# Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8C

## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8C.

Other Exercises

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. (2014)
Solution:
f(x) = x3 + 10x2 – 37x + 26
f(1) = (1)3 + 10(1)2 – 37(1) + 26 = 1 + 10 – 37 + 26 = 0
x = 1
x – 1 is factor of f(x) 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
and x – 2 = 0 then x = 2
f(2) = (2)3 + 3(2)2 – m(2) + 4 = 8 + 12 – 2m + 4 = 24 – 2m
Remainder = 24 – 2m
But, remainder is given m + 3
m + 3 = 24 – 2m
⇒ m + 2m = 24 – 3
⇒ 3m = 21
⇒ m = 7
Hence 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:
The number to be subtracted = Remainder obtained by dividing 3x3 – 8x2 + 4x – 3 by x + 2
Let f(x) = 3x3 – 8x2 + 4x – 3
and x + 2 = 0, then x = – 2
Remainder = f(-2) = 3 (-2)3 – 8 (-2)2 + 4 (-2) – 3 = -24 – 32 – 8 – 3 = -67
Hence the number to be subtracted = – 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 6. And then, factorise the given expression completely.
Solution:  Question 6.
If x – 2 is a factor of x2 – ax + b and a + b = 1, find the values of a and b.
Solution:
(x – 2) is a factor of x2 + ax + b
Let x – 2 = 0 ⇒ x = 2
Now x2 + ax + b = (2)2 + a x 2 + b = 4 + 2a + b = 2a + b + 4
x – 2 is the factor Remainder = 0 or 2a + b + 4 = 0
⇒ 2a + b = -4 …(i)
But a + b = 1 (given) …(ii)
Subtracting, we get : a = -5
Substituting the value of a in (ii)
-5 + b = 1 ⇒ b = 1 + 5 ⇒ b = 6
Hence a = -5, b = 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
Let x – 2 = 0, then x = 2
f(2) = m (2)3 + 2 (2)2 – 3 = 8m + 8 – 3 = 8m + 5
g(2) = (2)2 – mx2 + 4 = 4 – 2m + 4 = 8 – 2m
In both cases the remainder is same
8m + 5 = 8 – 2m
⇒ 8m + 2m = 8 – 5
⇒ 10m = 3
⇒ m = $$\frac { 3 }{ 10 }$$

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:   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 k be added to f(x), then f(x) = x2 + x + 3 + k
Let x + 3 = 0, then x = -3
f(-3) = (-3)2 + (-3) + 3 + k = 9 – 3 + 3 + k = 9 + k
f(x) is divisible by x + 3, then remainder will be 0.
9 + k = 0 ⇒ k = -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:
Let f(x) = x3 + 2x2 – 5ax – 1
and let x – 1 = 0, then x = 1
f(1) = (1)3 + 2(1)2 – 5a x 1 – 7 = 1 + 2 – 5a – 7 = -5a – 4
-5a – 4 = A ….(i)
Let g (x) = x3 + ax2 +12x + 16
and let x + 2 = 0, then x = -2
g (-2) = (-2)3 + a (-2)2 – 12 (-2) + 16 = -8 + 4a + 24 + 16 = 32 + 4a
32 + 4a = B ….(ii)
2A + B = 0
2 (-5a – 4) + 32 + 4a = 0
⇒ -10a – 8 + 32 + 4a = 0
⇒ -6a + 24 = 0
⇒ 6a = 24
⇒ a = 4
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’. For this value of ‘a’, factorise the given polynomial completely.
Solution:
Let f(x) = (a – 1) x3 + (a + 1) x2 – (2a + 1) x – 15  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:
When (x – 3) divides x3 – px2 + x + 6,
then Remainder = p(3) = (3)3 – p(3)2 + (3) + 6 = 27 – 9p + 9 = 36 – 9p
When (x – 3) divides 2x3 – x2 – (p + 3) x – 6,
then Remainder = p(3) = 2(3)3 – (3)2 – (p + 3) (3) – 6
= 54 – 9 – 3p – 9 – 6 = 30 – 3p
A.T.Q. both remainders are equal
⇒ 36 – 9p = 30 – 3p
⇒ 36 – 30 = -3p + 9p
⇒ 6 = 6p
⇒ p = 1

Question 14.
Use the Remainder Theorem to factorise the following expression : 2x3 + x2 – 13x + 6
Solution:
(a) By hit and trial, putting x = 2, we have
2 (8) + 4 – 26 + 6 = 0
⇒ (x – 2) is the factor of 2x3 + x2 – 13x + 6 Question 15.
Using remainder theorem, find the value of k if on dividing 2x3 + 3x2 – kx + 5 by x – 2, leaves a remainder 7. (2016)
Solution:
Let f(x) = 2x3 + 3x2 – kx + 5 By the remainder theorem,
f(2) = 7
⇒ 2(2)3 + 3(2)2 – k(2) + 5 = 7
⇒ 2(8) + 3(4) – k(2) + 5 = 7
⇒ 16 + 12 – 2k + 5 = 7
⇒ 2k = 16 + 12 + 5 – 7
⇒ 2k = 26
⇒ k = 13
The value of k is 13.

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