They call it remainder theorem?

This is indeed an interesting problem. When I first solved it correctly, I was very happy.

It is given to us that we need to find the remainder when polynomial $\text{P}(x)=2x^5+x^3-2x^2-4x+5$ is divided by $x-1+\sqrt{2}$ .

In order to solve this problem I will use a little trick to simplify it, instead of dividing $\text{P}(x)$ by $x-1+\sqrt{2}$ , I will divide it by $(x-1+\sqrt{2})(x-1-\sqrt{2})=x^2-2x-1$ .

Now,

$\text{P}(x)=2x^5+x^3-2x^2-4x+5=(x^2-2x-1)(2x^3+4x^2+11x+20)+55x+29$

$\text{P}(x)=(x^2-2x-1)(2x^3+4x^2+11x+20)+55x+29$

$\text{P}(x)=(x-1+\sqrt{2})(x-1-\sqrt{2})(2x^3+4x^2+11x+20)+55x+29$

Now place $x=1-\sqrt{2}$ ,

$\text{P}(1-\sqrt{2})=(1-\sqrt{2}-1+\sqrt{2})(1-\sqrt{2}-1-\sqrt{2})(2(1-\sqrt{2})^3+...+20)+55(1-\sqrt{2})+29$

$\text{P}(1-\sqrt{2})=(0)(1-\sqrt{2}-1-\sqrt{2})(2(1-\sqrt{2})^3+...+20)+55(1-\sqrt{2})+29$

$\text{P}(1-\sqrt{2})=55-55\sqrt{2}+29=84-55\sqrt{2}=a-b\sqrt{c}$

Hence,

$a+b+c=\boxed{141}$

I LITERALLY SOLVED IT USING REMAINDER THEOREM:
First of all, $(1-\sqrt{2})^2 = 3-2\sqrt{2}$
ALSO, $(1-\sqrt{2})^3 = 7-5\sqrt{2}$

Now, remainder is:
$P(1-\sqrt{2}) = 2(1-\sqrt{2})^2(1-\sqrt{2})^3 + (1-\sqrt{2})^3 - 2(1-\sqrt{2})^2 - 4(1-\sqrt{2}) + 5$
$= 2(7-5\sqrt{2})(3-2\sqrt{2}) + (7-5\sqrt{2}) - 2(3-2\sqrt{2}) - 4(1-\sqrt{2}) + 5$
$= 42-30\sqrt{2}- 28\sqrt{2}+ 40+7-5 \sqrt{2} -6 +4\sqrt{2}- 4 +4 \sqrt{2} + 5$
$=84 - 55 \sqrt{2}$

Therefore $a+b+c=\boxed{141}$

$(1 - b)^5= 1 - 5b^1 +10b^2 - 10b^3 + 5b^4 - b^5. \ \ \\\ \ \\ If\ b=\sqrt2,\ \ \ \ \color{#3D99F6}{ b^1=\sqrt2,\ \ b^3=2\sqrt2,\ \ b^5=4\sqrt2, \ \ \ \ b^2=2,\ \ b^4=4}\\ \ \ \\ \therefore\ \color{#EC7300}{2(1 - \sqrt2)^5}=( -10 - 40 - 8 )\sqrt2=\color{#3D99F6}{-58\sqrt2}\ \ \ \ \ \ and\ \ \ \ \ \ (2 + 40 + 40)=\color{#3D99F6}{82}\\ \ \ \\ (1 - b)^3= 1 - 3b^1 +3b^2 - b^3.\ \ \ \ \therefore\ \color{#EC7300}{(1 - \sqrt2)^3}=( -3 - 2 )\sqrt2=\color{#3D99F6}{-5\sqrt2}\ \ \ \ \ \ and\ \ \ \ \ \ (1 + 6)=\color{#3D99F6}{7}\\ \ \ \\ (1 - b)^2= 1 - 2b^1 +b^2.\ \ \ \ \therefore\ \color{#EC7300}{- 2(1 - \sqrt2)^2}=-2( - 2 )\sqrt2=\color{#3D99F6}{4\sqrt2}\ \ \ \ \ \ and\ \ \ \ \ \ - 2(1 + 2)=\color{#3D99F6}{- 6}\\ \ \ \\ \color{#EC7300}{-4(1 - \sqrt2)}=- 4(-1)\sqrt2=\color{#3D99F6}{4\sqrt2}\ \ \ \ \ \ and\ \ \ \ \ \ - 4(1)=\color{#3D99F6}{- 4}\ \ \ \ \ \ \ \ \ \ \ also\ \ \color{#EC7300}{5}=\color{#3D99F6}{5}\\ \ \ \\ \implies\ \ (- 58 - 5 + 4 + 4)\sqrt2=- 55\sqrt2.\ \ \ \ \ \ \ \ and \ \ \ \ \ \ \ \ (82 + 7 - 6 - 4 +5)=84.\\ \ \ \\ \therefore\ \ \ P(1 - \sqrt2)=84 - 55\sqrt2=a-b\sqrt c .\ \ \ \ \ \ \ \ \ a+b+c=\ \ \ \color{#D61F06}{141}$

Warning: This is not LaTex . Haha.

By remainder theorem, (P(x) mod (x - c) = P(c))

x - 1 + √2 = x - c

=> c = 1 - √2

=> c - 1 = -√2

=> c^2 = 2c + 1

=> P(c) = 2c^5 + c^3 - 2c^2 - 4c + 5

=> 2c^5 = 4c^4 + 2c^3

By substitution, the P(c) is simply equal to: 4c^4 + 3c^3 - 2c^2 - 4c + 5.

=> 4c^4 = 8c^3 + 4c^2

By substitution again, => P(c) = 11c^3 + 2c^2 - 4c + 5

=> 11c^3 = 22c^2 + 11c

By substitution again, => P(c) = 24c^2 + 7c + 5

=> 24c^2 = 48c + 24

By substitution again, => P(c) = 55c + 29

As c = 1 - √2 => P(c) = 55(1-√2) + 29 = 55 + 29 - 55√2 = 84 - 55√2 = a - b√c

So, a+b+c = 84+55+2 = 141.

Using Horner's method of synthetic division remainder can be evaluated easily

The remainder when dividing $P(x)$ by $x-(1-\sqrt{2})$ is $P(1-\sqrt{2})=84-55\sqrt{2}$ . The answer is $\boxed{141}$ .