Can someone help me solve this quartic equation?

Question

$$x^4+5x^3-18x^2-10x+4=0$$

I cannot solve this quartic equation - is there any way to solve it apart from the quartic equation? It has no integer roots, and a hint given on the worksheet says to first divide through by $x^2$. Can somebody help?

Answer 1

Like Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$

divide both sides by $x^2$ as $x\ne0$

The left hand side becomes

$$x^2+\left(\dfrac2x\right)^2+5\left(x-\dfrac2x\right)-18 =\left(x-\dfrac2x\right)^2+4+5\left(x-\dfrac2x\right)-18$$

So, we have a Quadratic Equation in $x-\dfrac2x$

Answer 2

Multiplying out $(x^2+ax+b)(x^2+cx+d)$ and comparing coefficients with the quartic polynomial gives $$ (x^2 + 7x - 2)(x^2 - 2x - 2). $$

Answer 3

Hint:

$$x^2+5x-18-\frac{10}x+\frac 4{x^2}=0$$

is $$x^2-4+\frac4{x^2}+5x-\frac{5\cdot 2}x-14=0$$

or $$\left(x-\frac 2x\right)^2+5\left(x-\frac 2x\right)-14=0.$$

Solve

$$x-\frac2x=-7,$$ $$x-\frac2x=2.$$

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In fact the hint is not really necessary. You can process the question as

$$x^4+5x^3-18x^2-10x+4=(x^2-2)^2+5x(x^2-2)-14x^2=(x^2-2+7x)(x^2-2-2x).$$