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How can I solve and expand the equation which have more number of brackets ? Since opening and performing the operation on single brackets becomes a tedious task. Is there any other method which can be used to expand the equations and finding the values of coefficients fast ? For eg. $s(s+2)(s+4)(s+8)(s+10)$

DaSnipeKid
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    Um... looks like the works been done for you. $s(s+2)(s+4)(s+8)(s+10)$ is as simple as it gets. Expanding things out makes things more complicated, not simpler, and harder to solve. (This one has roots of $s = 0, -2, -4, -8, -10$. I solved that just by looking at it. No pencil, paper, or brain cells) But if you mean how to expand. Just do it step by step. $s(s+2)(s+4)(s+8)(s+10) = s(s^2+6s+8)(s+8)(s+10)= s(s^3 + 6s^2 + 8s + 8s^2 + 48s + 64)(s+10)=s(s^3 +14s^2 +56s +64)(s+10)$ etc. No easy way. Just do it. – fleablood Apr 08 '17 at 19:48

2 Answers2

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First of all, expanding things out is not simplifying. It's complicating. You simplify things by breaking them down into brackets; not getting rid of brackets. And you solve things by making brackets.

So $s(s+2)(s+4)(s+8)(s+10)$ is as simple as it gets and it's trivial to solve the roots at $s = 0,-2,-4,-8,-10$.

You are asking how to expand

So $s(s+2)(s+4)(s+8)(s+10) =(s+0)(s+2)(s+4)(s+8)(s+10) s^5 + as^4 + bs^3 + cs^2 + d^s + e$ and we just have to figure out what $a,b,c,d,e$ are.

$e = 0*2*4*8*10 = 0$

$d = 0*2*4*8 + 0*2*4*10 + 0*2*8*10 + 0*4*8*10 + 2*4*8*10 = 2*4*8*10$.

Oh... let's ignore the 0. It's trivial.

$c = 2*4*8 + 2*4*10 + 2*8*10+4*8*10$

$b = 2*4+2*8 + 2*10 + 4*8 + 4*10 + 8*10$

$a = 2+4+8+10$

Basically the coefficient of $s^k$ is the sum of all products of $n-k$ of the terms.

Easier example: $(x+ 1)(x-1)(x+3) = x^3 + bx^2+ cx + d$ and

$b = 1-1+3 = 3$

$c = 1*(-1) + (1*3) + (-1*3) = -1$

and $d = 1*(-1)*3 = -3$

so $x^3 -x^2 - 3x -3$ and indeed

$(x+1)(x-1)(x+3) = (x^2 + 1*x -1*x -1*1)(x+3)= (x^2 + [1-1]x + [-1*1])(x+3)=$

$x^3 + [1-1]x^2 + [-1*1]x + 3x^2 + 3[1-1]x + 3[-1*1] =$

$x^3 + (1-1+3)x^2 + ((-1*1)+(3*1)+(3*-1))x + 3*-1*1$

fleablood
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Please keep from now on this central statement in mind:

If your equation is given in the form $$ \prod_{k=1}^n(x-a_k) = (x-a_1)(x-a_2)...(x-a_n) $$ the solutions are $a_1, a_2, ..., a_n$.

Why? Well, the 'solutions' are those $x$ for which the equation becomes 0. So if $x$ becomes one of those $a_1, a_2, ..., a_n$ at least one of these elements of the product becomes $0$ and therefore the hole product.

As mentioned before: Your equation is already in the perfect form to easily see the solutions!

Fabian Schn.
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