Prove that 2x^4-6x^3+3x^2+3x-2 is exactly divisible by x^2-3x+2 (i) by actual division (ii) without actual division
Answers
Answered by
1
Answer:
o nhii aundaa pra............
.................m...
Answered by
3
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..... :-)
Similar questions