Selina Maths Class 10 Solutions

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

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8C

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
-9 should be added

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.

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8C are helpful to complete your math homework.

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

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 8B.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8C

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.
(iv) 2x + 7 is a factor of 2x³ + 5x² – 11x – 14. Hence, factorise the given expression 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. (2004)
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) = o.
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 show that a = \(\frac { n – q }{ m – p }\)
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:

Question 9.
Find the value of ‘a’ if (x – a) is a factor of x³ – ax² + x + 2.
Solution:

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:

⇒ 9 – k = 0 ⇒ k = 9
Hence 9 should be subtracted.

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B are helpful to complete your math homework.

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

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 8A.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8C

Question 1.
Find in each case, the remainder when :
(i) x⁴ – 3x² + 2x + 1 is divided by x – 1.
(ii) x³ + 3x² – 12x + 4 is divided by x – 2.
(iii) x⁴ + 1 is divided by x + 1.
(iv) 4x³ – 3x² + 2x – 4 is divided by 2x + 1.
(v) 4x³ + 4x² – 21x + 16 is divided by 2x – 3.
(vi) 2x³ + 9x² – x – 15 is divided by 2x + 3.
Solution:

Question 2.
Show that:
(i) x – 2 is a factor of 5x² + 15x – 50.
(ii) 3x + 2 is a factor of 3x² – x – 2.
(iii) x + 1 is a factor of x³ + 3x² + 3x + 1.
Solution:

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
(iv) 3x – 2
(v) 2x – 3.
Solution:
(i) f(x) = 2×3 + 3×2 – 5x – 6

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:

(x + 3) is a factor of f(x)
f(-3) = 0
9a – 3b – 39 – 0
⇒ 3a – b – 13 = 0
⇒ 3a – b = 13
Adding (i) and (ii) we get:
5a = 15 ⇒ a = 3
Substituting the value of a in (i)
2(3) + b = 2
⇒ 6 + b = 2
⇒ b = 2 – 6 = – 4
Hence a = 3, b = – 4

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:

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 constants.
Solution:
f(x) = x³ + 2x² – kx + 4

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. [2005]
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 by x – 2, the remainder is 10.
Solution:

Question 15.
The polynomials 2x³ – 7x² + ax – 6 and x³ – 8x² + (2a + 1) x – 16 leave 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 7x³ + 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. (2015)
Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7D

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D

Question 1.
If a : b = 3 : 5, find : (10a + 3b) : (5a + 2b)
Solution:

Question 2.
If 5x + 6y : 8x + 5y = 8 : 9, find x : y
Solution:

Question 3.
If k (3x – 4y) : (2x – 3y) = (5x – 6y) : (4x – 5y), find x : y.
Solution:

Question 4.
Find the :
(i) duplicate ratio of 2√2 : 3√5
(ii) triplicate ratio of 2a : 3b,
(iii) sub-duplicate ratio of 9x² a⁴ : 25y⁶ b²
(iv) sub-triplicate ratio of 216 : 343
(v) reciprocal ratio of 3 : 5
(vi) ratio compounded of the duplicate ratio of 5 : 6, the reciprocal ratio of 25 : 42 and the sub-duplicate ratio of 36 : 49.
Solution:

Question 5.
Find the value of x, if :
(i) (2x + 3) : (5x – 38) is the duplicate ratio of √5 : √6
(ii) (2x + 1) : (3x + 13) is the sub-duplicate ratio of 9 : 25.
(iii) (3x -7): (4x + 3) is the sub-triplicate ratio of 8 : 27.
Solution:

Question 6.
What quantity must be added to each term of the ratio x : y so that it may become equal to c : d ?
Solution:

Question 7.
A woman reduces her weight in the ratio 7 : 5. What does her weight become if originally it was 84 kg?
Solution:
Ratio in reduction of weight = 7 : 5
Originally weight of the woman = 84 kg
Reduced weight = \(\frac { 84 x 5 }{ 7 }\) = 60 kg

Question 8.
If 15 (2x² – y²) = 7xy, find x : y, if x and y both are positive.
Solution:

Question 9.
Find the :
(i) fourth proportional to 2xy, x² and y².
(ii) third proportional to a² – b² and a + b.
(iii) mean proportion to (x – y) and (x³ – x²y)
Solution:
(i) Let a be the fourth proportional
Then 2xy : x² :: y² : a

Question 10.
Find two numbers such that the mean proportional between them is 14 and third proportional to them is 112.
Solution:
Let x and y be the two numbers.
Then 14 is the mean proportional between x and y
xy = 14² => xy = 196 ….(i)
and 112 is the third proportional to x, y
y² = 112 x ….(ii)

Question 11.
If x and y be unequal and x : y is the duplicate ratio of x + z and y + z, prove that z is mean proportional between x and y.
Solution:

Question 12.
If q is the mean proportional between p and r, prove that:

Solution:

Question 13.
If a, b and c are in continued proportion, prove that: a : c = (a² + b²) : (b² + c²).
Solution:

Question 14.

Solution:

Question 15.
If (4a + 9b) (4c – 9d) = (4a – 9b) (4c + 9d), prove that a : b = c : d.
Solution:

Question 16.

Solution:

Question 17.

Solution:

Question 18.
There are 36 members in a student council in a school and the ratio of the number of boys to the number of girls is 3 : 1. How many more girls should be added to the council so that the ratio of number of boys to the number of girls may be 9 : 5 ?
Solution:
Total number of members = 36
Ratio in boys and girls = 3 : 1

Question 19.
If 7x – 15y = 4x + y, find the value of x : y. Hence, use componendo and dividendo to find the values of:

Solution:

Question 20.

Solution:

Question 21.
If x, y, z arc in continued proportion, prove that

Solution:

Question 22.

Solution:

Question 23.

Solution:

Question 24.
Using componendo and dividendo, find the value

Solution:

Question 25.

Solution:

Question 26.

Solution:

Question 27.

Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D

Question 1.
If a : b = c : d, prove that:
(i) 5a + 7b : 5a – 7b = 5c + 7d : 5c – 7d.
(ii) (9a + 13b) (9c – 13d) = (9c + 13d) (9a – 13b).
(iii) xa + yb : xc + yd = b : d.
Solution:

Question 2.
If a : b = c : d, prove that: (6a + 7b) (3c – 4d) = (6c + 7d) (3a – 4b).
Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.
If (7a + 8b) (7c – 8d) = (7a – 8b) (7c + 8d), prove that a : b = c : d.
Solution:

Question 6.

Solution:

Question 7.
If (a + b + c + d) (a – b – c + d) = (a + b – c – d) (a – b + c – d), prove that a : b = c : d.
Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.
If a, b and c are in continued proportion, prove that

Solution:

Question 11.
Using properties of proportion, solve for x:

Solution:



But x = – 1 does not satisfy it
x = 1

Question 12.

Solution:

Question 13.
Using the properties of proportion, solve for x, given

Solution:

Question 14.

Solution:

Question 15.

Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.