If ax^3 + bx^2 + x - 6 has x + 2 has a factor and leaves a remainder 4 when divided by x - 1, find the values of a and b.

Guest Jan 30, 2022

#1**+1 **

x + 2 is a factor, so plugging in -2 into the function will get 0.

-8a + 4b - 8 = 0

when divided by x - 1, the remainder is 4, so plugging in 1 should get 4.

a + b - 5 = 4

We now have a system of equations:

-8a + 4b = 8

a + b = 9

Multiply the second equation by 8 to eliminate a.

-8a + 4b = 8

8a + 8b = 72

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12b = 80

b = **20/3**

a = 9 - 20/3 = **7/3**

CubeyThePenguin Jan 30, 2022