Selina Concise Class 10 Math Chapter 8 Remainder and Factor Theorems Solutions
Exercise – 8B
(Q1) Solution:-
(i) (x – 2) is a factor of x3 – 2x2 – 9x + 18. Hence, factorise the expression x3 – 2x2 – 9x + 18 completely.
(ii) (x + 5) is a factor of 2x3 + 5x2 – 28x – 15. Hence, factorise the expression 2x3 + 5x2 – 28x – 15 completely.
(iii) (3x + 2) is a factor of 3x3 + 2x2 – 3x – 2. Hence, factorise the expression 3x3 + 2x2 – 3x – 2 completely.
= (i) Given that, (x- 2) is a factor of x3 – 2x2 – 9x + 18
∴ x – 2 = 0
x = 2
Let, f(x) = x3 – 2x2 – 9x + 18
Put x = 2
f(2) = (2)3 – 2(2)2 – 9(2) + 18
= 8 – 2(4) – 18 + 18
= 8 – 8 – 18 + 18
= 0
∴ x – 2 is a factor of f(x)
Now, factorise the given polynomial and division of polynomial.
Remainder = 0
Quotient = x2 – 9
∴ x2 – 9 = 0
x2 = 9
x = ± 3
Or by using,
Dividend = divisor × Quotient + Remainder
x3 – 2x2 – 9x + 18 = (x – 2) × (x2 – 9) + 0
= (x – 2) (x + 3) (x – 3)
(ii) Given that
(x + 5) is a factor of 2x3 + 5x2 – 28x – 15.
∴ x + 5 = 0
x = -5
Let, f(x) = 2x3 + 5x2 – 28x – 15
Put x = -5
f(-5) = 2 (-5)3 + 5 (-5)2 – 28 (-5) – 15
= 2 (-125) + 5 (25) + 140 – 15
= – 250 + 125 + 140 – 15
= – 250 + 140 – 15 + 125
f(-5) = -265 + 265
f(-5) = 0
∴ x + 5 is a factor of given polynomial f(x).
Now, factorise the given polynomial and division of polynomial.
Or by using,
Dividend = divisor × Quotient + Remainder
2x3 + 5x2 – 28x – 15 = (x + 5) (2x2 – 5x – 3) + 0
= (x + 5) (2x2 – 5x – 3)
Factorise 2x2 – 5x – 3 = 0
2x2 – 3x – 2x – 3 = 0
x (2x – 3) + 1 (2x – 3) = 0
(2x – 3) (x + 1) = 0
∴ (x + 5) (2x – 3) (x + 1)
∴ The factor of f(x) is (x + 5) (2x – 3) (x + 1)
(iii) Given that,
(3x + 2) is a factor of 3x2 + 2x2 – 3x – 2
∴ 3x + 2 = 0
3x = – 2
x = -2/3
Let, f(x) = 3x3 + 2x2 – 3x – 2
f(-2/3) = 3 (-2/3)3 + 2 (-2/3)2 – 3 (-2/3) – 2
= 3 (-8/27) + 2 (4/9) + 6/3 – 2
= -8/9 + 8/9 + 2 – 2
f(-2/3) = 0
∴ 3x + 2 is a factor of f(x)
Now, factorise the given polynomial and division of polynomial.
Remainder = 0
Quotient = x2 – 1
Or by using
Dividend = divisor × Quotient + remainder
3x3 + 2x2 – 3x – 2 = (3x + 2) × (x2 – 1) + 0
= (3x + 2) (x + 1) (x – 1)
(Q2) Solution:-
(i) 3x3 + 2x2 – 19x + 6
= Let f(x) = 3x3 + 2x2 – 19x + 6
Put x = 0
f(0) = 3(0)3 + 2(0) – 19(0) + 6
= 0 + 0 – 0 + 6
f(0) = 6
f(0) is not factor of f(x)
put x = 1
f(1) = 3(1)3 + 2(1)2 – 19(1) + 6
= 3 + 2 – 19 + 6
= 6 – 19 + 6
= 12 – 19
f(1) = -7
x – 1 is not factor of f(x) Put x = 2
f(2) = 3(2)3 + 2 (2)2 – 19(2) + 6
= 3 (8) + 2(4) – 38 + 6
= 24 + 8 – 38 + 6
f(2) = 32 – 38 + 6
= 38 – 38
f(2) = 0
∴ x – 2 is a factor of f(x)
Now factorise the given polynomial and division of polynomial.
Remainder = 0
Quotient = 3x2 + 8x – 3
By using,
Dividend = divisor × Quotient + Remainder
3x3 + 2x2 – 19x + 6 = (x – 2) (3x2 + 8x – 3) + 0
= (x – 3) [3x2 + 8x – 3]
= (x – 3) [3x2 + 9x – x – 3]
= (x – 3) [3x (x + 3) – 1 (x + 30]
= (x – 3) [(x + 3) (3x – 1)]
(ii) 2x3 + x2 – 13x + 6
Let, f(x) = 2x3 + x2 – 13x + 6
Put x = 1
f(1) = 2(1)3 + (1)2 – 13(1) + 6
= 2 + 1 – 13 + 6
= 3 – 13 + 6
= – 10 + 6
= – 4
f(1) = – 4
x – 1 is not factor of f(x).
Put x = 2
f(2) = 2(2)3 + (2)2 – 13(2) + 6
= 2 (8) + 4 – 26 + 6
= 16 + 4 – 26 + 6
= 20 – 26 + 6
= 26 – 26
= 0
x – 2 is a factor of f(x)
Now, factorise the given polynomial and division of polynomial.
Remainder = 0
Quotient = 2x2 + 5x – 3
By using,
Dividend = divisor × Quotient + Remainder
2x3 + x2 – 13x + 6 = (x – 2) × (2x2 + 5x – 3) + 0
= (x – 2) [2x2 + 5x – 3]
+ 6 – 1
= (x – 2) [2x2 + 6x – x – 3] + 6 – 1
= (x – 2) [2x (x + 3) – 1 (x + 3)]
= (x – 2) [(x + 3) (2x – 1)]
(iii) 3x3 + 2x2 – 23x – 30
Let, f(x) = 3x3 + 2x2 – 23x – 30
Put x = 1
f(1) = 3(1)3 + 2(1)2 – 23(1) – 30
= 3 + 2 – 23 – 30
= 6 – 23 – 30
= 6 – 53
f(1) = -47
x – 1 is not factor of f(x)
Put x = 2
f(2) = 3(2)3 + 2(2)2 – 23(2) – 30
= 3(8) + 2(4) – 46 – 30
= 24 + 8 – 46 – 30
f(2) = – 44
x – 2 is not a factor of f(x)
Put x = 3
f(3) = 3 (3)3 + 2(3)2 – 23 (3) – 30
= 3(27) + 2(9) – 69 – 30
= 81 + 18 – 69 – 30
= 99 – 69 – 30
= 30 – 30
f(3) = 0
∴ x – 3 is a factor of f(x).
Now, factorise the given polynomial and division of polynomial.
Remainder = 0
Quotient = 3x2 + 11x + 10
By using,
Dividend = divisor × Quotient + Remainder
3x3 + 2x2 – 23x – 30 = (x – 3) × (3x2 + 11x + 10) + 0
= (x – 30) [3x2 + 11x + 10]
= (x – 3) [3x2 + 6x + 5x + 10]
= (x – 3) [3x (x + 2) + 5 (x + 2)]
= (x – 3) [(x + 2) (3x + 5)]
(iv) 4x3 + 7x2 – 36x – 63
Let, f(x) = 4x3 + 7x2 – 36x – 63
Put x = 1
f(1) = 4(1)3 + 7(1)2 – 36(1) – 63
= 4 + 7 – 36 – 63
= 11 – 36 – 63
= – 52 – 36
f(1)= – 88
x – 1 is not factor of f(x)
Put x = 2
f(2) = 4(2)3 + 7 (2)2 – 36 (2) – 63
= 4 (8) + 7 (4) – 72 – 63
= 32 + 28 – 72 – 63
f(2) = – 75
x – 2 is not factor of f(x)
Put x = 3
f(3) = 4 (3)3 + 7 (3)2 – 36 (3) – 63
= 4 (27) + 7 (9) – 108 – 63
= 108 + 63 – 108 – 63
= 0
x – 3 is a factor of f(x)
Now, factorise the given polynomial and division of polynomial.
Remainder = 0
Quotient = 4x2 + 19x + 21
By using,
Dividend = divisor × Quotient + Remainder
4x3 + 7x2 – 36x – 63 = (x – 3) × [4x2 + 19x + 21] + 0
= (x – 3) [4x2 + 19x + 21]
= (x – 3) [4x2 + 12x + 7x + 21]
= (x – 3) [4x (x + 3) + 7 (x + 3)]
= (x – 3) [(x + 3) (4x + 7)]
(v) x3 + x2 – 4x – 4
= Let, f(x) = x3 + x2 – 4x – 4
Put x = 1
f(1) = (1)3 + (1)2 – 4(1) – 4
= 1 + 1 – 4 – 4
= 2 – 8
= – 6
x – 1 is not factor of f(x).
Put x = 2
f(2) = (2)3 + (2)2 – 4(2) – 4
f(2) = + 4 – 8 – 4
= 0
x – 2 is a factor of f(x).
Now factorise the given polynomial and division of polynomial.
Remainder = 0, quotient = x2 + 3x + 2 by using,
Dividend = divisor × Quotient × Remainder
x3 + x2 – 4x – 4 = (x – 2) × (x2 + 3x + 2) + 0
= (x – 2) [x2 + 3x + 2]
= (x – 2) [x2 + 2x + x + 2]
= (x – 2) [x (x + 2) + 1 (x + 2)]
= (x – 2) [(x + 2) (x + 1)]
(Q3) Using the remainder theorem factories the expression 3x3 + 10x2 + x – 6 Hence, solve the equation 3x3 + 10x2 + x – 6 = 0
= Solution:
Given that,
3x3 + 10x2 + x – 6
Le f(x) = 3x3 + 10x2 + x – 6
Put x = 1
f(1) = 3(1)3 + 10(1)2 + 1 – 6
= 3 + 10 + 1 – 6
= 13 – 5
f (1) = 8
Put x = – 1
f(-1) = 3(-1)3 + 10 (-1)2 + (-1) – 6
= 3(-1) + 10 (1) + [-1-6]
= -3 + 10 – 7
= – 10 + 10
f(-1) = 0
∴ x + 1 is the factor of the given f(x)
Now factorise the given polynomial and long division of polynomial
∴ Remainder = 0
Quotient = 3x2 + 7x – 6
∴ f(x) = (x + 1) [3x2 + 7x – 6]
= (x + 1) [3x2 + 9x – 2x – 6]
= (x + 1) [3x (x + 3) – 2 (x + 3)]
= (x+ 1) [(x + 3) (3x – 2)]
∴ 3x3 + 10x2 + x – 6= 0
(x + 1) (x + 3) (3x – 2) = 0
∴ x + 1 = 0 or x + 3 = 0 or 3x – 2 = 0
x = -1, x = – 3, 3x = 2 = x = 2/3
(Q4) Factorise the expression f(x) = 2x3 – 7x2 – 3x + 18 Hence, find all possible values of x for which f(x) = 0
= Solution:
Given that,
f(x) = 2x3 – 7x2 – 3x + 18
Put x = 1
f(1) = 2(1)3 – 7(1)2 – 3(1) + 18
= 2 – 7 – 3 + 18
= – 5 – 3 + 18
= – 8 + 18
f(1) = 10
Put x = 2
f(2) = 2(2)3 – 7(2)2 – 3(2) + 18
= 2(8) – 7(4) – 6 + 18
= 16 – 28 – 6 + 18
f(2) = 16 – 28 + 12
= 16 – 16
f(2) = 0
∴ x – 2 is the factor of f(x) now, factorise the f(x) and long division of given polynomial.
∴ Remainder = 0
Quotient = 2x2 – 3x – 9
f(x) = (x – 2) [2x2 – 3x – 9]
= (x – 2) [2x2 – 6x + 3x – 9]
= (x – 2) [2x (x – 3) + 3 (x – 3)]
f(x) = (x – 2) [(x – 3) (2x + 3)]
∴ f (x) = 0
2x3 – 7x2 – 3x + 18 = 0
(x – 2) (x – 3) (2x + 3) = 0
x – 2 = 0 Or x – 3 = 0 Or 2x + 3 = 0
x = 2 or x = 3 or 2x = -3 = x = -3/2
Here is your solution of Selina Concise Class 10 Math Chapter 8 Remainder and Factor Theorems Exercise 8B Solutions
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