Simple polynomial division

Let $f(x) = x^3 + 4x^2 - 5x + 3$ and $r = Remainder$

$x - 2 = 0 \Rightarrow x = 2$

Due to the remainder theorem $f(x) = r$ so when $x = 2$

$f(2) = r$

Substituting $f(x) = x^3 + 4x^2 - 5x + 3$ gives us

$(2)^3 + 4(2)^2 - 5(2) + 3 = r$

This expands out into

$8 + 16 - 10 + 3 = r$

Evaluating this (using Bidmas of course) gives us

$17 = r$

So the remainder is $17$

You can use synthetic division: Pardon my poor illustration, I can't figure out how to do this in latex: $x-2$ becomes 2 and $x^{3}+4x^{2}-5x+3$ becomes 1 4 -5 3.

Therefore, our answer is $\boxed{17}$

x^3 + 4x^2 - 5x + 3 ÷ x-2 can be written as an identity as this form:

x^3 + 4x^2 - 5x + 3 = (x-2)(ax^2 + bx +c) +d (d being the remainder )

We can expand the second side out to get ax^3 + (b-2a)x^2 + (c-2b)x -2c +d

We can replace those values with the original equation so a=1, b-2a =4, c-2b=-5, -2c+d=3

We can rearrange and solve those to get a=1, b=6, c=7 and d=17, and since d is the remainder , the answer is 17

$Let\quad P(x)={ x }^{ 3 }+4{ x }^{ 2 }-5x+3\\ \\ The\quad Remainder\quad Theorem\quad states\quad that\\ the\quad remainder\quad when\quad P(x)\quad is\quad divided\quad by\quad x-a\\ is\quad P(a)\\ \\ Using\quad the\quad remainder\quad theorem,\\ The\quad remainder\quad when\quad dividing\quad P(x)\quad by\quad x-2\quad is\\ P(2)=8+16-10+3=24-7=17\\ \\ \qquad \qquad \qquad \qquad \boxed { P(2)=17 } \\ \\ \\$

put x-2=0 or x=2 in given equation x^3 + 4 x^2 -5 x +3 and we get 17 that will be the reminder