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

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?

oscar6721
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    It factors as the product of two quadratics. – lulu Jun 03 '20 at 10:36
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    Did you try the hint ? –  Jun 03 '20 at 10:38
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    Hint: after divide it by $x^2$, try express it as a quadratic polynomial in $y = x - \frac2x$. Solving the quadratic polynomial in $y$ will give you a factorization of the original quartic polynomial. – achille hui Jun 03 '20 at 10:40

3 Answers3

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

Dietrich Burde
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  • thank you - I hadn't thought of this – oscar6721 Jun 03 '20 at 10:39
  • And since roots are divided in pairs, each of which is symmetrical around half an integer, looking at a graph immediately shows that $a=7$ and $c=-2$. (The symmetry points are $-3/2$ and $1$.) Then only a linear system of equations remains for $b$ and $d$. – WimC Jun 03 '20 at 11:40
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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 $x-\dfrac2x$

lab bhattacharjee
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  • I have done this and ended up with roots of $1 ± (3√2)/2$ and $-3.5 ± (√57)/2$. None of these four roots, when inserted into the start equation, give a result of 0. I have checked to see where I have made a mistake but do not seem to have made any. Any other ways of finding x, or should I check again? – oscar6721 Jun 03 '20 at 13:26
  • @oscar6721, Could you follow my steps? If the process is correct, the correct result should follow – lab bhattacharjee Jun 03 '20 at 13:47
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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

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

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

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