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„2ā 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„2ā 5š„ā 6
š„ + 1 = 0
š„ = ā1
Put the value of š„ in given equation,
š(š„) = 2š„3 + 3š„2ā 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„2ā 5š„ā 6.
(ii) It is given that,
2š„ ā 1 is the factor of 2š„3 + 3š„2ā 5š„ā 6
2š„ ā 1 = 0
š„ = 1/2
(iii) It is given that,
š„ + 2 is the factor of 2š„3 + 3š„2ā 5š„ā 6
š„ + 2 = 0
š„ = ā2
Put the value of š„ in given equation,
š(š„) = 2š„3 + 3š„2ā 5š„ā 6
š (ā2) = 2(ā2)3 + 3(ā2)2ā 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„5ā 6š„4ā 2šš„3 + 6šš„2 + 4šš„ + 8.
Solution:
It is given that,
š(š„) = 2š„5ā 6š„4ā 2šš„3 + 6šš„2 + 4šš„ + 8 = 0
š„ā 2 is a factor of 2š„5ā 6š„4ā 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„2ā 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„2ā 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„3ā 7š„2 + šš„ā 6 and š„3ā 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„3ā 7š„2 + šš„ā 6
š„ ā 2 is a factor of 2š„3ā 7š„2 + šš„ ā 6
š„ ā 2 = 0
š„ = 2
Put the value of š„ in given equation,
š(2) = 2(2)3ā 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„3ā 8š„2 + (2š + 1)š„ā 16
š„ ā 2 is a factor of š„3ā 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„2ā 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„2ā 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„3ā 2š„2ā 9š„ + 18.
Hence, factories the expression š„3ā 2š„2ā 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„2ā 3š„ā 2.
Hence, factories the expression 3š„3 + 2š„2ā 3š„ā 2 completely.
Solution:
(i) It is given that,
(š„ā 2) is a factor of š„3ā 2š„2 ā 9š„ + 18
š„ ā 2 = 0
š„ = 2
Put the value of š„ in given equation,
š(2) = (2)3ā 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„3ā 2š„2ā 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„3ā 2š„2ā 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„2ā 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„3ā 2š„2ā 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„3ā 3š„ ā 2 is
(3š„ + 2)(š„3 ā 1)
(3š„ + 2)(š„ ā 1)(š„ + 1)
Hence, the factors of 3š„3 + 2š„3ā 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„3ā 13š„ + 6
(ššš) 3š„3 + 2š„3ā 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„2ā 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„2ā 19š„ + 6 is (š„ ā 2)(3š„ ā 1)(š„ + 3).
(šš) 2š„3 + š„2 ā 13š„ + 6
Assumed that, 2š„3 + š„2ā 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)2ā 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„2ā 13š„ + 6 is (š„ ā 2)(2š„ ā 1)(š„ + 3).
(ššš) 3š„3 + 2š„2ā 23š„ā 30
Assumed that, 3š„3 + 2š„2ā 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)2ā 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„2ā 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„2ā 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„2ā 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„2ā 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„3ā 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
š2 ā šš + š = 0
š2 = 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
š2 ā šš + š = 0
š2 = 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„2ā 3 and 2š„2ā 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„2ā 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„3ā 5š„ + š
2š„3ā 5š„ + š is divided by š„ā 4
š(4) = 2(4)3ā 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„3ā šš„2 + š„ + 2.
Solution:
Let us assumed that,
š(š„) = š„3ā šš„2 + š„ + 2
š„3ā 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
š3 ā š3 + 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¦2ā 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¦2ā 22š¦ + 15
3š¦3 + š¦2ā 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„3ā 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„2ā 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„2ā šš„ + 4
Assumed that, š„3 + 3š„2ā 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„3ā 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„3ā 8š„2 + 4š„ā 3
Assumed that, š„3 + 3š„2ā šš„ + 4 is divided by š„ + 2
š„ + 2 = 0
š„ = ā2
Put the value of š„ in given equation,
3(ā2)3ā 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„2ā (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„2ā (šā 2)š„ā 6
Given that, š„2 + (š + 1)š„2ā (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„2ā (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„2ā 5š„ā 6
Hence, it is proved that (š„ + 1)(š„ā 2) are factor of š„3 + 2š„2ā 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„2ā 3 and š„2ā šš„ + 4 leave the same remainder when each is divided by š„ā 2.
Solution:
Let us assumed that,
š(š„) = šš„2 + 2š„2ā 3
š(š„) = š„2ā 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)2ā 3 = (2)2ā 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„2ā 3š„ + š is completely divisible by š„2ā 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„2ā 3š„ + š
According to the question,
šš„3 + 4š„2ā 3š„ + š is divisible by (š„2ā 1) = (š„ + 1)(š„ā 1).
š(1) = 0 ššš š(ā1) = 0
š(1) = š(1)3 + 4(1)2ā 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)2ā 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„2ā 3š„ā 4.
Factorization of 3š„3 + 4š„2ā 3š„ā 4 is
(š„2 ā 1)(3š„ + 4)
(š„ ā 1)(š„ + 1)(3š„ + 4)
Hence, the factors of 3š„3 + 4š„2ā 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„2ā 5šš„ā 7 is divided by (š„ā 1), the remainder is A and when the polynomial š„3 + šš„2ā 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„2ā 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„2ā 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)2ā 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„2ā (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„2ā 9š„ā 15 is
(3š„ + 5)(š„2 ā 3)
(3š„ + 5)(š„ + ā3)(š„ ā ā3)
Hence, the factors of 3š„3 + 5š„2ā 9š„ā 15 is (3š„ + 5)(š„ + ā3)(š„ ā ā3).
Question 13. When divided by š„ā 3 the polynomials š„3ā šš„2 + š„ + 6 and 2š„3ā š„2ā (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„3ā šš„2 + š„ + 6,
š(3) = (3)3ā š(3)2 + 3 + 6
š(3) = 27ā 9š + 3 + 6
š(3) = 36ā 9š
(š„ ā 3) is factor of š(š„) = 2š„3ā š„2 + (š + 3) ā 6
š(3) = 2(3)3ā (3)2ā (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„2ā 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„2ā 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„2ā šš„ + 5 by š„ā 2, leaves a remainder 7.
Solution:
Let us assumed that,
š(š„) = 2š„3 + 3š„2ā šš„ + 5
By Remainder Theorem,
š(2) = 7
2(2)3 + 3(2)2ā 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„3ā 8š„2 + 4š„ + 7
It is given that 2š„ + 1 is a factor of 16š„3ā 8š„2 + 4š„ + 7.
2š„ + 1 = 0
2š„ = ā1
š„ = ā 1/2
Put the value š„ in given equation,
