Question

without actual division, prove that
$2x {}^{4} - 6x ^{3} + 3x {}^{2} + 3x - 2$
is exactly divisible by
$x {}^{2} - 3x + 2$

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Answer 1

First we factorize divisor x² - 3x+2

= x² -2x - x +2 = x(x-2) -1 ( x - 2)

= ( x-2) (x-1)

Now, dividend p(x) = 2x^4 - 6x^3 + 3x² + 3x -2

If p(x) is dividible by ( x-2) & (x-1) each, then p(x) is also divisible by its product.

& by remainder theorem , if p(2) = 0 & p(1) =0

Then, p(x) is divisible by the divisor.

P( 2) = 2* 2^4 - 6* 2^3 + 3* 2² + 3*2 -2

= 32 - 48 + 12 + 6 - 2 = 50 - 50

= 0 ….. (1)

Also p(1) = 2* 1^4 - 6* 1^3 + 3 *1² + 3*1 -2

= 2 - 6 +3 + 3 - 2

= 0 ….. (2).

By (1) & (2), we can conclude that

P(x) is exactly dividible by x² - 3x + 2