The polynomials f (x) =ax³+3x²-3 and g(x) =2x³-5x+a when divided by (x-4) leave remainders R1 and R2 respectively.Find the value of a when :-
(i) R1=R2
(ii) R1+R2 =0
(iii) 2R1-R2=0
Answers
Answer:
0
Step-by-step explanation:
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
step-by-step explanation
let f(x)= ax³+3x²-3
g(x)= 2x³-5x+a
x-4=0
x= 4
f(4)= a(4)³+3(4) ²-3
= a(64)+3(16)-3
= 64a+48-3
= 64a+45
therefore R1= 64a+45
g(4)= 2(4)³-5(4)+a
= 2(64)-20+a
= 128-20+a
= 108+a
therefore R2=108+a
(i) R1=R2
= 64a+45=108+a
= 64a-a= 108-45
= 63a=63
= a=63/63
a=1 (ans)
(ii) R1+R2=0
= (64a+45)+(108+a) =0
= 64a+45+108+a =0
= 64a+a+108+45 =0
= 65a+153 =0
= a=153/65 =0
=-153/65 (ans)
(iii)2R1-R2=0
= 2(64a+45)-(108+a) =0
= 128a+90-108+a =0
=128a-a-108-90 =0
= 127a-18 =0
= a=18/127 =0
= 18/127 (ans)
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