## 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.**

**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 2x^{3} + 3x^{2} – 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) = 2x^{3} + 3x^{2} – 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 2x^{2} + ax – 3, find the value of a.

(ii) Find the value of k, if 3x – 4 is a factor of expression 3x^{2} + 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 x^{3} + ax^{2} + bx – 12.

**Solution:**

**Question 6.**

find the value of k, if 2x + 1 is a factor of (3k + 2)x^{3} + (k – 1).

**Solution:**

**Question 7.**

Find the value of a, if x – 2 is a factor of 2x^{5} – 6x^{4} – 2ax^{3} + 6ax^{2} + 4ax + 8.

**Solution:**

f(x) = 2x^{5} – 6x^{4} – 2ax^{3} + 6ax^{2} + 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 x^{3} + (3m + 1) x^{2} + nx – 18.

**Solution:**

**Question 9.**

When x^{3} + 2x^{2} – 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 ax^{3} + 9x^{2} + 4x – 10 by x + 3 leaves a remainder 5.

**Solution:**

**Question 11.**

If x^{3} + ax^{2} + 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 2x^{3} + ax^{2} + 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 3x^{3} – 5x^{2} + 6x so that when resulting polynomial is divided by x – 3, the remainder is 8?

**Solution:**

**Question 14.**

What number should be subtracted from x^{3} + 3x^{2} – 8x + 14 so that on dividing it with x – 2, the remainder is 10.

**Solution:**

**Question 15.**

The polynomials 2x^{3} – 7x^{2} + ax – 6 and x^{3} – 8x^{2} + (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 2x^{3} + ax^{2} + 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 ax^{3} + 3x^{2} – 9 and 2x^{3} + 4x + a, leave the same remainder when divided by x + 3.

**Solution:**

### 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 x^{3} – 2x^{2} – 9x + 18. Hence, factorise the expression x^{3} – 2x^{2} – 9x + 18 completely.

(ii) (x + 5) is a factor of 2x^{3} + 5x^{2} – 28x – 15. Hence, factorise the expression 2x^{3} + 5x^{2} – 28x – 15 completely.

(iii) (3x + 2) is a factor of 3x^{3} + 2x^{2} – 3x – 2. Hence, factorise the expression 3x^{3} + 2x^{2} – 3x – 2 completely.

**Solution:**

**Question 2.**

Using the Remainder Theorem, factorise each of the following completely.

(i) 3x^{3 }+ 2x^{2} − 19x + 6

(ii) 2x^{3} + x^{2} – 13x + 6

(iii) 3x^{3} + 2x^{2} – 23x – 30

(iv) 4x^{3} + 7x^{2} – 36x – 63

(v) x^{3} + x^{2} – 4x – 4

**Solution:**

**Question 3.**

Using the Remainder Theorem, factorise the expression 3x^{3} + 10x^{2} + x – 6. Hence, solve the equation 3x^{3} + 10x^{2} + x – 6 = 0.

**Solution:**

**Question 4.**

Factorise the expression f (x) = 2x^{3} – 7x^{2} – 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) = x^{3} + 3x^{2} + ax + b; calculate the values of a and b. Hence, find all the factors of f(x).

**Solution:**

**Question 6.**

The expression 4x^{3} – bx^{2} + 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) = x^{2} + px + q and g(x) = x^{2} + mx + n;

**Solution:**

**Question 8.**

The polynomials ax^{3} + 3x^{2} – 3 and 2x^{3} – 5x + a, when divided by x – 4, leave the same remainder in each case. Find the value of a.

**Solution:**

Let f(x) = ax^{3} + 3x^{2} – 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) = 2x^{3} – 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 x^{3} – ax^{2} + x + 2.

**Solution:**

Let f(x) = x^{3} – ax^{2} + x + 2

It is given that (x – a) is a factor of f(x).

Remainder = f(a) = 0

a^{3} – a^{3} + a + 2 = 0

a + 2 = 0

a = -2

**Question 10.**

Find the number that must be subtracted from the polynomial 3y^{3} + y^{2} – 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) = 3y^{3} + y^{2} – 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 x^{3} – 7x^{2} + 14x – 8. Hence, completely factorise the given expression.

**Solution:**

**Question 2.**

Using Remainder Theorem, factorise:

x^{3} + 10x^{2} – 37x + 26 completely.

**Solution:**

**Question 3.**

When x^{3} + 3x^{2} – mx + 4 is divided by x – 2, the remainder is m + 3. Find the value of m.

**Solution:**

Let f(x) = x^{3} + 3x^{2} – 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 3x^{3} – 8x^{2} + 4x – 3, so that the resulting expression has x + 2 as a factor?

**Solution:**

Let the required number be k.

Let f(x) = 3x^{3} – 8x^{2} + 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 x^{3} + (a + 1)x^{2} – (b – 2)x – 6, find the values of a and b. And then, factorise the given expression completely.

**Solution:**

Let f(x) = x^{3} + (a + 1)x^{2} – (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) = x^{3} + 2x^{2} – 5x – 6

Now, (x + 1) and (x – 2) are factors of f(x). Hence, (x + 1) (x – 2) = x^{2} – x – 2 is a factor of f(x).

f(x) = x^{3} + 2x^{2} – 5x – 6 = (x + 1) (x – 2) (x + 3)

**Question 6.**

If x – 2 is a factor of x^{2} + ax + b and a + b = 1, find the values of a and b.

**Solution:**

Let f(x) = x^{2} + 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 x^{3} + 6x^{2} + 11x + 6 completely using factor theorem.

**Solution:**

**Question 8.**

Find the value of ‘m’, if mx^{3} + 2x^{2} – 3 and x^{2} – mx + 4 leave the same remainder when each is divided by x – 2.

**Solution:**

Let f(x) = mx^{3} + 2x^{2} – 3

g(x) = x^{2} – 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 px^{3} + 4x^{2} – 3x + q is completely divisible by x^{2} – 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) = px^{3} + 4x^{2} – 3x + q

It is given that f(x) is completely divisible by (x^{2} – 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) = 3x^{3} + 4x^{2} – 3x – 4

Given that f(x) is completely divisible by (x^{2} – 1).

**Question 10.**

Find the number which should be added to x^{2} + x + 3 so that the resulting polynomial is completely divisible by (x + 3).

**Solution:**

Let the required number be k.

Let f(x) = x^{2} + 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 x^{3} + 2x^{2} – 5ax – 7 is divided by (x – 1), the remainder is A and when the polynomial x^{3} + ax^{2} – 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 x^{3} + 2x^{2} – 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 x^{3} + ax^{2} – 12x + 16 is divided by (x + 2), the remainder is B.

x^{3} + ax^{2} – 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)x^{3} + (a + 1)x^{2} – (2a + 1)x – 15. Find the value of ‘a’, factorise the given polynomial completely.

**Solution:**

**Question 13.**

When divided by x – 3 the polynomials x^{3} – px^{2} + x + 6 and 2x^{3} – x^{2} – (p + 3) x – 6 leave the same remainder. Find the value of ‘p’.

**Solution:**

If (x – 3) divides f(x) = x^{3} – px^{2} + x + 6, then,

Remainder = f(3) = 3^{3} – p(3)^{2} + 3 + 6 = 36 – 9p

If (x – 3) divides g(x) = 2x^{3} – x^{2} – (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:

2x^{3} + x^{2} – 13x + 6

**Solution:**

**Question 15. **Using remainder theorem, find the value of k if on dividing 2x

^{3}+ 3x

^{2}– kx + 5 by x – 2, leaves a remainder 7.

**Solution:**

Let f(x) = 2x

^{3}+ 3x

^{2}– 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 16x

^{3}– 8x

^{2}+ 4x + 7 so that the resulting expression has 2x + 1 as a factor?

**Solution:**

Here, f(x) = 16x

^{3}– 8x

^{2}+ 4x + 7

Let the number subtracted be k from the given polynomial f(x).

Given that 2x + 1 is a factor of f(x).

Therefore 1 must be subtracted from 16x

^{3}– 8x

^{2}+ 4x + 7 so that the resulting expression has 2x + 1 as a factor.

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