x³-ax²+bx+7 and ax³+7x²-5x+b leaves the remainder -1 and 7 on division by x+1 and x-2 respectively, find a and b
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f(x)=x
3
+ax
2
+bx+6
(x−2) in a factor ⇒f(2)=0
f(2)⇒2
3
+a(2)
2
+2b+6=0
⇒8+4a+2b+b=0
⇒4a+2b+14=0
⇒2a+b+7=0 -(1)
f(x−3)=3 (Remainder)
⇒f(3)⇒3
3
+a(3)
2
+b×3+6=3
⇒27+9a+3b+b=3
⇒9a+3b+30=0
⇒3a+b+10=0 _____ (2)
From (1) & (2)
b=−2a−7 & b=−10−3a
−2a−7=−10−3a
3a−2a=−10+7
a=−3
from (3)
b=−2(a)−7=2(−3)−7
=6−7
b=−1
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