Use the remainder theorem!

Let $f(x) = 2x^4+hx^3+kx^2+34x-24$ . Since $x^2-5x+4=(x-1)(x-4)$ , This implies that:

$\begin{cases} f(1) = 2+h+k+34-24 =0 & \implies h+k = -12 &...(1) \\ f(4) = 512 + 64h + 16h + 136-24 = 0 & \implies 4h +k = -39 &...(2) \end{cases}$

$\begin{aligned} (2)-(1): \quad 3h & = -27 \\ \implies h & = -9 \\ (1): \quad \implies k & = -3 \end{aligned}$

$\begin{aligned} \implies f(x) & = 2x^4-9x^3-3x^2+34x-24 \\ & = (x^2-5x+4)(2x^2+x-6) \\ & = \boxed{(x-1)(x-4)(2x-3)(x+2)} \end{aligned}$

If we look at the option, then the coefficient of $x^4-(2)$ could be get in option 4 only.As in other option we would get $3,6,3$ as a coefficient of $x^4$

A solution that doesn't require you to find the values of $h$ and $k$ :

Since $x^2-5x+4$ is a factor of the expression, we can say that

$(x^2-5x+4)(ax^2+bx+c) = 2x^4+hx^3+kx^2+34x-24$

for some constants $a,\;b$ and $c$ . Expand the left hand side:

$ax^4+(b-5a)x^3+(4a-5b+c)x^2+(4b-5c)x+4c=2x^4+hx^3+kx^2+34x-24$

Compare the terms of the expressions:

$ax^4=2x^4 \implies a=2\\ 4c=-24 \implies c=-6\\ \big(4b-5(-6)\big)x=34x\\ 4b+30=34\\ 4b=4 \implies b=1$

Therefore,

$2x^4+hx^3+kx^2+34x-24\\ =(x^2-5x+4)(2x^2+x-6)\\ =\boxed{(x-1)(x-4)(2x-3)(x+2)}$

Let $f\left( x \right) =2{ x }^{ 4 }{ +hx }^{ 3 }+k{ x }^{ 2 }+34x-24$ .

We know,

$x^{ 2 }-5x+4=(x-1)(x-4)$

Since $x^{ 2 }-5x+4$ is a factor of $f\left( x \right)$ , it follows that $(x-1)$ and $(x-4)$ are factors of $f\left( x \right)$ .

Thus,

$f\left( 1 \right) =0\quad and\quad f\left( 4 \right) =0$ .

$f\left( 1 \right) =2(1)^{ 4 }+h(1)^{ 3 }+k(1)^{ 2 }+34(1)-24=0\quad \quad \quad \longrightarrow (1)\\ f\left( 4 \right) =2(4)^{ 4 }+h(4)^{ 3 }+k(4)^{ 2 }+34(4)-24=0\quad \quad \quad \longrightarrow (2)$

From $(1):\quad \quad \quad \quad \quad h+k=12=0 \quad \quad\longrightarrow (3)$

From $(2):\quad \quad \quad 64h+16k+624=0$

i.e., $\quad \quad \quad \quad \quad \quad \quad 4h+k+39=0 \quad \quad \longrightarrow (4)$

Solving $(3)$ and $(4)$ simultaneously we have, $h=-9$ and $k=-3$ .

Now,

$f\left( x \right) =2{ x }^{ 4 }{ -9x }^{ 3 }-3{ x }^{ 2 }+34x-24$

By long division, we have:

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 2x^{ 2 }\quad +\quad x\quad -\quad 6\quad \quad \\ x^{ 2 }\quad -5x\quad +4)\overline { 2{ x }^{ 4 }{ \quad -9x }^{ 3 }\quad -3{ x }^{ 2 }\quad +34x\quad -24 } \\ \quad \quad \quad \quad \quad \quad \quad \underline { 2{ x }^{ 4 }\quad{ -10x }^{ 3 }\quad +8{ x }^{ 2 }\quad \quad \quad \quad \quad \quad \quad } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad { \quad x }^{ 3 }\quad -11{ x }^{ 2 }\quad +34x\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \underline { { x }^{ 3 }\quad -\quad 5{ x }^{ 2 } +\quad 4x\quad \quad \quad \quad } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -\quad 6x^{ 2 }\quad +30x\quad -24\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \underline { -\quad 6x^{ 2 }\quad +30x\quad -24 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad0$

$\therefore f\left( x \right) =(x^{ 2 }-5x+4)(2x^{ 2 }+x-6)\\ \quad \quad \quad =\boxed{(x-1)(x-4)(2x-3)(x+2)}$

Let x^2 - 5x + 4= f(a) = ( x - 1)(x - 4)
.We have to find the remaining 2 factors. .In the given expression the first term is 2x^4. .This 2x^4 is formed by product of x^2 of f(a) and the first term of remaining factor. .Hence,the first term of remaining factor is 2x^2. .By options we find that (2x - 3)(x + 2) gives 2x^2 as first term. .Hence : (x - 1)(x - 4)(2x - 3)(x + 2)