Consider a polynomial f(x)=ax^3+bx^+x+2÷3, if x+3 is a factor of f(x) and if f (x) is divided by x+ 2 then we get remainder as5
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1
Let p(x)=ax
3
+bx
2
+x−6
Since (x+2) is a factor of p(x), then by Factor theorem p(−2)=0
⇒a(−2)
3
+b(−2)
2
+(−2)−6=0
⇒−8a+4b−8=0
⇒−2a+b=2 ...(i)
Also when p(x) is divided by (x-2) the remainder is 4, therefore by Remainder theorem p(2)=4
⇒a(2)
3
+b(2)
2
+2−6=4
⇒8a+4b+2−6=4
⇒8a+4b=8
⇒2a+b=2 ...(ii)
Adding equation (i) and (ii), we get
(−2a+b)+(2a+b)=2+2
⇒2b=4⇒b=2
Putting b=2 in (i), we get
−2a+2=2
⇒−2a=0⇒a=0
Hence, a=0 and b=2
Answered by
1
Answer:
a=143/108
b=457/108
Step-by-step explanation:
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