Question

Without actual division, prove that 2x^4–6x^3+3x^2+3x–2 is exactly divisible by x^2–3x+2

Answer 1

Hey!!!

$let \: f(x) = 2 {x}^{4} - 6 {x}^{3} + 3 {x}^{2} + 3x - 2$
$g(x) = {x}^{2} - 3x + 2$
$= {x}^{2} - 2x - x + 2$
$= x(x - 2) - 1(x - 2)$
$(x - 2)(x - 1)$

Putting x=2 in f(x)
$f(2) = 2( {2}^{4} ) - 6( {2}^{3} ) + 3( {2}^{2} + 3(2) - 2$
$= 32 - 48 + 12 + 6 - 2$
$= 0$

Now putting x= 1 in f(x)

$f(1)=2-6+3+3-2$
[tex] =0

Therefore, by Factor Theorem
(x-2) and (x-1) are factors of f(x)

Therefore,
f(x) is exactly divisible by g(x)

#BE BRAINLY

Answer 2

i hope you understand my answer.,.......