v) x³ + x² – 4x – 4
Solution:

Question 3.
Using the Remainder Theorem, factorise the expression 3x³ + 10x² + x – 6. Hence, solve the equation 3x³ + 10x² + x – 6 = 0.
Solution:

Question 4.
Factorise the expression f (x) = 2x³ – 7x² – 3x + 18. Hence, find all possible values of x for which f(x) = 0.
Solution:

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:

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:

Question 7.
If x + a is a common factor of expressions f(x) = x² + px + q and g(x) = x² + mx + n;

Solution:

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

Exercise 8(C)

Question 1.
Show that (x – 1) is a factor of x³ – 7x² + 14x – 8. Hence, completely factorise the given expression.
Solution:

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).

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:

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).

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:

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:

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