Prakkash's polynomial

From Vieta's Theorem, or by expanding $(x - \alpha)(x - \beta)(x - \gamma) = x^{3}+px^{2}+qx+2$ , we get $-\alpha\beta\gamma = 2$ .

$\alpha\beta+1=0\Rightarrow\alpha\beta=-1$ , we find $\gamma=2$ by substituting in the above equation.

$\gamma=2$ is a root, so

$2^{3}+2^{2}p+2q+2=0$

$4p+2q+10=0$

$2p+q+5=0$

$2p+q+100=95$

By Vieta's Formulas, the product of the roots is $\alpha\beta\gamma = -2.$ We know that $\alpha\beta = -1,$ so $\gamma = 2.$ Since $2$ is a root of the polynomial, we have $2^3 + p(2^2) + q(2) + 2 = 0,$ or $4p + 2q = -10.$ Dividing through by $2$ gives $2p+q=-5.$ Thus, $2p+q+100 = \boxed{95}.$

By Vieta's Formulas, $p=\alpha+\beta+\gamma$ , $q=\alpha \beta+ \alpha \gamma + \beta \gamma$ , and $2=\alpha \beta \gamma$ .

Since $\alpha \beta + 1 =0$ , or $\alpha \beta = -1$ , we have $2= (-1) \gamma$ , so $\gamma = -2$ .

From this, we have that $p=\alpha+\beta-2$ and $q=\alpha \beta - 2(\alpha + \beta)$ , so $2p+q+100 = 2(\alpha+\beta-2) + \alpha \beta - 2(\alpha + \beta)+100$ .

Simplifying and plugging in our value for $\alpha\beta$ , we get our answer to be $2(-2)-1+100 = \fbox{95}$ .

Its given that- αβ+1=0 => αβ= -1 Now, product of roots oof a cubic equals negative of coefficient of x^{0} divided by coefficient of x^{3}. Thus, αβγ= -(2) i.e. (-1)γ= -2 => γ=2 since its a solution of the cubic, 2^{3}+ 2^{2}p +q*2 +2 =0 => 2p+q= -5 so, 2p+q+100 = -5+100 = 95 :)

By using Vieta's Formula,we know that

$\alpha+\beta+\gamma=-p$

$\alpha\beta+\alpha\gamma+\beta\gamma=q$

$\alpha\beta\gamma=-2$

In addition,we know that $\alpha\beta=-1 \Rightarrow \gamma=2$

By substituting the value,we get

$\alpha+\beta=-p-2 \Rightarrow 2\alpha+2\beta=-2p-4$

and

$2\alpha+2\beta=q+1$

By equating the two equations,

$-2p-4=q+1\Rightarrow2p+q=-5\Rightarrow2p+q+100=95$

Expanding and factoring gives:

$p = -γ-β-α$

$q = βγ+αγ+αβ$

$-αβγ=2$

Given $αβ+1=0$ then $αβ=-1$ and $γ=2$

$2p+q+100$

$=2(-2-β-α)+2β+2α+αβ+100$

$=αβ+96=95$

Given that $\alpha \beta + 1 = 0$ or $\alpha \beta = - 1$ and using Vieta's formula, we have:

$\begin{cases} \alpha \beta = - 1 &...(0) \\ \alpha + \beta + \gamma = - p &...(1) \\ \alpha \beta + \beta \gamma + \gamma \alpha = q &...(2) \\ \alpha \beta \gamma = - 2 &...(3) \end{cases}$

$\begin{array} {rll} (0) \to (3): & \color{#3D99F6}{\alpha \beta} \gamma = \color{#3D99F6}{(-1)} \gamma =-2 & \Rightarrow \gamma = 2 \\ (1): & \alpha + \beta + \color{#3D99F6}{\gamma} = \alpha + \beta + \color{#3D99F6}{2} = - p & \Rightarrow \alpha + \beta = - p - 2 \\ (2): & \alpha \beta + \beta \gamma + \gamma \alpha = -1 + 2(\color{#3D99F6}{\alpha + \beta}) = - 1 + 2(-p-2) = q & \Rightarrow q = -5 - 2p \end{array}$

$\begin{aligned} q & = -5 - 2p \\ \Rightarrow 2p + q & = -5 \\ 2p+q + 100 & = 100 - 5 = \boxed{95} \end{aligned}$

x3+px2+qx+2 = (x-α)(x-β)(x-γ) so αβ=-1 ,or -p = α+β+γ , or q= αβ+βγ+αγ -αβγ=2 so γ=2 sub γ=2 into q= αβ+βγ+αγ and -p = α+β+γ q = -1+2(α+β) p = -2-(α+β) 2p+q+100 = 2(-2-(α+β))+-1+2(α+β)+100 = 95

notice aby = - 2

thus y = 2, ab = -1

ab + by + ay = q

thus -1 + 2 (a+b) = q

a+ b = -(p+2)

thus -1 - 2(p+2) = q

thus 2p-1-2p - 4 + 100 = 95 the answer

We know that the binomial of grade 0, in this case "2" is equal the the product of the roots (I will call them a,b and c). a \times b = -1 and a·b·c=2; so c=-2 . Then we get x^{3}+p·x^{2}+q·x+2=x^{3} -2·x^{2} -x+2; so p=-2 and q=1; then 2·(-2) -1 +100= 95

Well we know that AB=-1 and we can use the fact that ABY=-d/a to show that ABY=-2 and so we can deduce that Y=2 so one root is x=2. After subbing in the value of 2 into the polynomial set equal to 0 we get 4p+2q=-10. Divide by 2 and get 2p+q=-5 then add 100 to get 2p+q+100=95

Since we know that (alpha)(beta) = -1, alpha = -1, beta = 1.

Substituting these in the given equations, we get p+q = 3 and p-q = -1. Therefore, p = -2 and q = -1. Then we can substitute that in the required equation.

Firstly we have that $\alpha \beta +1 = 0 \Rightarrow \alpha \beta \gamma = -\gamma$

However, $\alpha \beta \gamma = -2) \Rightarrow \gamma = 2$

Next we have $\alpha \beta + \beta \gamma + \gamma \alpha = q \Rightarrow -1 + 2 \beta + 2 \alpha = q$

And that $\alpha + \beta + \gamma = p$

Therefore $2p + q +100 = -2(\alpha + \beta + \gamma) + 2 \beta +2 \alpha -1 +100$

$\Rightarrow -2(\alpha + \beta) - 4 +2 (\alpha + \beta) + 100$

$\Rightarrow 2p+q+100 = 95$

f(x): $x^{3}$ + p $x^{2}$ + qx + 2 and αβ = -1.

From basic properties of cubic polynomial, product of roots αβγ= -2 ...putting αβ = -1 , we get γ = 2. Now since γ is a root of f(x), f(γ) = 0 {again through basic properties of polynomials and their roots} ...So, f(2) => 8 + p(4) + q(2) + 2 = 0. => 4p + 2q + 10 = 0. => 2p + q + 5 = 0. => 2p + q + (95+5) = 0+95. Hence the value 95. :)

We know that the product of the roots of this equation must be -(2/1)

So , αβγ = -2

αβ+1=0

hence , γ = 2.. Now we know one root that is 2.

Substitute x=2 in the equation.

8+4p+2q+2=0

2p+q+5=0

2p+q+100=95

Using AB + 1 = 0, we can conclude that A and B are opposites, so replace B with -A. So now the roots are A, -A, and C.

Using the factor theorem, we know the polynomial can be written as (x - A)(x + A)(x - C).

Expanding the polynomial, we get:

x^3 - (C)(x^2) - (A^2)(x)+ (A^2)(C) = x^3 + (p)(x^2) + (q)(x^3) + 2

It follows that:

i) p = -C ii) q = -(A^2)

and

iii) 2 = (A^2)(C)

It follows from iii) that C > 0 because (A^2) > 0.

I then tried to test only integer valued numbers for A and C to make (A^2)(C) = 2, and the only combination that works is A = 1 or A = -1, and C = 2. Then i) p = -2 and ii) q = -1.

2p + q + 100 = 2(-2) -1 + 100 = 95.

The only problem is that I don't know why the roots have to be integers. Do they? It seems that they do because the problem has a unique solution.

Let a, b, c be the roots of the polynomial.

From vieta's formulas we know that a+b+c=-p , ab+ac+bc=q , abc=-2 and also were given that ab=-1

ab+ac+bc=q (ab=-1)

ac+bc=q+1

c(a+b)=q+1 (c=-2/ab)

-2(a+b) / (ab) = q+1 (ab=-1)

2(a+b) = q+1 (a+b=-p-c)

-2p - 2c = q+1

-2c = 2p + q +1 (Add 99 to both sides)

99-2c = 2p + q + 1 (we know that c=-2ab and ab=-1 so c=2)

99-2(2) = 2p + q + 1

Therefore 2p + q + 1 = 95.

The fact that it doesn't specify what $\alpha$ and $\beta$ has to be gives us an incentive to slack off. The simplest case would be to assume that a=1, b=-1. By vieta's, $\gamma = 2$ , and expanding (x-1)(x+1)(x-2) gives p=-2, q=-1, giving answer of 95.