Really sorry but idk how to translate these.

[lethal magnet]

you want a solution for it as well?

[smoky tree]

Yeah

[lethal magnet]

@smoky tree ?

k

you know the f(x) formula to solve this?

use r(x).(g(x)+1)

[smoky tree]

Yes.. but idk highest exponent of f(x)

[lethal magnet]

uk the g(X)=(x-1)(x-2)(x-3)..

so from this we need to get that the remainder must be degree 2 or less.

we have given q(x)=r(x)

[smoky tree]

What should I use between the identity in x or drawing graph?

[lethal magnet]

algebraic approach

x

[smoky tree]

Oh

It could be harder then

[lethal magnet]

it gives answer directly

did you get r(x)=a(x-1)(x-3)?

@smoky tree ?

then we substitute r(3) in the equation and the finals answer is 0

I hope you get it

[smoky tree]

Q(x) is ax^ + bx + c?

[lethal magnet]

in this question no

so whatever r is q is same

so where are stuck rn?

[smoky tree]

Q(x)= (x-1)(x-k) and I got lost after that

[lethal magnet]

yess

goood!

[smoky tree]

x=2??

[lethal magnet]

i mean now make q is r

r(X)

[smoky tree]

Got it

[lethal magnet]

then r(3)

[lethal magnet]

yes

[smoky tree]

Why r(3)??

[lethal magnet]

its given in the question

r(3)

[smoky tree]

Yes but how

[lethal magnet]

The result for this type of polynomial challenge is that it is a factor of the remainder.

so in last i want 0

get it?

[smoky tree]

Cuz of G(x?

Oh thanks

[lethal magnet]

Welcome. I am kind of bad at explaining but Glad you understood