If $ax^3+bx^2+x-6$ has $x+2$ as a factor and leaves remainder 4 when divided by $(x-2)$, find the values of a and b.
+128406 Using synthetic division
-2 [ a b 1 - 6 ]
-2a 4a -2b -8a + 4b - 2
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a b -2a 4a - 2b + 1 -8a + 4b - 6 - 2 = 0 ⇒ -8a +4b = 8 ⇒ 2a - b = -2 (1)
2 [ a b 1 -6 ]
2a 2b + 4a 8a + 4b + 2
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a b + 2a 4a + 2b + 1 8a + 4b - 4 = 4 ⇒ 8a + 4b = 8 ⇒ 2a + b = 2 (2)
Add (1) and (2)
4a = 0
a = 0
And
2(0) + b = 2
b = 2
