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I have been given this question after studying long division of polynomials and the factor theorem:

Find the set of values of $k$ for which the equation $3x^{4}+4x^3-12x^2+k=0$ has four real roots.

How do you find the set of values of $x$ in a polynomial of degree $4$?

I thought about using the formula $b^2-4ac$ from quadratics knowing very well it wasn't going to work, so I genuinely don't know how to go about this last question.

Could someone please help me with this question?

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    $\begingroup$ A study of the variations and local extrema of the associated function works here, no need for exact polynomial factorization. $\endgroup$ – KeiOh May 18 '20 at 9:43
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    $\begingroup$ Welcome to Math.SE! Even if you don't know how to approach a question, you should provide some context about it. In this case, you could say something about the topics and/or results are covered in the book/chapter from which the question comes. This information can help answerers tailor their responses to best serve you, without wasting time (theirs or yours) explaining things you already know or using techniques beyond your skill level. $\endgroup$ – Blue May 18 '20 at 10:27
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    $\begingroup$ Are you familiar with Descartes' rule of signs? If not, see en.wikipedia.org/wiki/Descartes%27_rule_of_signs $\endgroup$ – Barry Cipra May 18 '20 at 10:57
  • $\begingroup$ No I'm not, but I will certainly give it a go. Thank you. $\endgroup$ – mikejacob May 18 '20 at 11:09
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$f(x)=3x^4 +4x^3-12x^2 +k=0$

Differentiating, $$f’(x) = 12x^3 +12x^2 -24x =0 \implies x=-2,0,1$$ If you know anything about what a quartic looks like, you can deduce that there is a minimum, maximum and minimum at $-2,0,1$ respectively. For $f(x)$ to have four roots, we need $f(-2)\le 0 , f(0)\ge 0 , f(1)\le 0$. That is, $$48-32-48+k\le 0 \implies k\le32 \\ k\ge 0 \\ 3+4-12+k\le 0 \implies k\le 5$$ Taking the intersection of these values, we get $$0\le k\le 5$$

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    $\begingroup$ $k$ can be equal to $0$ or $5$. In that case, the equation will still have $4$ real roots but not distinct. But question wants only real roots. $\endgroup$ – SarGe May 18 '20 at 11:16
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    $\begingroup$ @Doubtnut Affirmative. $\endgroup$ – Tavish May 18 '20 at 11:18
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The challenge here is to find a solution that doesn't use calculus, i.e., doesn't take the derivative. The key insight is that, as you translate a polynomial graph up or down, the number of roots changes only when you get to a translate with a double root. So what we're looking for is values of $k$ such that

$$3x^4+4x^3-12x^2+k=(x-r)^2(3x^2+bx+c)$$

for some root $r$ and coefficients $b$ and $c$.

Re-expanding the right hand side to

$$\begin{align} (x-r)^2(3x^2+bc+c)&=(x^2-2rx+r^2)(3x^2+bx+c)\\ &=3x^4+(b-6r)x^3+(c-2br+3r^2)x^2+(br^2-2cr)x+cr^2 \end{align}$$

we see that

$$\begin{align} b-6r&=4\\ c-2br+3r^2&=-12\\ br^2-2cr&=0 \end{align}$$

Thus $b=4+6r$ and $c=2br-3r^2-12=2(4+6r)r-3r^2-12=9r^2+8r-12$ and

$$(4+6r)r^2-2(9r^2+8r-12)r=-12r^3-12r^2+24r=-12r(r-1)(r+2)=0$$

so the number of roots changes when $k=cr^2=(9r^2+8r-12)r^2$ with $r=0$, $1$, and $-2$, i.e, when $k=0$, $5$, and $32$. By thinking about the general nature of quartics, we see that $3x^4+4x^3-12x^2+k$ has two roots when $k\lt0$, four roots (counting multiplicities) when $0\le k\le5$, two roots again when $5\lt k\le32$, and no roots when $k\gt32$. In particular, the answer to the question posed is the range $0\le k\le5$.

Comparing this to the answers that use the derivative, the take-home lesson may be that it's well worth learning calculus!

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  • $\begingroup$ Thank you. I agree, using calculus is way easier for this question. I did not remember we could use it in this question which is why I was confused. $\endgroup$ – mikejacob May 19 '20 at 9:59
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More details: the derivative of $f$ is a cubic polynomial which is very easy to factor, and that has $3$ real roots. From this it is easy to infer the signs that $f'$ takes over $\mathbb{R}$ (for verification purposes: it's negative up to the first root, then positive, then negative, then positive again). Next, $f$ is a continuous function (since it is a polynomial) and goes to $+\infty$ at infinities. From this you have a pretty clear picture of what the graph of $f$ looks like. Then you want to find the possible $k$ such that $f$ will cross the abscissa line each time before it changes its variation (the intermediate value theorem will guarantee it). I did not finish the exercise, but the set of possible $k$ contains the interval $]0, 5[$.

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  • $\begingroup$ Is there a way to do this without using derivatives or graphs? Algebraically I mean. Forgive me for not understanding your notation, I am in high school. Do you have any working out as well? $\endgroup$ – mikejacob May 18 '20 at 10:30
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    $\begingroup$ For a quartic polynomial, there is probably an algebraic approach, but you do not want to take it (even more so if you are in high school) as the formula are going to be "untractable". To convince yourself, you can have a look at cubic polynomials. I guessed that you were a high school student, and I thought that qualitative studies of real functions was something you'd had done already. By "qualitative", I mean using the derivative to understand the variation, and so on. Glad that the other comments helped you getting mine! $\endgroup$ – KeiOh May 18 '20 at 11:38
  • $\begingroup$ I have been taught to use derivatives to find out information about a function, although we had not been asked to use derivatives for these questions which is why I was a bit confused about your approach. It was only the notation you used that I did not understand at first, but it's all good now. Thank you very much. $\endgroup$ – mikejacob May 18 '20 at 12:05
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No worries. Let's keep it simple. Now, if the equation has four real roots, it will cut the $x$-axis four times. Like this,

enter image description here (This graph is not of given equation but general.)

So, from the graph we can conclude two facts:

  1. Value $f'(x)$ of the graph should be zero three times (say at $x=a,\,b,\,c$; $a<b<c$).
  2. $f(a)\le 0,\,f(b)\ge 0,\,f(c)\le 0$.

By calculating all these stuffs you get is $k\in[0,5]$.

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  • $\begingroup$ Thank you so much for your help. Reading this answer lets me understand the other answers. I cannot upvote yet because I do not have enough reputation, but when I do I will upvote. $\endgroup$ – mikejacob May 18 '20 at 11:07

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