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In each of the following , using the remainder theorem, find the remainder when f(x) is divided by g(x) and verify the by actual division : (1 – 8)
Q1. f(x) = x^{3}+4x^{2}–3x+10 , g(x) = x + 4
Sol :
Here, f(x) = x^{3}+4x^{2}–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
Q2. f(x) = 4x^{4}–3x^{3}–2x^{2}+x–7 , g(x) = x – 1
Sol :
Here, f(x) = 4x^{4}–3x^{3}–2x^{2}+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)
Let , g(x) = 0
⇒ x – 1 = 0
⇒ x = 1
Substitute the value of x in f(x)
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
Q3. f(x) = 2x^{4}–6x^{3}+2x^{2}–x+2 , g(x) = x + 2
Sol :
Here, f(x) = 2x^{4}–6x^{3}+2x^{2}–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)
Let , g(x) = 0
⇒ x + 2 = 0
⇒ x = 2
Substitute the value of x in f(x)
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
Q4. f(x) = 4x^{3}–12x^{2}+14x–3 , g(x) = 2x – 1
Sol:
Here, f(x) = 4x^{3}–12x^{2}+14x–3
g(x) = 2x – 1
from, the remainder theorem when f(x) is divided by g(x) = the remainder is equal to f(1/2)
Let , g(x) = 0
⇒ 2x – 1 = 0
⇒ 2x = 1
⇒ x = 1/2
Substitute the value of x in f(x)
Taking L.C.M
Therefore, the remainder is
Q5. f(x) = x^{3}–6x^{2}+2x–4 , g(x) = 1 – 2x
Sol :
Here, f(x) = x^{3}–6x^{2}+2x–4
g(x) = 1 – 2x
from, the remainder theorem when f(x) is divided by g(x) =  the remainder is equal to f(1/2)
Let , g(x) = 0
⇒ 1 – 2x = 0
⇒ 2x = 1
⇒ 2x = 1
⇒ x = 1/2
Substitute the value of x in f(x)
Taking L.C.M
Therefore, the remainder is
Q6. f(x) = x^{4}–3x^{2}+4 , g(x) = x – 2
Sol :
Here, f(x) = x^{4}–3x^{2}+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)
let , g(x) = 0
⇒ x – 2 = 0
⇒ x = 2
Substitute the value of x in f(x)
f(2) = 24–3(2)^{2}+4
= 16 – (3* 4) + 4
= 16 – 12 + 4
= 20 – 12
= 8
Therefore, the remainder is 8
Q7. f(x) = 9x^{3}–3x^{2}+x–5 , g(x)
Sol :
Here, f(x) = 9x^{3}–3x^{2}+x–5
g(x) =
from, the remainder theorem when f(x) is divided by g(x) = the remainder will be equal to f(1/2) substitute the value of x in f(x)
Therefore, the remainder is 3
Q8.
Sol :
from remainder theorem when f(x) is divided by g(x) the remainder is equal to substitute the value of x in f(x)
= 0
Therefore, the remainder is 0
Q9. If the polynomial 2x^{3}+ax^{2}+3x–5 and x^{3}+x^{2}–4x+a leave the same remainder when divided by x – 2 , Find the value of a
Sol :
Given , the polymials are
f(x) = 2x^{3}+ax^{2}+3x–5
p(x) = x^{3}+x^{2}–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) + a4 + 6 – 5
= 16 + 4a + 1
= 4a + 17 —— 1
for, p(2)
substitute (x = 2) in p(x)
p(2) = 2^{3}+2^{2}–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
⇒
Q10. If polynomials ax^{3}+3x^{2}–3 and 2x^{3}–5x+a when divided by (x – 4) leave the remainders as R_{1 }and R_{2} respectively. Find the values of a in each of the following cases, if
1. R_{1} = R_{2}
2. R_{1}+R_{2} = 0
3. 2R_{1}–R_{2} = 0
Sol :
Here, the polynomials are
f(x) = ax^{3}+3x^{2}–3
p(x) = 2x^{3}–5x+a
let,
R_{1} is the remainder when f(x) is divided by x – 4
⇒ R_{1} = f(4)
⇒ R1 = a(4)^{3}+3(4)^{2}–3
= 64a + 48 – 3
= 64a + 45 ——— 1
Now , let
R_{2} is the remainder when p(x) is divided by x – 4
⇒ R_{2} = p(4)
⇒ R_{2} = 2(4)^{3}–5(4)+a
= 128 – 20 + a
= 108 + a ——– 2
1. Given , R_{1} = R_{2}
⇒ 64a + 45 = 108 + a
⇒ 63a = 63
⇒ a = 1
2. Given, R_{1}+R_{2} = 0
⇒ 64a + 45 + 108 + a = 0
⇒ 65a + 153 = 0
⇒
3.Given, 2R_{1}–R_{2} = 0
⇒ 2(64a + 45) – 108 – a = 0
⇒ 128a + 90 – 108 – a = 0
⇒ 127a – 18 = 0
⇒ a = 18/127
Q11. If the polynomials ax^{3}+3x^{2}–13 and 2x^{3}–5x+a when divided by (x – 2) leave the same remainder, Find the value of a
Sol :
Here , the polynomials are
f(x) = ax^{3}+3x^{2}–13
p(x) = 2x^{3}–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
⇒ a = 1
The value of a = 1
Q12. Find the remainder when x^{3}+3x^{3}+3x+1 is divided by,
1. x + 1
2.
3. x
4. x + π
5. 5 + 2x
Sol :
Here, f(x) = x^{3}+3x^{2}+3x+1
by remainder theorem
1. ⇒ x + 1 = 0
⇒ x = 1
substitute the value of x in f(x)
f(1) = (−1)^{3}+3(−1)^{2}+3(−1)+1
= 1 + 3 – 3 + 1
= 0
Here, f(x) = x^{3}+3x^{2}+3x+1
By remainder theorem
⇒
⇒
substitute the value of x in f(x)
= 27/8
3. x
Sol :
Here, f(x) = x^{3}+3x^{2}+3x+1
by remainder theorem
⇒ x = 0
substitute the value of x in f(x)
f(0) = 03+3(0)^{2}+3(0)+1
= 0 + 0 + 0 + 1
= 1
4. x + π
Sol :
Here, f(x) = x^{3}+3x^{2}+3x+1
by remainder theorem
⇒ x + π = 0
⇒ x = π
Substitute the value of x in f(x)
f(π) = (−π)^{3}+3(−π)^{2}+3(−π)+1
= –(π)^{3}+3(π)^{2}–3(π)+1
5. 5 + 2x
Sol :
Here, f(x) = x^{3}+3x^{2}+3x+1
by remainder theorem
5 + 2x = 0
2x = 5
x = 5/2
substitute the value of x in f(x)
Taking L.C.M
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