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How to solve $$x(3x+3)(x+5)(2x+12)+576 = 0?$$

This was a question on a test i recently took, And i wasn't able to solve it. I later tried to solve it using online calculators and it turns out this doesn't have any real solutions.

I know there's a general formula for quartic polynomials that can work but we were only taught two methods, modifying the equation to a quadratic one using substitution or guessing some of the solutions. Both of these didn't work for me.

Are there any ways to prove that this doesn't have real solutions without the quartic formula?

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  • $\begingroup$ I have used Newton-Raphson. I don't think there is any solution in Real Line. $\endgroup$
    – IamKnull
    Sep 7, 2019 at 8:08
  • $\begingroup$ I know this question is tagged algebra-precalculus, but using calculus it shouldn't be hard to show the minima of the function are above $0$. $\endgroup$
    – Toby Mak
    Sep 7, 2019 at 8:15

5 Answers 5

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It is equivalent to $x(x+1)(x+5)(x+6)+96 = 0$

Now $$(x^2+6x)(x^2+6x+5)+96=0$$

Let $t=x^2+6x$ and finish the job...

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  • $\begingroup$ how is it equivalent? can you please explain? $\endgroup$ Sep 7, 2019 at 8:08
  • $\begingroup$ Can you explain why you jumped to the second form? How did you know that were the way to go? $\endgroup$
    – skyking
    Sep 7, 2019 at 8:10
  • $\begingroup$ Since the RHS is 0, you can take out a factor of $6$ to get the first line. For the second line, $(x+1)(x+6) = x^2+6x$ and $(x+1)(x+5) = x^2+6x+5$. $\endgroup$
    – Toby Mak
    Sep 7, 2019 at 8:11
  • $\begingroup$ I didn't, but some experimenting and voila, try joing1.and 2. and the rest or 1. and 3.rd and the rest or 1.st and last and the rest... $\endgroup$
    – nonuser
    Sep 7, 2019 at 8:11
  • $\begingroup$ Very clever. +1 $\endgroup$
    – Deepak
    Sep 7, 2019 at 8:12
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As already said, $$x(3x+3)(x+5)(2x+12)+576 = 0$$ is equivalent to $$x(x+1)(x+5)(x+6)+96 = 0.$$ Note the symmetry of the set of numbers $0,1,5,6.$
Set $a=x+3,$ the equation is equivalent to $$\begin{aligned}(a-3)(a-2)(a+2)(a+3)+96=&0\\ (a^2-4)(a^2-9)+96=&0\\ a^4-13a^2+132=&0 \end{aligned}$$ With the methods you know it is easy to finish.

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Let $f(x)=x(3x+3)(x+5)(2x+12)$. Then $f'(x)=24x^3+216x^2+492x+180$, whose roots are $-3$ and $\frac12\left(-6\pm\sqrt{26}\right)$. But $f(-3)=216$ and $f\left(\frac12\left(-6\pm\sqrt{26}\right)\right)=-\frac{75}2$. Therefore, the absolute minimum of $f$ is $-\frac{75}2$ and so the absolute minimum of $x(3x+3)(x+5)(2x+12)+576$ is greater than $0$.

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  • $\begingroup$ +1 Didn't notice your answer at the bottom before I wrote my comment. $\endgroup$
    – Toby Mak
    Sep 7, 2019 at 8:16
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Another possible way is using the symmetry of $x(3x+3)(x+5)(2x+12) = 6x(x+1)(x+5)(x+6)$ around $x=\color{blue}{3}$:

$$6x(x+1)(x+5)(x+6) = 6(x+\color{blue}{3}-3)(x+\color{blue}{3}+3)(x+\color{blue}{3}-2)(x+\color{blue}{3}+2)$$ $$=6(\underbrace{(x+\color{blue}{3})^2}_{y:=}-9)((x+\color{blue}{3})^2-4)$$

The minimum value of $(y-9)(y-4)$ is $-\frac{25}{4}$. Hence, $$6x(x+1)(x+5)(x+6)+576 \geq 6(-\frac{25}{4})+576 >0$$

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Along with all other answers, you can also try the general method for any degree 4 polynomial.

If f(x) is monic and has degree 4, You can break it as:

$f(x) = (x^2 + px + q)(x^2 + rx + s)$

The idea is now to compare the coefficients. If

$f(x) = x^4 + ax^3 + bx^2 + cx + d$

this gives the relations:

a = p + r

b = pr + q + s

c = ps + qr

d = qs

And then Check for the solution of two quadratic equations.

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