Compelling Cubic

the function can be written as $f(x) = 2(x-\alpha)(x-\beta)(x-\gamma)$ Substituting $x=-1$ in $f(x)$ we get $f(-1)=-2(\alpha+1)(\beta+1)(\gamma+1)=-2+20-201+2013$ Thus the answer is $915$

Using Vieta's Forumulas we know the following:

$\alpha + \beta + \gamma = \frac{-20}{2} = -10$

$\alpha \beta + \alpha \gamma + \beta \gamma = \frac{201}{2}$

$\alpha \beta \gamma = \frac{-2013}{2}$

Then, expanding out the target value gives: $-( \alpha + 1 )( \beta + 1) ( \gamma + 1) = - [ \alpha \beta \gamma + \alpha \beta + \alpha \gamma + \beta \gamma + \alpha + \beta + \gamma + 1 ]$

Then substitute the values we know:

$-[ \frac{-2013}{2} + \frac{201}{2} + \frac{-20}{2} + 1] = -\frac{-1830}{2} = \boxed{915}$

The polynomial with roots $\alpha+1,\beta+1,\gamma+1$ can be obtain by tansforming f(x) to f(x-1) (why?) giving $f(x-1)=2(x-1)^3+20(x-1)^2+201(x-1)+2013.$ Since we are only interested in the product of the roots - i.e. $(\alpha+1)(\beta+1)(\gamma+1)$ - then by Viete's relations we only need the 0th degree term of f(x-1). This can be found easily as $2(-1)+20(1)+201(-1)+2013=1830$ , and so is $(\alpha+1)(\beta+1)(\gamma+1)=-1830/2=-915$ . But we wanted minus that, which is $\boxed{915}.$

The polynomial $f$ defined by $f(x)=2x^3+20x^2+201x+2013$ has roots $\alpha,\beta$ and $\gamma$ . That is, $f(\alpha)=0$ , $f(\beta)=0$ and $f(\gamma)=0$ .

To find $-(\alpha+1)(\beta +1)(\gamma +1)$ use the transformation $t=x+1$ on the polynomial to form a new polynomial. This new polynomial will have roots $(\alpha +1) ,(\beta +1 )$ and $(\gamma +1)$ . Then $-(\alpha+1)(\beta +1)(\gamma +1)$ will be the constant term divided by the leading coefficient.

So using $t=x+1$ , $f(t-1)=2(t-1)^3+20(t-1)^2+201(t-1)+2013$ . Which simplifies to $2t^3+14t^2+167t+1830$ . The desired quantity is $-(\alpha+1)(\beta +1)(\gamma +1) =\frac{1830}{2}=915$ .

A quicker way to use the transformation without lots of multiplying is: Write $2x^3+20x^2+201x+2013 \equiv k_1(x+1)^3+k_2(x+1)^2+k_3(x+1)+k_4$ (which is the same as $k_1(t^3)+k_2(t^2)+k_3(t)+k_4$ ). Now to find the coefficients, consider dividing the polynomial by $(x+1)$ , the remainder upon division by this will be $k_4$ but since the polynomials above are equivalent, $k_4$ will be equal to the remainder when $2x^3+20x^2+201x+2013$ is divided by $(x+1)$ . This can be repeated to quickly with synthetic division to find the rest of the coefficients.

on multiplying the expression we get -(αβ√+ αβ+ β√+ α√+ α+ β+ √+1) now calculating the values for cubic polynomial αβ√= −2013/2 αβ+ β√+ α√= 201/2 α+ β+ √= −10 by substitution we get 915

Expanding gives: $-(\alpha +1)(\beta +1)(\gamma +1)=-((\alpha \beta \gamma )+(\alpha \beta +\alpha \gamma +\beta \gamma )+(\alpha +\beta +\gamma )+1)$

By $Vieta's\:Formulas$ : $-((\alpha \beta \gamma )+(\alpha \beta +\alpha \gamma +\beta \gamma )+(\alpha +\beta +\gamma )+1)=-(\frac{-2013}{2}+\frac{201}{2}+\frac{-20}{2}+1)=\boxed{915}$

From the given equation ${a}={2}, {b}={20}, {c}={201}, {d}={2013}$

Thus $\alpha+\beta+\gamma=\frac{-b}{a}=\frac{-20}{2}={-10}$

$\alpha\beta\gamma=\frac{c}{a}=\frac{201}{2}$

and $\alpha\beta+\alpha\gamma+\beta\gamma=\frac{-d}{a}=\frac{-2013}{2}$

Expanding the given expression we get $-((\alpha+\beta+\gamma)+(\alpha\beta\gamma)+(\alpha\beta+\alpha\gamma+\beta\gamma)+{1})$

substituting we get $-(({-10})+(\frac{201}{2})+(\frac{-2013}{2}))$

which gives $-({-915})$

which equals $\boxed{915}$

As $\alpha$ , $\beta$ and $\gamma$ are roots of polynomial

So, $f(x)$ = $k(x-\alpha)$ $(x-\beta)$ ( $x-\gamma)$

where $k=2$ as the leading coefficient of the polynomial is $2$

Putting $x=(-1)$ ; we get $f(-1)$ = $-2(1+\alpha)$ $(1+\beta)$ ( $1+\gamma)$

$1830 = -2(1+\alpha)$ $(1+\beta)$ ( $1+\gamma)$

Thus, $-(1+\alpha)$ ( $1+\beta)$ ( $1+\gamma) = 1830/2 = \boxed {915}$

Note that $-(\alpha + 1)(\beta + 1)(\gamma +1) = -(\alpha\beta\gamma + (\alpha\beta + \alpha\gamma + \beta\gamma) + (\alpha + \beta + \gamma) +1)$ .

By the Vieta formulas of $f(x)$ , the values of $\alpha\beta\gamma$ , $(\alpha\beta + \alpha\gamma + \beta\gamma)$ and $(\alpha + \beta + \gamma)$ are $-\frac {2013}{2}$ , $\frac {201}{2}$ and $-\frac {20}{2} = -10$ .

Therefore, the value of $-(\alpha + 1)(\beta + 1)(\gamma +1)$ is:

$-(\alpha + 1)(\beta + 1)(\gamma +1) = -(\alpha\beta\gamma + (\alpha\beta + \alpha\gamma + \beta\gamma) + (\alpha + \beta + \gamma) +1)$

$= -(-\frac {2013}{2}+\frac {201}{2}-10+1 = -(-915) = \boxed {915}$

We are given the polynomial : $2x^3+20x^2+201x+2013$ .

Let the roots be $\alpha, \beta, \gamma$

We want to find $- (\alpha + 1) (\beta +1) (\gamma +1)$

Now we can expand $- (\alpha + 1) (\beta +1) (\gamma +1)$ to get :

$- ( \alpha\beta\gamma + \alpha\beta + \alpha\gamma + \beta\gamma +\alpha + \beta + \gamma + 1 )$

Now we use vieta's formula. (For more information visit : http://en.wikipedia.org/wiki/Vieta's_formulas )

Using vieta's we get that :

$\alpha\beta\gamma$ = -2013/2

$\alpha\beta + \alpha\gamma + \beta\gamma$ = 201/2

$\alpha + \beta + \gamma$ = -20/2 = -10

And combining them (and adding one, don't forget to add the one) and multiplying by negative one, we get $\boxed {915}$

Let alpha = a , beta = b , gamma = c

Now, -(a+1)(b+1)(c+1) = -(a+b+c)-(ab+bc+ca)-abc-1

We have,

   (a+b+c) = -coefficient of x^2/coefficient of x^3 = -20/2    Therefore,          -(a+b+c) =  20/2 

    (ab+bc+ca) = coefficient of x/coefficient of x^3 = 201/2         Therefore, - (ab+bc+ca) = -201/2                                                                                 

      (abc) =  -constant term/coefficient of x^3 =  -2013/2                       Therefore,      (-abc) =  2013/2

Therefore, -(a+b+c)-(ab+bc+ca)-abc-1
= (20/2) -(201/2)+(2013/2)-1
= (20-201+2013-2)/2 = 1830/2 = 915

            so, value of -(a+1)(b+1)(c+1) = 915

-(a+1)(b+1)(c+1)= -(abc+ab+ac+bc+a+b+c+1)

Usando as relações de Girard

abc=-2013/2
ab+ac+bc=201/2
a+b+c=-20/2

Substituindo na formula:

-(-2013/2+201/2-20/2+2/2)= -(-1830/2)= +915