Question

Using remainder theorem (long division method) factorise x^3-3x^2-x+3

Answer 1

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)

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Answer 2

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

1(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)

please mark as branliest