Question

Prove that 2x^4-6x^3+ 3x^2+3x-2 is exactly divisible by x^ 2-3x+2 by actual division

Answer 1

Answer:

Yes, it is exactly divisible by 2x^2 - 1.

Step-by-step explanation:

(2x^4-6x^3+ 3x^2+3x-2) ÷ ( x^ 2-3x+2)

⇒(2x^4-6x^3+ 3x^2+3x-2) ÷ ((x^ 2-3x+2)2x^2) ⇒ (-x^2+3x-2) (remainder)

⇒(-x^2+3x-2) ÷ ((x^ 2-3x+2)-1) ⇒ 0 (remainder)

Therefore, in the final, we can say that:-

  (2x^4-6x^3+ 3x^2+3x-2) ÷  ( x^ 2-3x+2)(2x^2 - 1) = 0

because,    (2x^4-6x^3+ 3x^2+3x-2) is exactly divisible with  ( x^ 2-3x+2) when multiplied with (2x^2 - 1).