If the polynomials ax³+3x²-3x and 2x³-5x+a when divided by (x-4) leave the remainder R1 and R2 respectively. Find the value of a in each of the following cases, if
(i) R₁ = R₂ (ii) R₁ + R₂=0(iii) 2R₁-R₂ = 0.
Answers
Given: If the polynomials ax³ + 3x² - 3 and 2x³ - 5x + a when divided by (x - 4) leave the remainder R1 and R2 respectively.
[MISTAKE IN THE QUESTION]
Let, p (x) = ax³ + 3x² - 3 and q (x) = 2x³ - 5x + a be the given polynomials.
Now,
R1 = Remainder when p (x) is divided by (x – 4)
⇒ R1 = p (4)
⇒ R1 = a (4)³ + 3 (4)² – 3
[∵ p (x) = ax³ + 3x² - 3]
⇒ R1 = 64a + 48 – 3
⇒ R1 = 64a + 45
And,
R2 = Remainder when q (x) is divided by (x – 4)
⇒ R2 = q (4)
⇒ R2 = 2 (4)³ – 5 (4) + a
[∵ q (x) = 2x³ - 5x + a]
⇒ R2 = 2 × 64 - 20 + a
⇒ R2 = 128 – 20 + a
⇒ R2 = 108 + a
(i) Given case is : R1 = R2
⇒ 64a + 45 = 108 + a
⇒ 63a – 63 = 0
⇒ 63a = 63
⇒ a = 63/63
⇒ a = 1
Hence, the value of 'a' is 1.
(ii) Given case is R1 + R2 = 0
⇒ 64a + 45 + 108 + a = 0
⇒ 65a + 153 = 0
⇒ 65a = -153
⇒ a = - 153/65
Hence, the value of 'a' is - 153/65.
(iii) Given case is 2R1 – R2 = 0 ⇒ 2 (64a + 45) – (108 + a) = 0
⇒ 128a + 90 – 108 – a
⇒ 127a – 18 = 0
⇒ 127a = 18
⇒ a = 18/127
Hence, the value of 'a' is 18/27.
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Answer:
When (x - 4) is divided by ax³+3x²-3:
x - 4 = 0
=> x = 4
p(4) = a(4)³+3(4)²-3(4) = R1
=> R1 = 64a + 48 - 3
=> R1 = 45 + 64a
When (x - 4) is divided by 2x³-5x+a:
x - 4 = 0
=> x = 4
p(4) = 2(4)³-5(4)+a = R2
=> R2 = 128 - 20 + a
=> R2 = 108 + a
(i) R1 = R2
=> 45 + 64a = 108 + a
=> 63 = 63a
=> a = 1
(ii) R1 + R2 = 0
=> 45 + 64a + 108 + a = 0
=> -153 = 65a
=> a = -153/65
(iii) 2R1 - R2 = 0
=> 2(45 + 64a) - (108 + a) = 0
=> 90 + 128a - 108 - a = 0
=> 18 = 127a
=> a = 18/127
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