From Quadratic to Quartic

We use long division on $\frac{ x^4 + 3x^3 } { x^2 + 2x - 4 }$ .

$\frac{ x^4 + 3x^2 } { x^2 + 2x - 4 } = x^2 + x + 2 + \frac{8}{ x^2 + 2x - 4 }$

Multiplying throughout by $x^2 + 2x - 4$ , we see that $x^4 + 3x^2 = (x^2+x+2)(x^2+2x-4) + 8$ .
Hence, when $x^2 + 2x - 4 = 0$ , we get that $x^4 + 3x^2 = 8$ .

With $x^{2} + 2x = 4 \Longrightarrow x^{2} = 4 - 2x$ we have that

$x^{4} + 3x^{3} = x^{2}(x^{2} + 3x) = (4 - 2x)(4 - 2x + 3x) = (4 - 2x)(4 + x) =$

$16 - 4x - 2x^{2} = 16 - 4x - 2(4 - 2x) = 16 - 4x - 8 + 4x = \boxed{8}$ .

We proceed by using the defining equation to obtain new equations that help us to solve the problem.

$\begin{aligned}x^2+2x=4\implies x^3+2x^2&=4x\\\iff x^3&=4x-2x^2\\&=4x-2\left (4-2x\right )\\&=4x-8+4x\\&=\color{#EC7300}{\boxed{8x-8}}\\\\x^3=8x-8\implies x^4&=8x^2-8x\\&=8\left ( 4-2x\right )-8x\\&=32-16x-8x\\&=\color{#EC7300}{\boxed{32-24x}}\end{aligned}$

$\therefore x^4+3x^3=\left [32-24x\right ]+\left [3(8x-8)\right ]=32-24x+24x-24=\color{#20A900}{\boxed{8}}$ .

Minor note: For the $({\impliedby})$ direction, we can divide by $x$ as $x^2+2x=4\implies x\neq 0$ .