Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
Using the remainder theorem, find the remainder when f(x) is divided by g(x) and verify the by actual division
f(x) = x3 + 4x2 − 3x + 10, g(x) = x + 4
Here, f(x) = x3 + 4x2 − 3x + 10
g(x) = x + 4
from, the remainder theorem when f(x) is divided by g(x) = x - (- 4) the remainder will be equal to f(- 4)
Let, g(x) = 0
⟹ x + 4 = 0
⟹ x = - 4
Substitute the value of x in f(x)
f(-4) = (- 4)3 + 4(- 4)2 - 3(- 4) + 10
= - 64 + (4*16) + 12 + 10
= - 64 + 64 + 12 + 10
= 12 + 10
= 22
Therefore, the remainder is 22
f(x) = 4x4 - 3x3 - 2x2 + x - 7, g(x) = x - 1
Here, f(x) = 4x4 - 3x3 - 2x2 + x -7
g(x) = x - 1
from, the remainder theorem when f(x) is divided by g(x) = x - (-1) the remainder will be equal to f(1)
⟹ x - 1 = 0
⟹ x = 1
f(1) = 4(1)4 - 3(1)3 - 2(1)2 + 1 - 7
= 4 - 3 - 2 + 1 - 7
= 5 - 12
= -7
Therefore, the remainder is 7
f(x) = 2x4 - 6x3 + 2x2 - x + 2, g(x) = x + 2
Here, f(x) = 2x4 - 6x3 + 2x2 - x + 2
g(x) = x + 2
from, the remainder theorem when f(x) is divided by g(x) = x - (-2) the remainder will be equal to f(-2)
⟹ x + 2 = 0
⟹ x = - 2
f(-2) = 2(−2)4 - 6(-2)3 + 2(-2)2 - (-2) + 2
= (2 * 16) - (6 * (-8)) + (2 * 4) + 2 + 2
= 32 + 48 + 8 + 2 + 2
= 92
Therefore, the remainder is 92
f(x) = 4x3 - 12x2 + 14x - 3, g(x) = 2x - 1
Here, f(x) = 4x3 - 12x2 + 14x - 3
g(x) = 2x - 1
from, the remainder theorem when f(x) is divided by g(x) = 2(x - 1/2), the remainder is equal to f(1/2)
⟹ 2x - 1 = 0
⟹ 2x = 1
⟹ x = 1/2
f(1/2) = 4(1/2)3 - 12(1/2)2 + 14(1/2) - 3
= 4(1/8) - 12(1/4) + 4(1/2) - 3
= (1/2) - 3 + 7 - 3
= (1/2) + 1
Taking L.C.M
= (3/2)
Therefore, the remainder is (3/2)
f(x) = x3 − 6x2 + 2x − 4, g(x) = 1 - 2x
Here, f(x) = x3 - 6x2 + 2x - 4
g(x) = 1 - 2x
from, the remainder theorem when f(x) is divided by g(x) = -2(x - 1/2), the remainder is equal to f(1/2)
⟹ 1 - 2x = 0
⟹ -2x = -1
f(1/2) = (1/2)³ - 6(1/2)² + 2(1/2) - 4
= 1/8 - 8(1/4) + 2(1/2) - 4
= 1/8 - (1/2) + 1 - 4
= 1/8 - (1/2) - 3
Therefore, the remainder is (-35)/8
f(x) = x4 − 3x2 + 4, g(x) = x - 2
Here, f(x) = x4 - 3x2 + 4
g(x) = x - 2
from, the remainder theorem when f(x) is divided by g(x) = x - 2 the remainder will be equal to f(2)
⟹ x - 2 = 0
⟹ x = 2
f(2) = 24 - 3(2)2 + 4
= 16 - (3* 4) + 4
= 16 - 12 + 4
= 20 - 12
= 8
Therefore, the remainder is 8
f(x) = 9x3 − 3x2 + x − 5, g(x) = x − 2/3
Here, f(x) = 9x3 − 3x2 + x − 5
g(x) = x − 2/3
from, the remainder theorem when f(x) is divided by g(x) = x - 2/3 the remainder will be equal to f(2/3)
substitute the value of x in f(x)
f(2/3) = 9(2/3) - 3(2/3)² + (2/3) - 5
= 9(8/27) − 3(4/9) + 2/3 − 5
= (8/3) − (4/3) + 2/3 − 5
= - 9/3
= - 3
Therefore, the remainder is - 3
f(x) = 3x4 + 2x3 − x3/3 − x/9 + 2/27, g(x) = x + 2/3
Here, f(x) = 3x4 + 2x3 − x3/3 − x/9 + 2/27
g(x) = x + 2/3
from remainder theorem when f(x) is divided by g(x) = x - (-2/3), the remainder is equal to f(- 2/3)
= 0
Therefore, the remainder is 0
If the polynomial 2x3 + ax2 + 3x − 5 and x3 + x2 − 4x + a leave the same remainder when divided by x - 2, Find the value of a
Given, the polymials are
f(x) = 2x3 + ax2 + 3x − 5
p(x) = x3 + x2 − 4x + a
The remainders are f(2) and p(2) when f(x) and p(x) are divided by x - 2
We know that,
f(2) = p(2) (given in problem)
we need to calculate f(2) and p(2)
for, f(2)
substitute (x = 2) in f(x)
f(2) = 2(2)3 + a(2)2 + 3(2) − 5
= (2 * 8) + 4a + 6 - 5
= 16 + 4a + 1
= 4a + 17 .... 1
for, p(2)
substitute (x = 2) in p(x)
p(2) = 23 + 22 - 4(2) + a
= 8 + 4 - 8 + a
= 4 + a .... 2
Since, f(2) = p(2)
Equate eqn 1 and 2
⟹ 4a + 17 = 4 + a
⟹ 4a - a = 4 - 17
⟹ 3a = -13
⟹ a = -13/3
The value of a = −13/3
If polynomials ax3 + 3x2 − 3 and 2x3 − 5x + a when divided by (x - 4) leave the remainders as R1 and R2 respectively. Find the values of a in each of the following cases, if
1. R1 = R2
2. R1 + R2 = 0
3. 2R1 − R2 = 0
Here, the polynomials are
f(x) = ax3 + 3x2 - 3
p(x) = 2x3 - 5x + a
Let,
R1 is the remainder when f(x) is divided by x - 4
⟹ R1 = f(4)
⟹ R1 = a(4)3 + 3(4)2 − 3
= 64a + 48 - 3
= 64a + 45 .... 1
Now, let
R2 is the remainder when p(x) is divided by x - 4
⟹ R2 = p(4)
⟹ R2 = 2(4)3 - 5(4) + a
= 128 - 20 + a
= 108 + a .... 2
1. Given, R1 = R2
⟹ 64a + 45 = 108 + a
⟹ 63a = 63
⟹ a = 1
2. Given, R1 + R2 = 0
⟹ 64a + 45 + 108 + a = 0
⟹ 65a + 153 = 0
⟹ a = −153/65
3. Given, 2R1 − R2 = 0
⟹ 2(64a + 45) - 108 - a = 0
⟹ 128a + 90 - 108 - a = 0
⟹ 127a - 18 = 0
⟹ a = 18/127
If the polynomials ax3 + 3x2 − 13 and 2x3 − 5x + a when divided by (x - 2) leave the same remainder, Find the value of a
f(x) = ax3 + 3x2 − 13
p(x) = 2x3 − 5x + a
equate, x - 2 = 0
x = 2
substitute the value of x in f(x) and p(x)
f(2) = (2)3 + 3(2)2 − 13
= 8a + 12 - 13
= 8a - 1 ..... 1
p(2) = 2(2)3 - 5(2) + a
= 16 - 10 + a
= 6 + a .... 2
f(2) = p(2)
⟹ 8a - 1 = 6 + a
⟹ 8a - a = 6 + 1
⟹ 7a = 7
The value of a = 1
Find the remainder when x3 + 3x3 + 3x + 1 is divided by,
1. x + 1
2. x - 1/2
3. x
4. x + π
5. 5 + 2x
Here, f(x) = x3 + 3x2 + 3x + 1
by remainder theorem
1 ⟹ x + 1 = 0
⟹ x = -1
f(-1) = (−1)3 + 3(−1)2 + 3(−1) + 1
= -1 + 3 - 3 + 1
By remainder theorem
⟹ x - 1/2 = 0
f(1/2) = (1/2)³ + 3(1/2)² + 3(1/2) + 1
= (1/2)3 + 3(1/2)2 + 3(1/2) + 1
= 1/8 + 3/4 + 3/2 + 1
= 27/8
⟹ x = 0
f(0) = 03 + 3(0)2 + 3(0) + 1
= 0 + 0 + 0 + 1
= 1
⟹ x + π = 0
⟹ x = - π
f(- π) = (- π)3 + 3(-π)2 + 3(-π) + 1
= -(π)3 + 3(π)2 − 3(π) + 1
5 + 2x = 0
2x = -5
x = -5/2
f(-5/2) = (-5/2)³ + 3(-5/2)² + 3(-5/2) + 1
= −125/8 + 3(25/4) + 3(-5/2) + 1
= −125/8 + 75/4 - 15/2 + 1
= −27/8
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Chapter 6: Factorization of Polynomials Exercise...