Question

please solve the 16th question :

Answer 1

Hi factorise the divisor
${x}^{2} - 3x + 2 = (x - 2)(x - 1)$
i.e. if (x-2) and (x-1) are divisible, then x^2-3x+2 will also be divisible.
To see if they are divisible, check if x=1 and x=2 satisfy the dividend.
i.e. put them one by one in
$2 {x}^{4} - 6 {x}^{3} + 3 {x}^{2} + 3x - 2$
You will see both are satisfying the above polynomial.
This implies (x-2) and (x-1) are factors.
Therefore x^2-3x+2 is divisible.

A simple example, if 24 is divisible by 2 and 3, clearly 24 will be divisible by 6 which is 2×3.
Hope this helps.

Answer 2

↑ Here is your answer ↓
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$x^2 - 3x + 2 = 0$

Find the zeroes,

$x^2 - 2x - x + 2 = 0$

$x(x-2) - 1 (x-2) = 0$

$(x - 1)(x-2) = 0$

$x = 1, x = 2$

Sub these zeroes in other equation,

$= 2x^4 - 6x^3 + 3x^2 + 3x - 2$

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

$= 2 - 6 + 3 + 3 - 2$

$= 0$

$= 2(2)^4 - 6(2)^3 + 3(2)^2 + 3(2) - 2$

$= 2(16) - 6(8) + 3(4) + 3(2) - 2$

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

$= 0$

Hence, proved.