Question

Q19. Prove that 2x4 - 6x3 + 3x2 + 3x – 2 is exactly divisible by x2 – 3x +2
i. By actual division


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Answer 1

Answer

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)

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Answer 2

Answer:

Step-by-step explanation:

Required to prove by actual division that 2x⁴- 6x³ + 3x² + 3x – 2 is exactly divisible by x² – 3x +2

Solution:

Doing the actual  division we get

                     

                     2x² + 0x -1

                     ______________________

x² – 3x +2 |   2x⁴- 6x³ + 3x² + 3x – 2

                     2x⁴- 6x³ + 4x²

                     _____________________

                                    - x² + 3x – 2

                                     - x² + 3x – 2

                    _________________________

                                                  0  →remainder

Since the remainder is '0', we can conclude that 2x⁴- 6x³ + 3x² + 3x – 2 is exactly divisible by x² – 3x +2

Then by division lemma we can write

2x⁴- 6x³ + 3x² + 3x – 2 = (2x² + 0x -1)(x² – 3x +2) +0

2x⁴- 6x³ + 3x² + 3x – 2 = (2x² + 0x -1)(x² – 3x +2)

That is 2x⁴- 6x³ + 3x² + 3x – 2  can be expressed an the product of two factors, hence

we can conclude that  2x⁴- 6x³ + 3x² + 3x – 2 is exactly divisible by x² – 3x +2

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