The polynomials ax³ + 3x2 -3 and 2x³ = 5x + a when divided by (x − 4) leave the remainders R₁ and R₂ respectively. Find the values of a in each of the following cases, if (i) R₁ = R₂ (ii) R₁ + R₂ = 0 (iii) 2R₁ R₂ = 0.
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Answered by
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Answer:
Step-by-step explanation:
Let the remainder be R1 and R2 as given :
R1 = a(4)3+ 3(4)2 − 3 = 64a + 45 …(i)
R2 = 2(4)3 − 5(4) + a = 128 − 20 +a = 108 + a …(ii)
Given: 2R1−R2= 0
2(64a+45) − (108+a) = 0
(from (i) and (ii))
⇒ 128a + 90 − 108 − a = 0
⇒ 127a = 18
a = 18/127
Answered by
2
Answer:
it is your answer
Let the remainder be R1 and R2 as given :
R1 = a ( 4 ) 3+ 3 ( 4 ) 2 - 3 = 64a + 45 ... ( i )
R2 = 2 ( 4 ) 3 – 5 ( 4 ) + a = 128 - 20 + a = 108 + a ... ( ii )
Given : 2R1 - R2 = 0
Given : 2R1 - R2 = 0 2 ( 64a + 45 ) - ( 108 + a ) = 0
Given : 2R1 - R2 = 0 2 ( 64a + 45 ) - ( 108 + a ) = 0 ( from ( i ) and ( ii ) )
Given : 2R1 - R2 = 0 2 ( 64a + 45 ) - ( 108 + a ) = 0 ( from ( i ) and ( ii ) ) = 128a + 90 – 108 - a = 0
Given : 2R1 - R2 = 0 2 ( 64a + 45 ) - ( 108 + a ) = 0 ( from ( i ) and ( ii ) ) = 128a + 90 – 108 - a = 0 = 127a = 18
Given : 2R1 - R2 = 0 2 ( 64a + 45 ) - ( 108 + a ) = 0 ( from ( i ) and ( ii ) ) = 128a + 90 – 108 - a = 0 = 127a = 18i hope help you
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