Question

without actual division show that f(x) = 2x^4-6x^3+3x-2 is exactly divisble by x^2-3x+2​

Answer 1

Hii friend,

Let P(X) = 2X⁴-6X³+3X²+3X-2

Let G(X) = (X²-3X+2) = (X²-2X-X+2)

=> X(X-2) -1(X-2)

=> (X-2) (X-1).

Now, P(X) will be exactly divisible by G(X) if it is exactly divisible by (X-2) as well as (X-1).

Putting X = 2 in P(X).

P(X) = 2X⁴-6X³+3X²+3X-2

P(2) = ( 2 × 2⁴ - 6 × 2³ + 3 × 2² + 3 × 2 -2)

=> (32-48+12+6-2) = (50-50) = 0

And,

P(1) = (2 × 1⁴ -6 × 1³ + 3 × 1² + 3 × 1 -2)

=> (2-6+3+3-2) = (8-8) = 0

Therefore,

P(X) is exactly divisible by (X-2) and (X-1)

So , P(X) is exactly divisible by (X²-3X+2)

Hence,

P(X) is exactly divisible by (X²-3X+2)

HOPE IT WILL HELP YOU..... :-)