**Question **1. Find in each case the remainder when:

(i) š„^{4} ā 3š„^{2} + 2š„ + 1 is divided by š„ ā 1.

(ii) š„^{3} ā 3š„^{2} ā 12š„ + 4 is divided by š„ ā 2.

(iii) š„^{4} + 1 is divided by š„ + 1.

Solution:

(i) š„^{4} ā 3š„^{2} + 2š„ + 1 is divided by š„ ā 1.

š„ ā 1 = 0

š„ = 1

Put the value of x in given equation.

š„^{4 }ā 3š„^{2} + 2š„ + 1

(1)^{4} ā 3(1)^{2} + 2(1) + 1

1 ā 3(1) + 2 + 1

1 ā 3 + 2 + 1

4 ā 3 = 1

Hence, the reminder is 1.**(ii) š„ ^{3} ā 3š„^{2} ā 12š„ + 4 is divided by x ā 2.**

š„ ā 2 = 0

š„ = 2

Put the value of x in given equation.

š„

^{3 }ā 3š„

^{2}ā 12š„ + 4

(2)

^{3 }+ 3(2)

^{2}ā 12(2) + 4

8 + 3(4) ā 24 + 4

8 + 12 ā 24 + 4

24 ā 24 = 0

Hence, the reminder is 0.

(iii) š„

^{4 }+ 1 is divided by š„ + 1.

š„ + 1 = 0

š„ = ā1

Put the value of x in given equation.

š„

^{4}+ 1

(ā1)4 + 1

1 + 1 = 2

Hence, the reminder is 2.

**Question **2. Show that:**(i) š„ ā 2 is a factor of 5š„ ^{2} + 15š„ ā 50.(ii) 3š„ + 2 is a factor of 3š„^{2} ā š„ ā 2.Solution:**

**(i) It is given that,**

š„ ā 2 is the factor of 5š„

^{2}+ 15š„ ā 50

So,

š„ ā 2 = 0

š„ = 2

Put the value of š„ in given equation,

š(š„) = 5š„

^{2}+ 15š„ ā 50

š(2) = 5(2)

^{2}+ 15(2) ā 50

š(2) = 5 Ć 4 + 30 ā 50

š(2) = 20 + 30 ā 50

š(2) = 50 ā 50

š(2) = 0

Hence, it is proved that š„ ā 2 is a factor of 5š„

^{2}+ 15š„ ā 50.

**(ii) It is given that,**

3š„ + 2 is the factor of 3š„

^{2}ā š„ ā 2

So,

3š„ + 2 = 0

š„ = ā2/3

**Question **3. Use the Remainder Theorem to find which of the following is a factor of 2š„^{3} + 3š„^{2}ā 5š„ā 6.

(i) š„ + 1

(ii) 2š„ā 1

(iii) š„ + 2

Solution:

(i) It is given that,

š„ + 1 is the factor of 2š„^{3} + 3š„^{2}ā 5š„ā 6

š„ + 1 = 0

š„ = ā1

Put the value of š„ in given equation,

š(š„) = 2š„^{3} + 3š„^{2}ā 5š„ā 6

š(ā1) = 2(ā1)3 + 3(ā1)2 ā 5(ā1) ā 6

š(ā1) = 2 Ć ā1 + 3 Ć 1 ā 5 Ć (ā1) ā 6

š(ā1) = ā2 + 3 + 5 ā 6

š(ā1) = ā8 + 8

š(ā1) = 0

Hence, it is proved that š„ + 1 is a factor of 2š„^{3} + 3š„^{2}ā 5š„ā 6.**(ii) It is given that,**

2š„ ā 1 is the factor of 2š„^{3} + 3š„^{2}ā 5š„ā 6

2š„ ā 1 = 0

š„ = 1/2

**(iii) It is given that,**

š„ + 2 is the factor of 2š„^{3} + 3š„^{2}ā 5š„ā 6

š„ + 2 = 0

š„ = ā2

Put the value of š„ in given equation,

š(š„) = 2š„^{3} + 3š„^{2}ā 5š„ā 6

š (ā2) = 2(ā2)^{3} + 3(ā2)^{2}ā 5(ā2)ā 6

š (ā2) = ā16 + 12 + 10ā 6

š (ā2) = 0

Thus, (x + 2) is a factor of the polynomial f(x).

**Question **4. (i) If 2š„ + 1 is a factor of 2š„^{2} + šš„ā 3, find the value of a.

(ii) Find the value of k, if 3š„ā 4 is a factor of expression 3š„^{2} + 2š„ā š.

Solution:**(i) It is given that,**

2š„ + 1 is a factor of 2š„^{2} + šš„ā 3

So,

2š„^{2} + šš„ā 3 = 0

We know that,

2š„ + 1 = 0

2š„ = ā1

š„ = ā 1/2

Put the value of š„ in above equation,

2š„^{2} + šš„ā 3 = 0

1 ā š = 3 Ć 2

1 ā š = 6

āš = 6 ā 1

āš = 5

š = ā5

Hence, the value of š is ā5.**(ii) It is given that,**

3š„ ā 4 is a factor of 3š„^{2} + 2š„ā š

So,

3š„^{2} + 2š„ā š = 0

We know that,

3š„ ā 4 = 0

3š„ = 4

š„ = 4/3

Put the value of š„ in above equation,

3š„^{2} + 2š„ā š = 0

**Question **5. Find the values of constants a and b when š„ā 2 and š„ + 3 both are the factors of expression š„^{3} + šš„^{2} + šš„ā 12.

Solution:

It is given that,

š(š„) = š„ + šš„^{2} + šš„ā 12

š„ ā 2 is the factor of š„ + šš„^{2} + šš„ā 12

š„ ā 2 = 0

š„ = 2

Put the value of š„ in given equation,

(2)^{3} + š(2)^{2} + š(2)ā 12 = 0

8 + 4š + 2šā 12 = 0

4š + 2šā 4 = 0

2(2š + šā 2) = 0

2š + šā 2 = 0

2š + š = 2 **_**(1)

š„ + 3 is the factor of š„ + šš„

^{2}+ šš„ā 12

š„ + 3 = 0

š„ = ā3

Put the value of š„ in given equation,

(ā3)

^{3}+ š(ā3)

^{2}+ š(ā3)ā 12 = 0

ā27 + š Ć 9 ā š Ć 3ā 12 = 0

ā27 + 9š ā 3šā 12 = 0

9š ā 3šā 39 = 0

3(3š ā šā 13) = 0

3š ā šā 13 = 0

3š ā š = 13_____________(2)

From equation 1 we get,

2š + š = 2

**Question **6. Find the value of k, if 2š„ + 1 is a factor of (3š + 2)š„^{3} + (šā 1).

Solution:

It is given that,

2š„ + 1 is a factor of (3š + 2)š„^{3} + (šā 1)

2š„ + 1 = 0

š„ = ā 1/2

Put the value of š„ in given equation,

(3š + 2)š„^{3}+ (šā 1) = 0

ā3š ā 2 + 8šā 8 = 0

5šā 10 = 0

5š = 10

š = 10/5

š = 2

Hence, the value of š is 2.

**Question **7. Find the value of a, if š„ā 2 is a factor of 2š„^{5}ā 6š„^{4}ā 2šš„^{3} + 6šš„^{2} + 4šš„ + 8.

Solution:

It is given that,

š(š„) = 2š„^{5}ā 6š„^{4ā} 2šš„^{3} + 6šš„^{2} + 4šš„ + 8 = 0

š„ā 2 is a factor of 2š„^{5}ā 6š„^{4}ā 2šš„^{3} + 6šš„^{2} + 4šš„ + 8

š„ā 2 = 0

š„ = 2

Put the value of š„ in given equation,

2(2)^{5} ā 6(2)^{4} ā 2š(2)^{3 }+ 6š(2)^{2 }+ 4š(2) + 8 = 0

2(32) ā 6(16) ā 2š(8) + 6š(4) + 4š(2) + 8 = 0

64ā 96ā 16š + 24š + 8š + 8 = 0

ā24 + 16š = 0

16š = 24

š = 1.5

Hence, the value of š is 1.5.

**Question **8. Find the values of m and n so that š„ā 1 and š„ + 2 both are factors of š„^{3} + (3š + 1)š„^{2} + šš„ā18. **Solution:**

It is given that,

š(š„) = š„^{3} + (3š + 1)š„^{2} + šš„ ā 18

š„ā 1 is a factor of š„^{3} + (3š + 1)š„^{2} + šš„ ā 18

š„ā 1 = 0

š„ = 1

Put the value of š„ in given equation,

š„^{3} + (3š + 1)š„^{2} + šš„ ā 18 = 0

(1)^{3} + (3š + 1)(1)^{2} + š(1) ā 18 = 0

1 + (3š + 1)1 + š ā 18 = 0

1 + 3š + 1 + š ā 18 = 0

3š + š ā 16 = 0____________(1)

š„ + 2 is a factor of š„^{3} + (3š + 1)š„^{2} + šš„ ā 18

š„ + 2 = 0

š„ = ā2

Put the value of š„ in given equation,

(ā2)^{3 }+ (3š + 1)(ā2)^{2} + š(ā2) ā 18 = 0

ā8 + (3š + 1)(4) ā 2š ā 18 = 0

ā8 + 12š + 4 ā 2š ā 18 = 0

12š ā 2š ā 22 = 0

6š ā š ā 11 = 0____________(2)

From equation (1) we get, the value of š

3š + š ā 16 = 0

š = 16āš /3 ** _**(3)

Put the value of š in equation (2)

6 (16āš/3 ) ā š ā 11 = 0

2(16 ā š) ā š ā 11 = 0

32 ā 2š ā š ā 11 = 0

21 ā 3š = 0

ā3š = ā21

š = ā21/ā3

š = 7

Put the value of š in equation (3)

š = 16ā7/3

š = 9/3

š = 3

Hence, the value of š is 3 and š is 7.

**Question **9. When š„^{3 }+ 2š„^{2}ā šš„ + 4 is divided by š„ā 2, the remainder is k. Find the value of constant k.

Solution:

It is given that,

š(š„) = š„^{3 }+ 2š„^{2} + šš„ + 4 = 0

š„ ā 2 is a factor of š„^{3} + 2š„^{2} ā šš„ + 4

š„ ā 2 = 0

š„ = 2

Put the value of š„ in given equation,

š„^{3 }+ 2š„^{2} ā šš„ + 4 = š

(2)3 + 2(2)^{2} ā š(2) + 4 = š

8 + 2 Ć 4 ā š(2) + 4 = š

8 + 8 ā 2š + 4 = š

20 ā 2š = š

20 = š + 2š

20 = 3š

š = 20/3

š = 6 (2/3)

Hence, the value of k is 6 (2/3).

**Question **10. Find the value of š, if the division of šš„^{3} + 9š„^{2} + 4š„ā 10 by š„ + 3 leaves a remainder 5.

Solution:

It is given that,

š(š„) = šš„^{3} + 9š„^{2} + 4š„ ā 10 = 5

š„ + 3 is a factor of š„^{3} + 2š„^{2} ā šš„ + 4

š„ + 3 = 0

š„ = ā3

Put the value of š„ in given equation,

šš„^{3} + 9š„^{2} + 4š„ ā 10 = 5

š(ā3)^{3} + 29 + 4(ā3) ā 10 = 5

ā27š + 9 Ć 9 ā 12 ā 10 = 5

ā27š + 81 ā 12 ā 10 = 5

ā27š + 54 = 0

ā27š = ā54

š = ā54/ā27

š = 2

Hence, the value of š is 2.

**Question **11. If š„^{3} + šš„^{2} + šš„ + 6 has š„ā 2 as a factor and leaves a remainder 3 when divided by š„ā 3, find the values of a and b.

Solution:

It is given that,

š(š„) = š„^{3} + šš„^{2} + šš„ + 6 = 0

š„ + 2 is a factor of š„^{3} + šš„^{2} + šš„ + 6

š„ + 2 = 0

š„ = ā2

Put the value of š„ in given equation,

(2)^{3} + š(2)^{2} + š(2) + 6 = 0

8 + š(4) + š(2) + 6 = 0

8 + 4š + 2š + 6 = 0

4š + 2š + 14 = 0

2(2š + š + 7) = 0

2š + š + 7 = 0 **_**(1)

š„ ā 3 is a factor of š„

^{3}+ šš„

^{2}+ šš„ + 6 = 3

š„ ā 3 = 0

š„ = 3

Put the value of š„ in given equation,

(3)3 + š(3)

^{2}+ š(3) + 6 = 3

27 + š(9) + š(3) + 6 = 3

27 + 9š + 3š + 6 = 3

9š + 3š + 33 ā 3 = 0

9š + 3š + 30 = 0

3(3š + š + 10) = 0

3š + š + 10 = 0

**(2)**

*_*From equation (1) we get,

2š + š + 7 = 0

š = ā7 ā 2š

**(3)**

**__**Put the value of b in equation (2)

3š + (ā7 ā 2š) + 10 = 0

3š ā 7 ā 2š + 10 = 0

3š ā 2š + 3 = 0

š = ā3

Put the value of a in equation (3)

š = ā7 ā 2(ā3)

š = ā7 + 6

š = ā1

Hence, the value of a is -3 and value of b is -1.

**Question **12. The expression 2š„^{3 }+ šš„^{2} + šš„ā 2 leaves remainder 7 and 0 when divided by 2š„ā 3 and š„ + 2 respectively. Calculate the values of a and b.

Solution:

It is given that,

š(š„) = 2š„^{3} + šš„^{2 }+ šš„ ā 2 = 0

2š„ ā 3 is a factor of 2š„^{3} + šš„^{2} + šš„ ā 2

2š„ ā 3 = 0

š„ = 3/2

Put the value of š„ in given equation,

27 + 9š + 6š = 9 Ć 4

27 + 9š + 6š = 36

9š + 6š = 36 ā 27

9š + 6š = 9

3(3š + 2š) = 9

3š + 2š = 9/3

3š + 2š = 3__________(1)

š„ + 2 is a factor of 2š„^{3} + šš„^{2} + šš„ ā 2 = 0

š„ + 2 = 0

š„ = ā2

Put the value of š„ in given equation,

2(ā2)^{3} + š(ā2)^{2} + š(ā2) ā 2 = 0

2 Ć (ā8) + š(4)^{2} + š(ā2) ā 2 = 0

ā16 + 4š ā 2š ā 2 = 0

4š ā 2š ā 18 = 0 ** _**(2)

**From equation (1) we get,**

3š + 2š = 3

š = 3ā2š/3 ___(3)

Put the value of a in equation (2)

4š ā 2š ā 18 = 0

**Question **13. What number should be added to 3š„^{3} ā 5š„^{2} + 6š„ so that when resulting polynomial is divided by š„ā 3, the remainder is 8?**Solution:**

Let us assumed that,

š¾ is added to given polynomial

So, the polynomial is,

š(š„) = 3š„^{3} ā 5š„^{2} + 6š„ + š

š„ ā 3 is a factor of 3š„^{3} ā 5š„^{2} + 6š„ + š and remainder is 8.

š„ ā 3 = 0

š„ = 3

Put the value of š„ in given equation,

3š„^{3} ā 5š„^{2} + 6š„ + š = 8

3(3)^{3 }ā 5(3)^{2} + 6(3) + š = 8

3(27) ā 5(9) + 6(3) + š = 8

81 ā 45 + 18 + š = 8

54 + š = 8

š = 8 ā 54

š = ā46

Hence, the value of k is -46.

**Question **14. What number should be subtracted from š„^{3} + 3š„^{2}ā 8š„ + 14 so that on dividing it with š„ā 2, the remainder is 10.

Solution:

Let us assumed that,

š is subtracted to given polynomial

So, the polynomial is,

š(š„) = š„^{3} + 3š„^{2} ā 8š„ + 14 ā š

š„ ā 2 is a factor of š„^{3} + 3š„^{2} ā 8š„ + 14 and remainder is 10.

š„ ā 2 = 0

š„ = 2

Put the value of š„ in given equation,

š„^{3} + 3š„^{2} ā 8š„ + 14 ā š = 10

(2)^{3} + 3(2)^{2} ā 8(2) + 14 ā š = 10

8 + 3(4) ā 8(2) + 14 ā š = 10

8 + 12 ā 16 + 14 ā š = 10

18 ā š = 10

āš = 10 ā 18

āš = ā8

š = 8

Hence, the value of k is 8.

**Question **15. The polynomials 2š„^{3}ā 7š„^{2} + šš„ā 6 and š„^{3}ā 8š„^{2} + (2š + 1)š„ā 16 leaves the same remainder when divided by š„ā 2. Find the value of āaā.

Solution:

It is given that,

š(š„) = 2š„^{3}ā 7š„^{2} + šš„ā 6

š„ ā 2 is a factor of 2š„^{3}ā 7š„^{2} + šš„ ā 6

š„ ā 2 = 0

š„ = 2

Put the value of š„ in given equation,

š(2) = 2(2)^{3}ā 7(2)^{2} + š(2)ā 6

š(2) = 2(8)ā 7(4) + š(2)ā 6

š(2) = 16ā 28 + 2šā 6

š(2) = 2šā 18

Again,

š(š„) = š„^{3}ā 8š„^{2} + (2š + 1)š„ā 16

š„ ā 2 is a factor of š„^{3}ā 8š„^{2} + (2š + 1)š„ ā 16

š„ ā 2 = 0

š„ = 2

Put the value of š„ in given equation,

š(2) = (2)^{3} ā 8(2)^{2} + (2š + 1)(2) ā 16

š(2) = 8 ā 8(4) + 4š + 2 ā 16

š(2) = 8 ā 32 + 4š + 2 ā 16

š(2) = 4š ā 38

It is given that,

Polynomial leaves the same remainder

š(2) = š(2)

2šā 18 = 4š ā 38

2š ā 4š = ā38 + 18

ā2š = ā20

š = 20/2

š = 10

Hence, the value of a is 10.

**Question **16. If (š„ā 2) is a factor of the expression 2š„^{3} + šš„^{2} + šš„ā 14 and when the expression is divided by (š„ā 3), it leaves a remainder 52, find the values of a and b.

Solution:

It is given that,

š(š„) = 2š„^{3} + šš„^{2} + šš„ā 14

š„ ā 2 is a factor of 2š„^{3} + šš„^{2} + šš„ā 14

š„ ā 2 = 0

š„ = 2

2(2)^{3 }+ š(2)^{2} + š(2)ā 14 = 0

16 + 4š + 2šā 14 = 0

4š + 2š + 2 = 0

2š + š + 1 = 0

2š + š = ā1_________(š)

(š„ ā 3) is a factor of 2š„^{3} + šš„^{2 }+ šš„ā 14 and remainder is 52,

2(3)^{3 +} š(3)^{2} + š(3)ā 14 = 52

54 + 9š + 3šā 14 = 52

9š + 3š + 40 = 52

9š + 3š = 12

3š + š = 4___________(šš)

From equation (1) we get the value of a,

2š + š = ā1

**Question **17. Find āaā if the two polynomials šš„^{3} + 3š„^{2}ā 9 and 2š„^{3} + 4š„ + š, leave the same remainder when divided by š„ + 3.

Solution:

It is given that,

Two polynomial have same remainder.

š„ + 3 = 0

š„ = ā3

Value of polynomial šš„^{3} + 3š„^{2}ā 9 at š„ = ā3 is same as value of polynomial 2š„^{3} + 4š„ + š šš” š„ = ā3

š(ā3)^{3} + 3(ā3)^{2} ā 9 = 2(ā3)^{3} + 4(ā3) + š

ā27š + 27 ā 9 = ā54 ā 12 + š

ā27š + 18 = ā66 + š

ā27š ā š = ā66 ā 18

ā28š = ā84

š = ā84/ā28

š = 3

Hence, the value of a is 3.

### Exercise 8B

**Question **1. Using the Factor Theorem, show that:

(i) (š„ā 2) is a factor of š„^{3}ā 2š„^{2}ā 9š„ + 18.

Hence, factories the expression š„^{3}ā 2š„^{2}ā 9š„ + 18 completely.

(ii) (š„ + 5) is a factor of 2š„^{3} + 5š„^{2} ā 28š„ā 15.

Hence, factories the expression 2š„^{3} + 5š„^{2} ā 28š„ā 15 completely.

(iii) (3š„ + 2) is a factor of 3š„^{3} + 2š„^{2}ā 3š„ā 2.

Hence, factories the expression 3š„^{3} + 2š„^{2}ā 3š„ā 2 completely.

Solution:

(i) It is given that,

(š„ā 2) is a factor of š„^{3}ā 2š„^{2} ā 9š„ + 18

š„ ā 2 = 0

š„ = 2

Put the value of š„ in given equation,

š(2) = (2)^{3}ā 2(2)^{2} + 9(2) + 18

š(2) = (8)ā 2(4) + 9(2) + 18

š(2) = 8ā 8 + 18 + 18

š(2) = 0

Now,

Factorization of š„^{3}ā 2š„^{2}ā 9š„ + 18 is

(š„ ā 2)(š„^{2} ā 9)

(š„ ā 2)(š„^{2} ā (3)2)

(š„ ā 2)(š„ ā 3)(š„ + 3)

Hence, the factors of š„^{3}ā 2š„^{2}ā 9š„ + 18 is (š„ ā 2)(š„ ā 3)(š„ + 3).

(ii) It is given that,

(š„ + 5) is a factor of 2š„^{3} + 5š„^{2} ā 28š„ā 15

š„ + 5 = 0

š„ = ā5

Put the value of š„ in given equation,

š(ā5) = 2(ā5)^{3} + 5(ā5)^{2} + 28(ā5) ā 15

š(ā5) = 2(ā125) + 5(25) + 28(ā5) ā 15

š(ā5) = ā250 + 125 + 140 ā 15

š(ā5) = ā265 + 265

š(ā5) = 0

Now,

Factorization of 2š„^{3 }+ 5š„^{2}ā 28š„ ā 15 is

(š„ + 5)(2š„^{2} ā 5š„ ā 3)

(š„ + 5)(2š„^{2} ā 6š„ + š„ ā 3)

(š„ + 5)[2š„(š„ ā 3) + 1(š„ ā 3)]

(š„ + 5)(2š„ + 1)(š„ ā 3)

Hence, the factors of š„^{3}ā 2š„^{2}ā 9š„ + 18 is (š„ + 5)(2š„ + 1)(š„ ā 3).

(iii) It is given that,

(3š„ + 2) is a factor of 3š„^{3} + 2š„^{2} ā 3š„ā 2

3š„ + 2 = 0

š„ = ā 2/3

Factorization of 3š„^{3 }+ 2š„^{3}ā 3š„ ā 2 is

(3š„ + 2)(š„^{3} ā 1)

(3š„ + 2)(š„ ā 1)(š„ + 1)

Hence, the factors of 3š„^{3} + 2š„^{3}ā 3š„ ā 2 is (3š„ + 2)(š„ ā 1)(š„ + 1).

**Question **2. Using the Remainder Theorem, factorise each of the following completely.

(š) 3š„^{3} + 2š„^{3} ā 19š„ + 6

(šš) 2š„^{3} + š„^{3}ā 13š„ + 6

(ššš) 3š„^{3} + 2š„^{3}ā 23š„ā 30

(šš£) 4š„^{3} + 7š„^{3} ā 36š„ ā 63

(š£) š„^{3} + š„^{3} ā 4š„ ā 4

Solution:

(š) 3š„^{3} + 2š„^{3} ā 19š„ + 6

Assumed that, 3š„^{3} + 2š„^{3} ā 19š„ + 6 is divided by š„ ā 2

š„ ā 2 = 0

š„ = 2

Put the value of š„ in given equation,

3š„^{3} + 2š„^{3} ā 19š„ + 6 = 0

3(2)^{3} + 2(2)^{2} ā 19(2) + 6 = 0

3(8) + 2(4) ā 19(2) + 6 = 0

24 + 8 ā 38 + 6 = 0

38 ā 38 = 0

0 = 0

Factorization of 3š„^{3} + 2š„^{2}ā 19š„ + 6 is

(š„ ā 2)(3š„^{2 }ā 8š„ ā 3)

(š„ ā 2)(3š„^{2} + 9š„ ā š„ ā 3)

(š„ ā 2)[3š„(š„ + 3) ā 1(š„ + 3)]

(š„ ā 2)(3š„ ā 1)(š„ + 3)

Hence, the factors of 3š„^{3} + 2š„^{2}ā 19š„ + 6 is (š„ ā 2)(3š„ ā 1)(š„ + 3).

(šš) 2š„^{3} + š„^{2} ā 13š„ + 6

Assumed that, 2š„^{3} + š„^{2}ā 13š„ + 6 is divided by š„ ā 2

š„ ā 2 = 0

š„ = 2

Put the value of š„ in given equation,

š(š„) = 2(2)^{3} + (2)^{2}ā 13(2) + 6

š(š„) = 2(8) + 4ā 26 + 6

š(š„) = 16 + 4ā 26 + 6

š(š„) =ā 26 + 26

š(š„) = 0

Factorization of 2š„^{3} + š„^{2} ā 13š„ + 6 is

(š„ ā 2)(2š„^{2} + 5š„ ā 3)

(š„ ā 2)(2š„^{2} + 6š„ ā š„ ā 3)

(š„ ā 2)[2š„(š„ + 3) ā 1(š„ + 3)]

(š„ ā 2)(2š„ ā 1)(š„ + 3)

Hence, the factors of 2š„^{3} + š„^{2}ā 13š„ + 6 is (š„ ā 2)(2š„ ā 1)(š„ + 3).

(ššš) 3š„^{3} + 2š„^{2}ā 23š„ā 30

Assumed that, 3š„^{3} + 2š„^{2}ā 23š„ + 30 is divided by š„ ā 2

š„ + 2 = 0

š„ = ā2

Put the value of š„ in given equation,

š(š„) = 3(ā2)^{3} + 2(ā2)^{2}ā 23(ā2) + 30

š(š„) = 3(ā8) + 2(4) + 46 + 30

š(š„) = 16 + 4ā 26 + 6

š(š„) =ā 26 + 26

š(š„) = 0

Factorization of 3š„^{3} + 2š„^{2}ā 23š„ ā 30 is

(š„ + 2)(3š„^{2} ā 4š„ ā 15)

(š„ + 2)(3š„^{2}+ 5š„ ā 9š„ ā 3)

(š„ + 2)[š„(3š„ + 5) ā 3(3š„ + 5)]

(š„ + 2)(3š„ + 5)(š„ ā 3)

Hence, the factors of 3š„^{3} + 2š„^{2}ā 23š„ ā 30 is (š„ + 2)(3š„ + 5)(š„ ā 3).

(šš£) 4š„^{3} + 7š„^{2} ā 36š„ ā 63

Assumed that, 3š„^{3} + 2š„^{2}ā 23š„ + 30 is divided by š„ ā 2

š„ ā 3 = 0

š„ = 3

Put the value of š„ in given equation,

š(š„) = 4(3)^{3} + 7(3)^{2} ā 36(3) ā 63

š(š„) = 4(27) + 7(9) ā 36(3) ā 63

š(š„) = 108 + 63ā 108 ā 63

š(š„) = 0

Factorization of 4š„^{3} + 7š„^{2}ā 36š„ ā 63 is

(š„ + 3)(4š„^{2} ā 5š„ ā 21)

(š„ + 3)(4š„^{2} ā 12š„ + 7š„ ā 21)

(š„ + 3)[4š„(š„ ā 3) + 7(š„ ā 3)]

(š„ + 3)(4š„ + 7)(š„ ā 3)

Hence, the factors of 4š„^{3} + 7š„^{2} ā 36š„ ā 63 is (š„ + 3)(4š„ + 7)(š„ ā 3).

(š£) š„^{3} + š„^{2} ā 4š„ ā 4

Assumed that, š„^{3} + š„^{2} ā 4š„ ā 4 is divided by š„ + 1

š„ + 1 = 0

š„ = ā1

Put the value of š„ in given equation,

š(ā1) = (ā1)^{3} + (ā1)^{2} ā 4(ā1) ā 4

š(ā1) = (ā1) + (1) ā 4(ā1) ā 4

š(ā1) = ā1 + 1 + 4 ā 4

š(š„) = 0

Factorization of š„^{3} + š„^{2} ā 4š„ ā 4 is

(š„ + 1)(š„^{2} ā 4)

(š„ + 1)(š„^{2} ā (2)2)

(š„ + 1)(š„ ā 2)(š„ + 3)

Hence, the factors of š„^{3} + š„^{2} ā 4š„ ā 4 is (š„ + 1)(š„ ā 2)(š„ + 3).

**Question **3. Using the Remainder Theorem, factories the expression 3š„^{3} + 10š„^{2} + š„ā 6. Hence, solve the equation 3š„^{3} + 10š„^{2} + š„ā 6 = 0.

Solution:

It is given that,

3š„^{3} + 10š„^{2} + š„ ā 6

Assumed that, 3š„^{3} + 10š„^{2} + š„ā 4 is divided by š„ + 1

š„ + 1 = 0

š„ = ā1

Put the value of š„ in given equation,

š(ā1) = 3(ā1)^{3} + 10(ā1)^{2} + (ā1) ā 6

š(ā1) = 3(ā1) + 10(1) ā 1 ā 6

š(ā1) = ā3 + 10 ā 1 ā 6

š(ā1) = 0

Factorization of 3š„^{3 }+ 10š„^{2} + š„ā 6 is

(š„ + 1)(3š„^{2} + 7š„ ā 6)

(š„ + 1)(3š„^{2} + 9š„ ā 2š„ ā 6)

(š„ + 1)[3š„(š„ + 3) ā 2(š„ + 3)]

(š„ + 1)(š„ + 3)(3š„ ā 2)

Hence, the factors of 3š„^{3} + 10š„^{2} + š„ ā 6 is (š„ + 1)(š„ + 3)(3š„ ā 2).

**Question **4. Factories the expression š(š„) = 2š„^{3} ā 7š„^{2} ā 3š„ + 18. Hence, find all possible values of x for which š(š„) = 0.

Solution:

It is given that,

2š„^{3} ā 7š„^{2} ā 3š„ + 18

Assumed that, 2š„^{3} ā 7š„^{2} ā 3š„ + 18 is divided by š„ ā 2

š„ ā 2 = 0

š„ = 2

Put the value of š„ in given equation,

š(2) = 2(2)^{3} ā 7(2)^{2} ā 3(2) + 18

š(2) = 2(8) + 7(4) ā 6 + 18

š(2) = 16 ā 28 ā 6 + 18

š(2) = 0

Factorization of 2š„^{3} ā 7š„^{2} ā 3š„ + 18 is

(š„ ā 2)(2š„^{2} ā 3š„ ā 9)

(š„ ā 2)(2š„^{2} ā 6š„ + 3š„ ā 9)

(š„ ā 2)[2š„(š„ ā 3) + 3(š„ ā 3)]

(š„ ā 2)(š„ ā 3)(2š„ + 3)

Hence, the factors of 2š„^{3} ā 7š„^{2} ā 3š„ + 18 is (š„ ā 2)(š„ ā 3)(2š„ + 3).

**Question **5. Given that š„ā 2 and š„ + 1 are factors of š(š„) = š„^{3 }+ 3š„^{2} + šš„ + š; calculate the values of a and b. Hence, find all the factors of š(š„).

Solution:

It is given that,

š„^{3 }+ 3š„^{2} + šš„ + š

Assumed that, š„^{3 }+ 3š„^{2} + šš„ + š is divided by š„ ā 2

š„ ā 2 = 0

š„ = 2

Put the value of š„ in given equation,

š(2) = (2)^{3} + 3(2)^{2} + š(2) + š

š(2) = 8 + 3(4) + 2š + š

š(2) = 8 + 12 + 2š + š

š(2) = 20 + 2š + š

2š + š + 20 = 0______________(i)

Assumed that, š„^{3 }+ 3š„^{2} + šš„ + š is divided by š„ + 1

š„ + 1 = 0

š„ = ā1

Put the value of š„ in given equation,

š(ā1) = (ā1)^{3} + 3(ā1)^{2} + š(ā1) + š

š(ā1) = ā1 + 3(1) ā š + š

š(ā1) = ā1 + 3 ā š + š

š(ā1) = 2 ā š + š

2 ā š + š = 0______________(ii)

From equation (i)

2š + š + 20 = 0

Factorization of š„^{3} + 3š„^{2} ā 6š„ ā 8 is

(š„ + 1)(š„^{2} + 2š„ ā 8)

(š„ + 1)(š„^{2} + 4š„ ā 2š„ ā 8)

(š„ + 1)[š„(š„ + 4) ā 2(š„ + 4)]

(š„ + 1)(š„ ā 2)(š„ + 4)

Hence, the factors of š„^{3} + 3š„^{2} ā 6š„ ā 8 is (š„ + 1)(š„ ā 2)(š„ + 4).

**Question **6. The expression 4š„^{3}ā šš„^{2} + š„ā š leaves remainders 0 and 30 when divided by š„ + 1 and 2š„ā 3 respectively. Calculate the values of b and c. Hence, factorize the expression completely.

Solution:

It is given that,

4š„^{3} ā šš„^{2} + š„ ā š is divided by š„ + 1

š„ + 1 = 0

š„ = ā1

Put the value of š„ in given equation,

š(ā1) = 4(ā1)^{3} ā š(ā1)^{2} + (ā1) ā š

š(ā1) = 4(ā1) ā š(1) ā 1 ā š

š(ā1) = ā4 ā š ā 1 ā š

š(ā1) = 5 + š + š

5 + š + š = 0______________(i)

Also,

4š„^{3} ā šš„^{2} + š„ ā š is divided by 2š„ ā 3 and leaves remainder 30.

2š„ ā 3 = 0

54 ā 9š + 6 ā 4š = 30 Ć 4

60 ā 9š ā 4š = 120

60 ā 9š ā 4š ā 120 = 0

ā9š ā 4š ā 60 = 0

ā(9š + 4š + 60) = 0

9š + 4š + 60 = 0______________(ii)

From equation (1) we get the value of š

5 + š + š = 0

š = ā5 ā š**_**(iii)

Put the value of c in equation in equation (ii)

9š + 4š + 60 = 0

9š + 4(ā5 ā š) + 60 = 0

9š ā 20 ā 4š + 60 = 0

5š + 40 = 0

š = ā 40/5

š = ā8

Put the value of b in equation (iii)

š = ā5 ā (ā8)

š = ā5 + 8

š = 3

So, the required equation is 4š„

^{3}+ 8š„

^{2}+ š„ ā 3

Factorization of 4š„^{3} + 8š„^{2} + š„ ā 3 is

(š„ + 1)(4š„^{2} + 4š„ ā 3)

(š„ + 1)(4š„^{2} + 6š„ ā 2š„ ā 3)

(š„ + 1)[2š„(2š„ + 3) ā (2š„ + 3)]

(š„ + 1)(2š„ ā 1)(2š„ + 3)

Hence, the factors of 4š„^{3} + 8š„^{2} + š„ ā 3 is (š„ + 1)(2š„ ā 1)(2š„ + 3).

**Question **7. If š„ + š is a common factor of expressions š(š„) = š„^{2} + šš„ + š and š(š„) = š„^{2} + šš„ + š; Show that: š = šāš/šāš

Solution:

It is given that,

š(š„) = š„^{2} + šš„ + š

š„^{2} + šš„ + š is divided by š„ + š

š„ + š = 0

š„ = āš

(āš)^{2} + š(āš) + š = 0

š^{2} ā šš + š = 0

š^{2} = šš ā š(1)

š(š„) = š„^{2} + šš„ + š

š„^{2} + šš„ + š is divided by š„ + š

š„ + š = 0

š„ = āš

(āš)^{2 }+ š(āš) + š = 0

š^{2} ā šš + š = 0

š^{2} = šš ā š(1)

From equation (1) and (2) we get,

šš ā š = šš ā š

š ā š = šš ā šš

š ā š = š(š ā š)

š = šāš/šāš

Hence proved.

**Question **8. The polynomials šš„^{3} + 3š„^{2}ā 3 and 2š„^{2}ā 5š„ + š, when divided by š„ā 4, leave the same remainder in each case. Find the value of a.

Solution:

Let us assumed that,

š(š„) = šš„^{3 }+ 3š„^{2}ā 3

šš„^{3} + 3š„^{2} ā 3 is divided by š„ā 4

š(4) = š(4)^{3} + 3(4)^{2} ā 3

š(4) = š(64) + 3(16)ā 3

š(4) = 64š + 45_________(1)

Let us assumed that,

š(š„) = 2š„^{3}ā 5š„ + š

2š„^{3}ā 5š„ + š is divided by š„ā 4

š(4) = 2(4)^{3}ā 5(4) + š

š(4) = 2(64)ā 20 + š

š(4) = 128ā 20 + š

š(4) = š + 108_________(2)

From equation (1) and (2)

64š + 45 = š + 108

64š ā š = 108 ā 45

63š = 63

š = 1

Hence, the value of š is 1.

**Question **9. Find the value of āaā, if (š„ā š) is a factor of š„^{3}ā šš„^{2} + š„ + 2.

Solution:

Let us assumed that,

š(š„) = š„^{3}ā šš„^{2} + š„ + 2

š„^{3}ā šš„^{2} + š„ + 2 is divided by š„ā š

š„ + š = 0

š„ = āš

Put the value of š„ in given equation,

(š)^{3} + š(š)^{2} + š + 2 = 0

š^{3} ā š^{3} + š + 2 = 0

š + 2 = 0

š = ā2

Hence, the value of š is ā2.

**Question **10. Find the number that must be subtracted from the polynomial 3š¦^{3} + š¦^{2}ā 22š¦ + 15, so that the resulting polynomial is completely divisible by š¦ + 3.

Solution:

Let us assumed that,

The number to be subtracted from the given polynomial be k.

š(š¦) = 3š¦^{3} + š¦^{2}ā 22š¦ + 15

3š¦^{3} + š¦^{2}ā 22š¦ + 15 is divisible by (š¦ + 3).

š¦ + 3 = 0

š¦ = ā3

Put the value of š¦ is ā3.

3(ā3)3 + (ā3)2ā 22(ā3) + 15ā k = 0

3(ā27) + 9 + 66 + 15ā k = 0

ā81 + 9 + 66 + 15 ā k = 0

9 ā k = 0

k = 9

Hence, the number to be subtracted from the polynomial be 9.

### Exercise 8C

**Question **1. Show that (š„ā 1) is a factor of š„^{3}ā 7š„^{2} + 14š„ā 8. Hence, completely factorize the given expression.

Solution:

It is given that,

š„^{3} ā 7š„^{2} + 14š„ ā 8

Assumed that, š„^{3} ā 7š„^{2} + 14š„ ā 8 is divided by š„ ā 1

š„ ā 1 = 0

š„ = 1

Put the value of š„ in given equation,

š(1) = (1)^{3} ā 7(1)^{2} + 14(1) ā 8

š(1) = 1 ā 7 ā 14 ā 8

š(1) = 0

š„^{3} ā 7š„^{2} + 14š„ ā 8 is divided by š„ ā 1

Factorization of š„^{3} ā 7š„^{2} + 14š„ ā 8 is

(š„ ā 1)(š„^{2} ā 6š„ + 8)

(š„ ā 1)(š„^{2} ā 2š„ ā 4š„ + 8)

(š„ ā 1)[š„(š„ ā 2) ā 4(š„ ā 2)]

(š„ ā 1)(š„ ā 2)(š„ ā 4)

Hence, the factors of š„^{3} ā 7š„^{2} + 14š„ ā 8 is (š„ ā 1)(š„ ā 2)(š„ ā 4).

**Question **2. Using Remainder Theorem, factorize: š„^{3} + 10š„^{2}**ā 37š„ + 26 completely.Solution:**

It is given that,

š„

^{3}+ 10š„

^{2}ā 37š„ + 26

Assumed that, š„

^{3}+ 10š„

^{2 }ā 37š„ + 26 is divided by š„ ā 1

š„ ā 1 = 0

š„ = 1

Put the value of š„ in given equation,

š(1) = (1)

^{3}+ 10(1)

^{2}ā 37(1) + 26

š(1) = 1 + 10 ā 37 + 26

š(1) = 37 ā 37

š(1) = 0

š„

^{3}+ 10š„

^{2}ā 37š„ + 26 is divided by š„ ā 1

Factorization of š„^{3} + 10š„^{2} ā 37š„ + 26 is

(š„ ā 1)(š„^{2} + 11š„ ā 26)

(š„ ā 1)(š„^{2} + 13š„ ā 2š„ ā 26)

(š„ ā 1)[š„(š„ + 13) ā 2(š„ + 13)]

(š„ ā 1)(š„ ā 2)(š„ + 13)

Hence, the factors of š„^{3} + 10š„^{2} ā 37š„ + 26 is (š„ ā 1)(š„ ā 2)(š„ + 13).

**Question **3. When š„^{3} + 3š„^{2}ā šš„ + 4 is divided by š„ā 2, the remainder is š + 3. Find the value of m.

Solution:

Let us assumed that,

š(š„) = š„^{3} + 3š„^{2}ā šš„ + 4

Assumed that, š„^{3 }+ 3š„^{2}ā šš„ + 4 is divided by š„ ā 1 and remainder is š + 3

š„ ā 2 = 0

š„ = 2

Put the value of š„ in given equation,

(2)^{3} + 3(2)^{2}

ā š(2) + 4 = š + 3

8 + 12ā 2š + 4 = š + 3

24ā 3 = š + 2š

3š = 21

š = 7

Hence, the value of š is 7.

**Question **4. What should be subtracted from 3š„^{3}ā 8š„^{2} + 4š„ā 3, so that the resulting expression has š„ + 2 as a factor?

Solution:

Let us assumed that,

The required number be k.

š(š„) = 3š„^{3}ā 8š„^{2} + 4š„ā 3

Assumed that, š„^{3} + 3š„^{2}ā šš„ + 4 is divided by š„ + 2

š„ + 2 = 0

š„ = ā2

Put the value of š„ in given equation,

3(ā2)^{3}ā 8(ā2)^{2} + 4(ā2)ā 3ā š = 0

3(ā8)ā 8(4) + 4(ā2)ā 3ā š = 0

ā24 ā 32ā 8ā 3ā š = 0

ā67ā š = 0

š = ā67

Hence, the required number is -67.

**Question **5. If (š„ + 1) and (š„ā 2) are factors of š„^{3} + (š + 1)š„^{2}ā (šā 2)š„ā 6, find the values of a and b. And then, factorize the given expression completely.

Solution:

Let us assumed that,

š(š„) = š„^{2} + (š + 1)š„^{2}ā (šā 2)š„ā 6

Given that, š„^{2} + (š + 1)š„^{2}ā (šā 2)š„ā 6 is divided by š„ + 1

š„ + 1 = 0

š„ = ā1

Put the value of š„ in given equation,

(ā1)^{3} + (š + 1)(ā1)^{2} ā (šā 2)(ā1)ā 6 = 0

ā1 + (š + 1) + (šā 2)ā 6 = 0

š + šā 8 = 0__________(š)

Given that, š„^{3} + (š + 1)š„^{2}ā (šā 2)š„ā 6 is divided by š„ + 2

š„ + 2 = 0

š„ = ā2

Put the value of š„ in given equation,

(2)^{3} + (š + 1)(2)^{2} ā (šā 2)(2)ā 6 = 0

8 + 4š + 4ā 2š + 4ā 6 = 0

4šā 2š + 10 = 0

2šā š + 5 = 0___________(šš)

From equation (i) we get,

š = 8 ā š** __**(ššš)

Put the value of a in equation (ii)

2(8 ā š)ā š + 5 = 0

16 ā 2šā š + 5 = 0

ā2šā š = ā5 ā 16

ā3š = ā21

š = ā21/ā3

š = 7

Put the value of b in equation (iii)

š = 8 ā š

š = 8 ā 7

š = 1

The required equation is,

š(š„) = š„

^{3}+ (1 + 1)š„

^{2}ā (7ā 2)š„ā 6

š(š„) = š„

^{3}+ 2š„

^{2}ā 5š„ā 6

Hence, it is proved that (š„ + 1)(š„ā 2) are factor of š„

^{3 }+ 2š„

^{2}ā 5š„ā 6.

Factorization of š„^{3} + 10š„^{2} ā 37š„ + 26 is

(š„ ā 1)(š„ ā 2)(š„ + 3)

Hence, the factors of š„^{3} + 2š„^{2} ā 5š„ ā 6 is (š„ + 1)(š„ ā 2)(š„ + 3).

**Question **6. If š„ā 2 is a factor of š„^{2} + šš„ + š and a + b = 1, find the values of a and b.

Solution:

It is given that,

š + š = 1_______(š)

Also,

Given that, š„^{2} + šš„ + š is divided by š„ ā 2

š„ ā 2 = 0

š„ = 2

Put the value of š„ in given equation,

(2)^{2 }+ š(2) + š = 0

4 + 2š + š = 0

2š + š = ā4 **_**(šš)

From equation (i),

š + š = 1

š = 1 ā š(ššš)

Put the value of a in equation (ii),

2(1 ā š) + š = ā4

2 ā 2š + š = ā4

āš = ā4 ā 2

āš = ā6

š = 6

Put the value of a in equation (iii),

š = 1 ā š

š = 1 ā 6

š = ā5

Hence, the value of š is -5 and š is 6

**Question **7. Factorise š„^{3} + 6š„^{2} + 11š„ + 6 completely using factor theorem.

Solution:

It is given that,

š„^{3} + 6š„^{2} + 11š„ + 6

Assumed that, š„^{3} + 6š„^{2} + 11š„ + 6 is divided by š„ + 1

š„ + 1 = 0

š„ = ā1

Put the value of š„ in given equation,

š(ā1) = (ā1)^{3} + 6(ā1)^{2} + 11(ā1) + 6

š(ā1) = ā1 + 6 ā 11 + 6

š(ā1) = 12 ā 12

š(ā1) = 0

š„^{3} + 6š„^{2} + 11š„ + 6 is divided by š„ + 1

Factorization of š„^{3} + 6š„^{2} + 11š„ + 6 is

(š„ + 1)(š„^{2} + 5š„ + 6)

(š„ + 1)(š„^{2} + 2š„ + 3š„ + 6)

(š„ + 1)[š„(š„ + 2) + 3(š„ + 2)]

(š„ + 1)(š„ + 2)(š„ + 3)

Hence, the factors of š„^{3} + 6š„^{2} + 11š„ + 6 is (š„ + 1)(š„ + 2)(š„ + 3).

**Question **8. Find the value of āšā, if šš„^{2} + 2š„^{2}ā 3 and š„^{2}ā šš„ + 4 leave the same remainder when each is divided by š„ā 2.

Solution:

Let us assumed that,

š(š„) = šš„^{2} + 2š„^{2}ā 3

š(š„) = š„^{2}ā šš„ + 4

According to the question,

š(š„) and š(š„) leave the same remainder if divided by (š„ā 2).

š„ā 2 = 0

š„ = 2

We have,

š(2) = š(2)

š(2)^{3} + 2(2)^{2}ā 3 = (2)^{2}ā š(2) + 4

š(8) + 2(4)ā 3 = 4ā 2š + 4

8š + 8ā 3 = 4ā 2š + 4

8š + 2š = 4 + 4 ā 8 + 3

10š = 3

š = 3/10

Hence, the value of š is 3/10.

**Question **9. The polynomial šš„^{3} + 4š„^{2}ā 3š„ + š is completely divisible by š„^{2}ā 1; find the values of p and q. Also, for these values of p and q factorize the given polynomial completely.

Solution:

Let us assumed that,

š(š„) = šš„^{3} + 4š„^{2}ā 3š„ + š

According to the question,

šš„^{3} + 4š„^{2}ā 3š„ + š is divisible by (š„^{2}ā 1) = (š„ + 1)(š„ā 1).

š(1) = 0 ššš š(ā1) = 0

š(1) = š(1)^{3} + 4(1)^{2}ā 3(1) + š = 0

š(1) = š + 4ā 3 + š = 0

š + š + 1 = 0 (š)

From (š„ + 1) we get,

š(ā1) = š(ā1)^{3} + 4(ā1)^{2}ā 3(ā1) + š = 0

š(ā1) = š(ā1) + 4(1)ā 3(ā1) + š = 0

š(ā1) = āš + 4(1) + 3 + š = 0

āš + š + 7 = 0__________(šš)

From equation (i) we get,

š + š + 1 = 0

š = ā1 ā š**__**(ššš)

Put the value of p in equation (ii)

āš + š + 7 = 0

ā(ā1 ā š) + š + 7 = 0

1 + š + š + 7 = 0

2š + 8 = 0

š = ā4

Put the value of q in equation (iii)

š = ā1 ā š

š = ā1 ā (ā4)

š = ā1 + 4

š = 3

Hence, the required equation is š(š„) = 3š„^{3} + 4š„^{2}ā 3š„ā 4.

Factorization of 3š„^{3} + 4š„^{2}ā 3š„ā 4 is

(š„^{2} ā 1)(3š„ + 4)

(š„ ā 1)(š„ + 1)(3š„ + 4)

Hence, the factors of 3š„^{3} + 4š„^{2}ā 3š„ā 4 is (š„ ā 1)(š„ + 1)(3š„ + 4).

**Question **10. Find the number which should be added to š„^{2} + š„ + 3 so that the resulting polynomial is completely divisible by (š„ + 3).

Solution:

Let us assumed that, the required number be k.

š(š„) = š„^{2} + š„ + 3 + š

According to the question,

š„^{2} + š„ + 3 + š is divisible by (š„ + 3) and remainder is 0.

š„ + 3 = 0

š„ = ā3

Put the value of x in given equation,

(ā3)2 + (ā3) + 3 + š = 0

9ā 3 + 3 + š = 0

9 + š = 0

š = ā9

Hence, the required number is -9.

**Question **11. When the polynomial š„^{3} + 2š„^{2}ā 5šš„ā 7 is divided by (š„ā 1), the remainder is A and when the polynomial š„^{3} + šš„^{2}ā 12š„ + 16 is divided by (š„ + 2), the remainder is B. Find the value of āaā if 2š“ + šµ = 0.

Solution:

According to question,

š„^{3} + 2š„^{2}ā 5šš„ā 7 is divided by (š„ā 1), the remainder is A.

š„ā 1 = 0

š„ = 1

Put the value of š„ in given equation,

š„^{3} + 2š„^{2}ā 5šš„ā 7 = š“

1 + 2ā 5š ā 7 = š“

ā 5šā 4 = š“**____**(š)

š„

^{3}+ šš„

^{2}ā 12š„ + 16 is divided by (š„ + 2), the remainder is B.

š„ + 2 = 0

š„ = ā2

Put the value of š„ in given equation,

š„

^{3}+ šš„

^{2}ā 12š„ + 16 = šµ

(ā2)3 + š(ā2)

^{2}ā 12(ā2) + 16 = šµ

ā8 + 4š + 24 + 16 = šµ

4š + 32 = šµ

**(šš)**

*_*It is given that 2A + B = 0

From equation (i) and (ii), we get,

2(ā5šā 4) + 4š + 32 = 0

ā10šā 8 + 4š + 32 = 0

ā6š + 24 = 0

6š = 24

š = 4

Hence, the value of š is 4.

**Question **12. (3š„ + 5) is a factor of the polynomial (šā 1)š„^{3} + (š + 1)š„^{2}ā (2š + 1)š„ā 15. Find the value of āaā, factorize the given polynomial completely.

Solution:

Let us assumed that,

š(š„) = (š ā 1)š„^{3 }+ (š + 1)š„^{2} ā (2š + 1)š„ ā 15

According to question,

(3š„ + 5) is a factor of š(š„) and remainder = 0

ā125(š ā 1) + 75(š + 1) + 45(2š + 1) ā 405 = 0

40š ā 160 = 0

40š = 160

š = 4

š(š„) = (š ā 1)š„^{3} + (š + 1)š„^{2 }ā (2š + 1)š„ ā 15

š(š„) = (4 ā 1)š„^{3} + (4 + 1)š„^{2} ā (2(4) + 1)š„ ā 15

š(š„) = 3š„^{3 }+ 5š„^{2} ā 9š„ ā 15

Factorization of 3š„^{3} + 5š„^{2}ā 9š„ā 15 is

(3š„ + 5)(š„^{2} ā 3)

(3š„ + 5)(š„ + ā3)(š„ ā ā3)

Hence, the factors of 3š„^{3} + 5š„^{2}ā 9š„ā 15 is (3š„ + 5)(š„ + ā3)(š„ ā ā3).

**Question **13. When divided by š„ā 3 the polynomials š„^{3}ā šš„^{2} + š„ + 6 and 2š„^{3}ā **š„ ^{2}**ā (š + 3)š„ā 6 leave the same remainder. Find the value of āpā.

Solution:

It is given that,

(š„ā 3) is factor of š(š„) = š„

^{3}ā šš„

^{2}+ š„ + 6,

š(3) = (3)

^{3}ā š(3)

^{2}+ 3 + 6

š(3) = 27ā 9š + 3 + 6

š(3) = 36ā 9š

(š„ ā 3) is factor of š(š„) = 2š„

^{3}ā š„

^{2}+ (š + 3) ā 6

š(3) = 2(3)

^{3}ā (3)

^{2}ā (š + 3)(3)ā 6

š(3) = 2(27)ā 9ā (š + 3)(3)ā 6

š(3) = 54ā 9ā 3š + 9ā 6

š(3) = 30 ā 3š

It is also given that both the sides have equal remainder,

š(3) = š(3)

36ā 9š = 30ā 3š

6š = ā6

š = 1

Hence, the value of š is 1.

**Question **14. Use the Remainder Theorem to factorise the following expression: 2š„^{3} + š„^{2}ā 13š„ + 6.

Solution:

Let us assumed that,

š(š„) = 2š„^{3} + š„^{2} ā 13š„ + 6

By Remainder theorem,

Put the š„ = 2, we get,

š(2) = 2(2)^{3} + (2)^{2} ā 13(2) + 6

š(2) = 2(8) + 4 ā 26 + 6

š(2) = 16 + 4 ā 26 + 6

š(2) = 0

Factorization of 2š„^{2} + 5š„ ā 3 is

(š„ ā 2)(2š„^{2} + 6š„ ā š„ ā 3)

(š„ ā 2)[2š„(š„ + 3) ā 1(š„ + 3)]

(š„ ā 2)(2š„ ā 1)(š„ + 3)

Hence, the factors of 2š„^{3} + š„^{2}ā 13š„ + 6 is (š„ ā 2)(2š„ ā 1)(š„ + 3).

**Question **15. Using remainder theorem, find the value of k if on dividing 2š„^{3} + 3š„^{2}ā šš„ + 5 by š„ā 2, leaves a remainder 7.

Solution:

Let us assumed that,

š(š„) = 2š„^{3} + 3š„^{2}ā šš„ + 5

By Remainder Theorem,

š(2) = 7

2(2)^{3 }+ 3(2)^{2}ā š(2) + 5 = 7

2(8) + 3(4) ā 2š + 5 = 7

16 + 12 ā 2š + 5 = 7

33 ā 2š = 7

2š = 26

š = 13

Hence, the value of š is 13.

**Question **16. What must be subtracted from 16š„^{3} ā 8š„^{2} + 4š„ + 7 so that the resulting expression has 2š„ + 1 as a factor?

Solution:

Let us assumed that,

š(š„) = 16š„^{3}ā 8š„^{2} + 4š„ + 7

It is given that 2š„ + 1 is a factor of 16š„^{3}ā 8š„^{2} + 4š„ + 7.

2š„ + 1 = 0

2š„ = ā1

š„ = ā 1/2

Put the value š„ in given equation,