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Find all roots of $$(x+1)(x+3)(x+4)(x+6)=720$$

I would like to solve it without brute-force, that is, not multiplying the factorization out.

I've done some similar problems like this before, for example: $$(x+1)(x+2)(x+3)(x+6)=k$$ LHS can be regroupen as $$(x^2+6x+6)(x^2+5x+6)=k$$

Let $y=x^2+6$ and we're solving a quadratic in terms of $y$.

I tried to implement that on this problem, but there doesn't seem to be a way to regroup them.

Cyh1368
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1 Answers1

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Actually you can use your method: $(x^2+7x+6)(x^2+7x+12) = 720$ and now let $x^2+7x+9 = u$.

Then $(u-3)(u+3) = 720$ which gives two values for $u$. This is easier than the other method I suggested.

Toby Mak
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  • That works, but it's also a virtual duplicate of this answer under the linked question. – dxiv Feb 26 '22 at 08:47
  • I've already answered it in the comments so why not? – Toby Mak Feb 26 '22 at 08:50
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    I don't see the point in voting to close the question as a duplicate, then also posting a duplicate answer. – dxiv Feb 26 '22 at 08:53
  • The point is to show that expanding the brackets first and then exploiting the symmetry is much easier, as compared to directly substituting $u = x + 3.5$. The duplicate doesn't make clear which method is more appropriate for this question. – Toby Mak Feb 26 '22 at 11:59