Selina Concise Class 10 Math Chapter 8 Remainder and Factor Theorems Solutions
Remainder and Factor Theorems
Exercise – 8A
(Q1) Find, in each case, the remainder when ?
(i) x4 – 3x2 + 2x + 1 s divided by x – 1
= Solution:
Given that x4 – 3x2 + 2x + 1
is divided by x – 1
Let, f(x) = x4 – 3x4 + 2x + 1
Put x = 1 in given equation
f(1) = (1)4 – 3(1)4 + 2 (1) + 1
= 1 – 3(1) + 2 + 1
= 1 – 3 + 2 + 1
= – 2 + 3
= 1
∴ The remainder is 1.
(ii) x3 + 3x2 – 12x + 4 is divided by x – 2
= Solution:
Given that x3 + 3x2 – 12x + 4
Is divided by x – 2
Let, f(x) = x3 + 3x2 – 12x + 4
Put x = 2 in given equation.
f(2) = (2)3 + 3(2)3 – 12(2) + 4
= 8 + 3(4) – 24 + 4
= 8 + 12 – 24 + 4
= 20 – 20
= 0
∴ The remainder is 0
(iii) x4 + 1 is divided by x + 1
= Solution:
Given that x4 + 1 is divided by x + 1
Let, f(x) = x4 + 1
Put x = -1 in given equation
f(-1) = (-1)4 + 1
= 1 + 1
f(-1) = 2
∴ The remainder is 2.
(Q2) Show that
(i) x – 2 is a factor of 5x2 + 15x – 50
(ii) 3x + 2 is a factor of 3x2 – x – 2
= Solution:
x – 2 = 0
x = 2
Let, f(x) = 5x2 + 15x – 50
Put x = 2
f(2) = 5(2)2 + 15(2) – 50
= 5(4) + 30 – 50
= 20 + 30 – 50
= 50 – 50
f (2) = 0
If given factor of value is substituted in given polynomial then the value of polynomial is zero, So, x – 2 is a factor of given polynomial.
(ii) 3x + 2 = 0
3x = -2
x = -2/3
Let, f(x) = 3x2 – x – 2
Put x = -2/3
f (-2/3) = 3x (-2/3)2 – (-2/3) -2
= 3 × 4/9 + 2/3 – 2
= 4/3 + 2/3 – 2
= 6/3 – 2
= 6-6/3
= 0/3
= 0
If given factor of value is substituted in given polynomial then the value of polynomial is zero, So, 3x + 2 is a factor of given polynomial.
(Q3) Use the remainder theorem to find which of the following is a factor of 2x3 + 3x2 – 5x – 6
(i) x + 1
(ii) 2x -1
(iii) x + 2
= (i) Given polynomial is,
2x3 + 3x2 – 5x – 6
We know that remainder theorem,
When a polynomial f(x) is divided by x – a, then the remainder is f(a).
Divident = divisor × quotient + remainder
∴ x + 1 = 0
x = -1
Let, f(x) = 2x3 + 3x2 – 5x – 6
Put x = -1
f(-1) = 2 (-1)3 + 3 (-1)2 – 5 (-1) – 6
= 2 (-1) + 3(1) – 5(-1) – 6
= -2 + 3 + 5 – 6
= 1 – 1
= 0
∴ x + 1 is a factor of given polynomial
(ii) 2x – 1
Given polynomial is
2x3 + 3x2 – 5x – 6
We know that remainder theorem, when a polynomial f(x) is divide by x-a, then remainder is f(a)
2x – 1 = 0
2x = 1
x = 1/2
Let, f(x) = 2x3 + 3x2 – 5x – 6
Put x = ½
F (1/2) = 2 (1/2)3 + 3 (1/2)2 – 5x – 6
= 2 (1/8) + 3 (1/4) – 5 (1/2) – 6
= 2 (1/8) + 3/4 – 5/2 – 6
= ¼ + 3/4 – 5/2 – 6
f (1/2) = 4/4 – 5/2 – 6
= 1 – 5/2 – 6
= -5 – 5/2
= -10 – 5/2
= -15/2
If given factor of value is substituted in given polynomial then the value of polynomial is zero.
So, In this polynomial the value is not zero.
So, 2x – 1 is not a factor of this given polynomial.
(iii) x + 2
Given polynomial is,
2x3 + 3x2 – 5x – 6
We know that remainder theorem,
When a polynomial f(x) is divided by x – a, then the remainder is f(a)
x + 2 = 0
x = -2
Let, f(x) = 2x3 + 3x2 – 5x – 6
Put x = -2
f(-2) = 2 (-2)3 + 3 (-2)2 – 5 (-2) – 6
= 2 (-8) + 3(4) + 10 – 6
= -16 + 12 + 10 – 6
= – 4 + 10 – 6
= 6 – 6
= 0
∴ x + 2 is a factor of given polynomial.
(Q4) (i) If 2x + 1 is a factor of 2x2 + ax – 3, find the value of a.
(ii) Find of expression 3x2 + 2x – k
= (i) Given that,
2x + 1 is a factor of
2x2 + ax – 3
∴ 2x + 1 = 0
2x = – 1
x = -1/2
Let, f(x) = 2x2 + ax – 3
Put x = -1/2
f(-1/2) = 2 (1/2)2 + a (-1/2) – 3
= 2 (1/4) – a/2 – 3
= ½ – a/2 -3
= 1-a/2 – 3
= 1 – a – 6/2
f(-1/2) = -a-5/2
∴ f(-1/2) = 0 because given that 2x + 1 is a factor of given polynomial.
0 = – a – 5/2
2×0 = – a – 5
0 = – a – 5
a = -5
(ii) Given that, 3x – 4 is a factor of 3x2 + 2x – k
∴ 3x – 4 = 0
3x = 4
x = 4/3
Let, f(x) = 3x2 + 2x – k
Put x = 4/3
f(4/3) = 3 (4/3)2 + 2 (4/3) – k
= 3 (16/9) + 8/3 – k
= 16/3 + 8/3 – k
f(4/3) = 24/3 – k
f(4/3) = 8 – K
∴ f(4/3) = 0
0 = 8 – K
K = 8
(Q5) Find the values of constants a and b when x – 2 and x + 3 both are the factors of expression x3 + ax2 + bx – 12
= Solution:
Given that, x – 2 and x + 3 both are the factor of expression x3 + ax2 + bx – 12
∴ x – 2 = 0 and x + 3 = 0
x = 2 and x = -3
Let f(x) = x3 + ax2 + bx – 12
Put x = 2
f(2) = (2)3 + a(2)2 + b(2) – 12
= 8 + 4a + 2b – 12
= -4 + 4a + 2b
∴ f(2) = 0 because given that x – 2 is a factor of given polynomial.
0 = – 4 + 4a + 2b
∴ 4a + 2b – 4 = 0
4a + 2b = 4 ——- (1)
Let f(x) = x3 + ax2 + bx – 12
Put = x = -3
f(-3) = (-3)3 + a(-3)2 +b (-3) – 12
= -27 + a(9) – 3b – 12
= -27 + 9a – 3b – 12
f(-3) = -39 + 9a – 3b
∴ f(-3) = 0
0 = -39 + 9a – 3b
9a – 3b – 39 = 0
9a – 3b = +39 —— (2)
Solving equation (1) and (2) simultaneously
Equation (1) = 4a + 2b = 4
Divide 2 on both side,
4a/2 + 2b/2 = 4/2
2a + b = 2 —— (A)
Equation (2) = 9a – 3b = -15
Divide 3 on both side,
9a/3 – 3b/3 = -39/3
3a – b = -13 ——— (B)
Adding equation (A) and (B),
2a + b = 12
3a – b = 13
5a = 15
a = 15/5
a = 3
Put a = 3 in equation (A)
2(3) + b = 2
6 + b = 2
b = 2 – 6
b = – 4
(Q6) Find the value of K, if 2x + 1 is a factor of (3K + 2) x3 + (K – 1)
= Solution:
Given that,
2x + 1 is factor of (3K + 2) x3 + 1 (K – 1)
∴ 2x + 1 = 0
2x = -1
x = -1/2
Let f(x) = (3K + 2) x3 + (K – 1)
Put x = -1/2
f(-1/2) = (3K + 2) (-1/2)3 + (K – 1)
= (3k + 2) (-1/8) + (k – 1)
= -3k – 2/8 + (k – 1)
= -3k + 2 + 8(k – 1)/8
= -3k – 2 + 8k – 8/8
f(-1/2) = 5k – 10/8
f(-1/2) = 5k – 10/8
∴ f(-1/2) = 0 because given that 2x + 1 is a factor of given polynomial.
0 = 5k – 10/8
8×0 = 5k – 10
0 = 5k – 10
10 = 5k
10 = 5k
10/5 = k
K = 2
(Q7) Find the value of a, if x – 2 is a factor of 2x5 + 6ax2 + 4ax + 8
= Solution:
Given that,
x – 2 is a factor of 2x5 – 6x4 – 2ax3 + 6ax2 + 4ax + 8
∴ x – 2 = 0
x = 2
Let, f(x) = 2x5 – 6x4 – 2ax3 + 6ax2 + 4ax + 8
Put x = 2
f(2) = 2(2)5 – 6 (2)4 – 2a (2)3 + 6a(2)2 + 4a (2) + 8
= 2 (2×2×2×2×2) – 6 (2×2×2×2) – 2a (2×2×2) + 6a (2×2) + 8a + 8
= 2(32) – 6 (16) – 2a (8) + 6a (4) + 8a + 8
= 64 – 96 – 16a + 24a + 8a + 8
= -32 – 16a + 24a + 8a + 8
= -24 – 16a + 24a + 8a
f(2) = -24 + 16a
∴ f(2) = 0 because given that x – 2 is a factor of given polynomial.
0 = -24 + 16a
24 = 16a
24/16 = a
a = 1.5
(Q8) Find the values of m and n so that x – 1 and x + 2 both are factors of x3 + (3m + 1) x2 + nx – 18
= Solution:
Given that x – 1 and x + 2
Are factors of x3 + (3m + 1) x2 + nx – 18
∴ x – 1 = 0 and x + 2 = 0
x = 1 and x = – 2
Let f(x) = x3 + (3m + 1) x2 + nx – 18
Put x = 1
f(1) = (1)3 + (3m + 1) (1)2 + n(1) – 18
= 1 + 3m + 1 + n – 18
= 2 + 3m + n – 18
f(1) = – 16 + 3m + n
∴ f(1) = 0 because given that x – 1 is a factor of given polynomial.
0 = -16 + 3m + n
3m + n = 16 —— (1)
Let, f(x) = x3 + (3m + 1) x2 + nx – 18
Put x = – 2
f(-2) = (-2)3 + (3m + 1) (-2)2 + n (-2) – 18
= -8 + (3m + 1) (+4) + (-2n) – 18
= -8 + 12m + 4 – 2n – 18
= -26 + 12m+ 4 – 2n
f(-2) = -22 + 12m – 2n
∴ f(-2) = 0 because given that x – 2 is a factor of given polynomial.
0 = -22 + 12m – 2n
12m – 2n = 22
Divide by 2 on both sides,
12m/2 – 2n/2 = 22/2
6m – n = 11 ——- (2)
Adding equation (1) and (2), we get
3m + n = 16
6m – n = 11
9m = 27
m = 27/9
m = 3
Put m = 3 in equation (1), we get.
3 (3) + n = 16
9 + n = 16
n = 16 – 9
n = 7
Here is your solution of Selina Concise Class 10 Math Chapter 8 Remainder and Factor Theorems Exercise 8.1 Solutions
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