Question

find the remainder (4x^3-3x-2)÷x+1 using long division method and remainder theorem.​

Answer 1

Answer: (x-1)(x+1)(x-2)

Step-by-step explanation:(i) Let take f(x) = x3 - 2x2 - x + 2

The constant term in f(x) is are ±1 and  ±2

Putting x = 1 in f(x), we have

f(1) = (1)3 - 2(1)2 -1 + 2

= 1 - 2 - 1 + 2 = 0

According to remainder theorem f(1) = 0 so that  (x - 1) is a factor of x3 - 2x2 - x + 2

Putting x = - 1 in f(x), we have

f(-1) = (-1)3 - 2(-1)2 –(-1) + 2

= -1 - 2 + 1 + 2 = 0

According to remainder theorem f(-1) = 0 so that  (x + 1) is a factor of x3 - 2x2 - x + 2

Putting x =  2 in f(x), we have

f(2) = (2)3 - 2(2)2 –(2) + 2

= 8 -82  - 2 + 2 = 0

According to remainder theorem f(2) = 0 so that  (x – 2 ) is a factor of x3 - 2x2 - x + 2

Here maximum power of x is 3 so that its can have maximum 3 factors

So our answer is (x-1)(x+1)(x-2)

Answer 2

Answer:

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Step-by-step explanation:

hope it's helpful :)