the real number $a, b, c, d, e$ have some equations as follows
\begin{array}{lcl} ac=1 \\ ad+bc=-1 \\ ae+bd=1 \\ be=-6 \end{array}
how can I find the value of $a+b+c+d+e$?
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(Fill in the gaps as needed. If you're stuck, write out your working and thought process to demonstrate where you're at.)
Hint: $x^3 - x^2 + x - 6 = (x-2) ( x^2 + x + 3)$.
Calvin Lin
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I know that (a+b)=5, (c+d+e)=-1 so the answer is 4.. But I dont understand why – Butter Nov 24 '20 at 20:52
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@Butter How would you use the hint? How is it helpful? How does it relate to the initial conditions? E.g. where is the 6? – Calvin Lin Nov 24 '20 at 21:48
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(ac)x^3+(ad+bc)x^2+(ae+bd)x+(be) and do I have to factorize this equation? – Butter Nov 25 '20 at 03:53
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@Butter Right. The idea is to factorize it over the reals (and hope that there is a unique factorization) – Calvin Lin Nov 25 '20 at 04:14
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wow Thank you so much...This is a brilliant idea. You're a genius!. But to be honest I don't know what to do after this...factorization (ax+b)(cx^2+dx+e) How can I figure the a+b+c+d+e? does it mean that a=1, b=-2, c=1, d=1, e=3? How did u get this idea in thought? this questions seems to have no connection at all with factorization. How did u got the clue? – Butter Nov 25 '20 at 05:37
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@Butter A) Note that the factorization is unique (over the reals) because the quadratic has no real roots. Hence we must have $ a = 1, b = -2, \ldots $. B) From experience, and looking at the cross terms. C) As Cosmos pointed out, from the given equations, we have $(a+b) \times (c+d+e) = 1 -1 + 1 - 6 = -5$. This doesn't fix the values are yet (even if we assume they are integers), so how can we further manipulate this to get what we want? One possible way is to define $ (ax+b) \times (cx^2 + dx + e)$. – Calvin Lin Nov 25 '20 at 11:06
$signs. – saulspatz Nov 24 '20 at 19:09