I was doing some questions related to quadratic equations and I stumbled upon this question: Solving the equation $$(x+1)(x+3)(x+5)(x+7)= 5760.$$

One of the hints was to convert it into a quadratic equation.

I tried to but couldn't get anything meaningful. Is there any way in which I can solve bi-quadratic equations fast?

(Sorry if this strikes as basic, I am just a highschooler.)

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    $\begingroup$ Hint: Use the substitution $y-4=x$. Then group differences of squares. Finally, let $z=y^2$. $\endgroup$ – Michael Burr May 20 '18 at 9:41
  • $\begingroup$ I general, solving quartics is possible, but not "easy". The formula, if written out, is very long. $\endgroup$ – Michael Burr May 20 '18 at 9:44

In general, quartics have closed form solutions, but they are not easy. For example, this image on wikipedia writes out the formulas for quartics.

In this case, if you multiply everything out, you could use the rational root theorem to factor it as @Dr.SonnhardGraubner does. This requires some arithmetic and doesn't really use the structure of the problem.

In this problem, if you look at the factors on the left, you see that the terms being added are symmetric around $4$ (in other words, $1$ and $7$, $3$ and $5$). Therefore, it makes sense to shift the variable $x$ by $4$. In other words, let $$ x=y-4. $$ After substitution, you get $$ (y-3)(y-1)(y+1)(y+3)=5760. $$ Rearranging this as differences of squares, you get $$ (y^2-9)(y^2-1)=5760. $$ Next, use the substitution $z=y^2$ to reduce the quartic to a quadratic. In other words, you get $$ (z-9)(z-1)=5760. $$ This can be solved with the quadratic formula after multiplying everything out. Alternatively, use @ypercube's approach of using symmetry again to make this even easier is a good idea too.


$$ (x+1) (x+3) (x+5) (x+7) = 5760 $$

The equation has a symmetry around 4. We can use that to replace with $ y = x + 4 $, it becomes:

$$ (y-3) (y-1) (y+1) (y+3) = 5760 $$


$$ (y^2-9) (y^2-1) = 5760 $$

Now, replacing with $ z = (y^2-5) $, it becomes:

$$ (z-4) (z+4) = 5760 $$


$$ z^2-16 = 5760 $$

$$ z^2 = 5776 $$

So we find that: $z = 76$ or $ z = -76 $

and then work backwards, to find $y$ and $x$,


There is no need for symmetry, since $$(x+1)(x+7)(x+3)(x+5)= 5760$$ so $$(x^2+8x+7)(x^2+8x+15)= 5760$$ Put $t=x^2+8x+11$ and we get $$(t-4)(t+4) = 5760\implies t= \pm \sqrt{5776} = \pm 76$$ so $t^2+8t+11 = \pm 76$ so...

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    $\begingroup$ Of course there is no need for symmetry. There are many ways to solve a polynomial equation. It just makes the solution simpler. By the way, 11 is in the middle between 7 and 15, so you did use symmetry ;) $\endgroup$ – ypercubeᵀᴹ May 20 '18 at 11:44
  • $\begingroup$ Yes it is, but I could put only $t=x^2+8x$ or something similary and then procedee. But you are right of course. $\endgroup$ – Aqua May 20 '18 at 11:46
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    $\begingroup$ Sure. Combining (x+1) with (x+7) and (x+3) with (x+5) was also a well hidden use of symmetry. You wouldn't get similar polynomials if 7-5 was not equal to 3-1. $\endgroup$ – ypercubeᵀᴹ May 20 '18 at 11:47

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