## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 6 Factorization Ex 6

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 6 Factorization Ex 6

**More Exercises**

- ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 6 Factorization EX 6
- ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 6 Factorization MCQS
- ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 6 Factorization Chapter Test

**Question 1.**

**Find the remainder (without divisions) on dividing f(x) by x – 2, where**

**(i) f(x) = 5x ^{2} – 1x + 4**

**(ii) f (x) = 2x**

^{3}– 7x^{2}+ 3**Solution:**

Let x – 2 = 0, then x = 2

(i) Substituting value of x in f(x)

f(x) = 5x

^{2}– 7x + 4

⇒ f(2) = 5(2)

^{2}– 7(2) + 4

**Question 2.**

**Using remainder theorem, find the remainder on dividing f(x) by (x + 3) where**

**(i) f(x) = 2x ^{2} – 5x + 1**

**(ii) f(x) = 3x**

^{3}+ 7x^{2}– 5x + 1**Solution:**

Let x + 3 = 0

⇒ x = -3

Substituting the value of x in f(x),

**Question 3.**

**Find the remainder (without division) on dividing f(x) by (2x + 1) where**

**(i) f(x) = 4x ^{2} + 5x + 3**

**(ii) f(x) = 3x**

^{3}– 7x^{2}+ 4x + 11**Solution:**

Let 2x + 1 = 0, then x = \(– \frac { 1 }{ 2 } \)

Substituting the value of x in f(x):

(i) f(x) = 4x

^{2}+ 5x + 3

**Question 4.**

**(i) Find the remainder (without division) when 2x ^{3} – 3x^{2} + 7x – 8 is divided by x – 1 (2000)**

**(ii) Find the remainder (without division) on dividing 3x**

^{2}+ 5x – 9 by (3x + 2)**Solution:**

(i) Let x – 1 = 0, then x = 1

Substituting value of x in f(x)

**Question 5.**

**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:**

f(x) = 2x

^{2}+ 3x

^{2}– kx + 5

g(x) = x – 2, if x – 2 = 0, then x = 2

Dividing f(x) by g(x) the remainder will be

**Question 6.**

**Using remainder theorem, find the value of a if the division of x ^{3} + 5x^{2} – ax + 6 by (x – 1) leaves the remainder 2a.**

**Solution:**

Let x – 1 = 0, then x = 1

Substituting the value of x in f(x)

**Question 7.**

**(i) What number must be subtracted from 2x ^{2} – 5x so that the resulting polynomial leaves the remainder 2, when divided by 2x + 1 ?**

**(ii) What number must be added to 2x**

^{3}– 7x^{2}+ 2x so that the resulting polynomial leaves the remainder – 2 when divided by 2x – 3?**Solution:**

(i) Let a be subtracted from 2x

^{2}– 5x,

Dividing 2x

^{2}– 5x by 2x + 1,

**Question 8.**

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

**(ii) Find ‘a’ if the two polynomials ax**

^{3}+ 3x^{2}– 9 and 2x^{3}+ 4x + a, leaves the same remainder when divided by x + 3.**Solution:**

By dividing

x

^{3}– px

^{2}+ x + 6

and 2x

^{3}– x

^{2}– (p + 3) x – 6

**Question 9.**

**By factor theorem, show that (x + 3) and (2x – 1) are factors of 2x ^{2} + 5x – 3.**

**Solution:**

Let x + 3 = 0 then x = – 3

Substituting the value of x in f(x)

**Question 10.**

**Show that (x – 2) is a factor of 3x ^{2} – x – 10 Hence factorise 3x^{2} – x – 10.**

**Solution:**

Let x – 2 = 0, then x = 2

Substituting the value of x in f(x),

**Question 11.**

**Show that (x – 1) is a factor of x ^{3} – 5x^{2} – x + 5 Hence factorise x^{3} – 5x^{2} – x + 5.**

**Solution:**

Let x – 1 = 0, then x = 1

Substituting the value of x in f(x),

**Question 12.**

**Show that (x – 3) is a factor of x ^{3} – 7x^{2} + 15x – 9. Hence factorise x^{3} – 7x^{2} + 15 x – 9**

**Solution:**

Let x – 3 = 0, then x = 3,

Substituting the value of x in f(x),

**Question 13.**

**Show that (2x + 1) is a factor of 4x ^{3} + 12x^{2} + 11 x + 3 .Hence factorise 4x^{3} + 12x^{2} + 11x + 3.**

**Solution:**

Let 2x + 1 = 0,

then x = \(– \frac { 1 }{ 2 } \)

Substituting the value of x in f(x),

**Question 14.**

**Show that 2x + 7 is a factor of 2x ^{3} + 5x^{2} – 11x – 14. Hence factorise the given expression completely, using the factor theorem. (2006)**

**Solution:**

Let 2x + 7 = 0, then 2x = -7

x = \(\\ \frac { -7 }{ 2 } \)

substituting the value of x in f(x),

**Question 15.**

**Use factor theorem to factorise the following polynominals completely.**

**(i) x ^{3} + 2x^{2} – 5x – 6**

**(ii) x**

^{3}– 13x – 12.**Solution:**

(i) Let f(x) = x

^{3}+ 2x

^{2}– 5x – 6

**Question 16.**

**(i) Use the Remainder Theorem to factorise the following expression : 2x ^{3} + x^{2} – 13x + 6. (2010)**

**(ii) Using the Remainder Theorem, factorise completely the following polynomial:**

**3x**

^{2}+ 2x^{2}– 19x + 6 (2012)**Solution:**

(i) Let f(x) = 2x

^{3}+ x

^{2}– 13x + 6

Factors of 6 are ±1, ±2, ±3, ±6

Let x = 2, then

**Question 17.**

**Using the Remainder and Factor Theorem, factorise the following polynomial: x ^{3} + 10x^{2} – 37x + 26.**

**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

**Question 18.**

**If (2 x + 1) is a factor of 6x ^{3} + 5x^{2} + ax – 2 find the value of a**

**Solution:**

Let 2x + 1 = 0, then x = \(– \frac { 1 }{ 2 } \)

Substituting the value of x in f(x),

f(x) = 6x

^{3}+ 5x

^{2}+ ax – 2

**Question 19.**

**If (3x – 2) is a factor of 3x ^{3} – kx^{2} + 21x – 10, find the value of k.**

**Solution:**

Let 3x – 2 = 0, then 3x = 2

⇒ x = \(\\ \frac { 2 }{ 3 } \)

Substituting the value of x in f(x),

f(x) = 3x

^{3}– kx

^{2}+ 21x – 10

**Question 20.**

**If (x – 2) is a factor of 2x ^{3} – x^{2} + px – 2, then**

**(i) find the value of p.**

**(ii) with this value of p, factorise the above expression completely**

**Solution:**

(i) Let x – 2 = 0, then x = 2

Now f(x) = 2x

^{3}– x

^{2}+ px – 2

**Question 21.**

**Find the value of ‘K’ for which x = 3 is a solution of the quadratic equation, (K + 2)x ^{2} – Kx + 6 = 0.**

**Also, find the other root of the equation.**

**Solution:**

(K + 2)x

^{2}– Kx + 6 = 0 …(1)

Substitute x = 3 in equation (1)

**Question 22.**

**What number should be subtracted from 2x ^{3} – 5x^{2} + 5x so that the resulting polynomial has 2x – 3 as a factor?**

**Solution:**

Let the number to be subtracted be k and the resulting polynomial be f(x), then

f(x) = 2x

^{3}– 5x

^{2}+ 5x – k

Since, 2x – 3 is a factor of f(x),

Now, converting 2x – 3 to factor theorem

**Question 23.**

**Find the value of the constants a and b, if (x – 2) and (x + 3) are both factors of the expression x ^{3} + ax^{2} + bx – 12.**

**Solution:**

Let x – 2 = 0, then x = 0

Substituting value of x in f(x)

f(x) = x

^{3}+ ax

^{2}+ bx – 12

**Question 24.**

**If (x + 2) and (x – 3) are factors of x ^{3} + ax + b, find the values of a and b. With these values of a and b, factorise the given expression.**

**Solution:**

Let x + 2 = 0, then x = -2

Substituting the value of x in f(x),

f(x) = x

^{3}+ ax + b

**Question 25.**

**(x – 2) is a factor of the expression x ^{3} + ax^{2} + bx + 6. When this expression is divided by (x – 3), it leaves the remainder 3. Find the values of a and b. (2005)**

**Solution:**

As x – 2 is a factor of

f(x) = x

^{3}+ ax

^{2}+ bx + 6

**Question 26.**

**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:**

f(x) = 2x

^{3}+ ax

^{2}+ bx – 14

∴ (x – 2) is factor of f(x)

f(2) = 0

**Question 27.**

**If ax ^{3} + 3x^{2} + bx – 3 has a factor (2x + 3) and leaves remainder – 3 when divided by (x + 2), find the values of a and 6. With these values of a and 6, factorise the given expression.**

**Solution:**

Let 2x + 3 = 0 then 2x = -3

⇒ x = \(\\ \frac { -3 }{ 2 } \)

Substituting the value of x in f(x),

f(x) = ax

^{3}+ 3x

^{2}+ 6x – 3

**Question 28.**

**Given f(x) = ax ^{2} + bx + 2 and g(x) = bx^{2} + ax + 1. If x – 2 is a factor of f(x) but leaves the remainder – 15 when it divides g(x), find the values of a and b. With these values of a and b, factorise the expression. f(x) + g(x) + 4x^{2} + 7x.**

**Solution:**

f(x) = ax

^{2}+ bx + 2

g(x) = bx

^{2}+ ax + 1

x – 2 is a factor of f(x)

Let x – 2 = 0

⇒ x = 2

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 6 Factorization Ex 6 are helpful to complete your math homework.

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