Selina Concise Class 10 Math Chapter 8 Remainder and Factor Theorems Solutions
Exercise – 8C
(Q1) Show that (x – 1) is a factor of x3 – 7x2 + 14x – 8. Hence, completely factorise the given expression.
= Solution:
Let f(x) = x3 – 7x2 + 14x – 8
To show: (x – 1) is a factor of f(x)
∴ Put x = 1 in given polynomial.
f(1) = (1)3 – 7 (1)2 + 14(1) – 8
= 1 – 7 + 14 – 8
= – 6 + 14 – 8
= – 14 + 14
f(1) = 0
∴ Remainder is zero.
So, x – 1 is a factor of f(x).
Now factorise the f(x) and long division of given polynomial.
Remainder = 0
Quotient = x2 – 6x + 8
f(x) = x3 – 7x2 + 14x – 8
(x – 1) [(x2 – 6x + 8)]
(x – 1) [x2 – 4x – 2x + 8]
(x – 1) [x (x – 4) – 2 (x – 4)]
(x – 1) [(x – 4) (x – 2)]
∴ f(x) = (x – 1) (x – 4) (x – 2)
(Q2) Using remainder theorem factorise:
x3 + 10x2 – 37x + 26 completely
= Solution:
Let, f(x) = x3 + 10x2 – 37x + 26
Put x = 1
f(1) = (1)3 + 10(1)2 – 37(1) + 26
= 1 + 10 – 37 + 26
= 11 – 37 + 26
= 37 – 37
f(1) = 0 [∵ Using remainder theorem]
Now, factorise the f(x) and long division of given polynomial.
0
f(x) = x3 + 10x2 – 37x + 26
= (x – 1) [x2 + 11x – 26]
= (x – 1) [x2 + 13x – 2x – 26]
= (x – 1) [x (x + 13) – 2 (x + 13)]
= (x – 1) [(x + 13) (x – 2)]
f(x) = (x – 1) (x + 13) (x – 2)
(Q3) When x3 + 3x2 – mx + 4 is divided by x – 2, the remainder is m + 3. Find the value of m.
= Solution:
Let f(x) = x3 + 3x2 – mx + 4
Given that x – 2 is a factor of f(x)
∴ Put x = 2
f (2) = (2)3 + 3(2)2 – m(2) + 4
= 8 + 3(4) – 2m + 4
= 8 + 12 – 2m + 4
= 20 – 2m + 4
f(2) = 24 – 2m
Given that m + 3 is the remainder.
∴ m + 3 = 24 – 2m
m + 2m = 24 – 3
m + 2m = 21
3m = 21
m = 21/3
m = 7
(Q4) What should be subtracted from 3x3 – 8x2 + 4x – 3, So that the resulting expression has x + 2 as a factor?
= Solution:
Let, f(x) = 3x3 – 8x2 + 4x – 3
Let us consider that the required number be ‘t’.
∴ f(x) = 3x3 – 8x2 + 4x – 3 – t
Given that x + 2 is a factor of f(x)
∴ Put x = -2
f(-2) = 3(-2)3 – 8 (-2)2 + 4 (-2) – 3 – t
= 3 (-8) – 8 (4) – 8 – 3 – t
= – 24 – 32 – 8 – 3 – t
f(-2) = – 56 – 8 – 3 – t
= – 64 – 3 – t
f(-2) = – 67 – t
∴ f(-2)= 0
∴ 0 = – 67 – t
67 = – t
∴ t = – 67
∴ – 67 is subtracted from given polynomial.
(Q5) If (x + 1) and (x – 2) are factors of x3 + (a + 1) x2 – (b – 2) x – 6, find the values of a and b, and then, factorise the given expression completely.
= Solution:
Let, f(x) = x3 + (a + 1) x2 – (b – 2) x – 6
Given that (x + 1) and (x – 2) are factorise of f(x).
Put x = – 1
f(-1) = (-1)3 + (a + 1) (-1)2 – (b – 2) (-1) – 6
= – 1 + (a + 1) (1) – (b – 2) (-1) – 6
= – 1 + a + 1 + b – 2 – 6
f(-1) = a + b – 8
∴ f(-1) = 0
a + b – 8 = 0
a + b = 8 ——- (1)
Put x = 2
f(2) = (2)3 + (a + 1) (2)2 – (b – 2) (2) – 6
= 8 + (a + 1) (4) – (b – 2) (2) – 6
= 8 + 4a + 4 – 2b + 4 – 6
= 12 + 4a – 2b – 2
= 10 + 4a – 2b
f(2) = 0
4a – 2b + 10 = 0
4a – 2b = – 10 —— (2)
Solving equation (1) and (2) simultaneously, adding equation (1) and (2), we get
a + b = 8
4a – 2b = – 10
_____________
5a – b = – 2 —— (3)
Adding equation (1) and (3),
a + b = 8
5a – b = – 2
__________
6a = 6
a = 6/6
a = 1
Substitute a = 1 in equation (1), we get,
1 + b = 8
b = 8 – 1
b = 7
∴ (x + 1) and (x – 2) are also a factor of given polynomial f(x).
∴ (x + 1) (x – 2) = x2 – 2x + x – 2
= x2 – x – 2
Now, factorise the f(x) and long division of the given polynomial
x3 + 2x2 – 5x – 6 / x2 – x – 2
[∵ a = 1 and b = 7 put in given polynomial
∴ x3 + (a + 1) x2 – (b – 2) x – 6
= x3 + (1 + 1) x2 – (7 – 2) x – 6
= x3 + 2x2 – 5x – 6
∴ Remainder = 0
Quotient = x + 3
∴ f(x) = x3 + 2x2 – 5x – 6 = (x + 1) (x – 2) (x + 3)
(Q6) If x – 2 is a factor of x2 + ax + b and a + b = 1, find the values of a and b.
= Solution:
Let, f(x) = x2 + ax + b
Given, x – 2 is a factor of f(x).
Put x = 2
f(2) = (2)2 + a (2) + b
f(2) = 4 + 2a + b
∴ f(2) = 0
2a + b + 4 = 0
2a + b = – 4 —– (1)
And also given that,
a + b = 1 —— (2)
Find the values of a and b, solve equation (1) and (2) simultaneously.
Subtracting equation (1) – (2),
We get,
2a + b = – 4
a + b = 1
– – –
_________
a = – 5
Put a = -5 in equation (1)
2(-5) + b = -4
– 10 + b = – 4
b = – 4 + 10
b = 6
∴ a = – 5 and b = 6
(Q7) Factorise x3 + 6x2 + 11x + 6 completely using factor theorem
= Solution:
Let, f(x) = x3 + 6x2 + 11x + 6
Put x = 1
f(1) = (1)3 + 6(1)2 + 11x + 6
= 1 + 6 + 11(1) + 6
= 7 + 11 + 6
= 18 + 6
f(1) = 24
Put x = 2
f(2) = (2)3 + 6(2)2 + 11 (2) + 6
= 8 + 6(4) + 22 + 6
= 8 + 24 + 22 + 6
= 32 + 28
f(2) = 60
Given polynomial is in only positive sign.
If, we have to put only positive, numbers then solution the numbers increases only.
So, we have to put negative number.
Put x = 1
f(-1) = (-1)3 + 6(-1)2 + 11 (-1) + 6
= – 1 + 6(1) – 11 + 6
= – 1 + 6 – 11 + 6
= 5 – 11 + 6
= – 6 + 6
f(-1) = 0
∴ x + 1 is the factor of f(x).
Now, factorise the f(x) and long division of the given polynomial
Remainder = 0
Quotient = x2 + 5x + 6
∴ f(x) = x3 + 6x2 + 11x + 6
= (x + 1) [x2 + 5x + 6]
= (x + 1) [x2 + 3x + 2x + 6]
= (x + 1) [x (x + 3) + 2 (x + 3)]
f(x) = (x + 1) [(x + 3) (x + 2)]
∴ f(x) = (x + 1) (x + 3) (x + 2)
Here is your solution of Selina Concise Class 10 Math Chapter 8 Remainder and Factor Theorems Exercise 8C
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