If p(x)=x³-ax²+bx+3 leaves a remainder -19 when divided by (x+2) and a remainder 17 when divided by (x-2) . Therefore (a+b) is
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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
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