## 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 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. (2014)**

**Solution:**

f(x) = x

^{3}+ 10x

^{2}– 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 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

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 3x ^{3} – 8x^{2} + 4x – 3, so that the resulting expression has x + 2 as a factor ?**

**Solution:**

The number to be subtracted = Remainder obtained by dividing 3x

^{3}– 8x

^{2}+ 4x – 3 by x + 2

Let f(x) = 3x

^{3}– 8x

^{2}+ 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 x ^{3} + (a + 1) x^{2} – (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 x ^{2} – ax + b and a + b = 1, find the values of a and b.**

**Solution:**

(x – 2) is a factor of x

^{2}+ ax + b

Let x – 2 = 0 ⇒ x = 2

Now x

^{2}+ 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 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

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}– mx

^{2}+ 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 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:**

**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 k be added to f(x), then f(x) = x

^{2}+ 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 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:**

Let f(x) = x

^{3}+ 2x

^{2}– 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) = x

^{3}+ ax

^{2}+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) x ^{3} + (a + 1) x^{2} – (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) x

^{3}+ (a + 1) x

^{2}– (2a + 1) x – 15

**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:**

When (x – 3) divides x

^{3}– px

^{2}+ x + 6,

then Remainder = p(3) = (3)

^{3}– p(3)

^{2}+ (3) + 6 = 27 – 9p + 9 = 36 – 9p

When (x – 3) divides 2x

^{3}– x

^{2}– (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 : 2x ^{3} + x^{2} – 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 2x

^{3}+ x

^{2}– 13x + 6

**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. (2016)**

**Solution:**

Let f(x) = 2x

^{3}+ 3x

^{2}– 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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