Question

Divide f(x) = 2x3

– 3x2

+ 2x – 1 by g(x) = x – 1 by long division and verify the

remainder through remainder theorem​

Answer 1

Solution:

Given –

→ f(x) = 2x³ - 3x² + 2x - 1

→ g(x) = x - 1

Using long division method, find out f(x)/g(x) –

x - 1 ) 2x³ - 3x² + 2x - 1 ( 2x² - x + 1

        2x³ - 2x²

     (–)    (+)

———————————————

               -x² + 2x - 1

               -x² +   x

             (+)  (–)

———————————————

                        x - 1

                        x - 1

                   (–)  (+)

———————————————

                        0

So, remainder obtained = 0

Now,

→ g(x) = x - 1

→ x - 1 = 0

→ x = 1

So, when f(x) is divided by g(x),

→ Remainder = f(1)

= 2 × (1)³ - 3 × (1)² + 2 × 1 - 1

= 2 - 3 + 2 - 1

= 4 - 4

= 0

So, remainder = 0

Therefore, remainder theorem is verified.