Forming a quartic equation by given roots

Question

Find the equation whose roots are given by adding 2 to the roots of the equation

$\ x^4+3x^3-13x^2-51x-36=0 $

Note:- I am aware that the question can be simply be solved by substituting x=X+2 or by figuring out the roots of this equation, add 2 in them and then multiply them. However the question forbids me from using those method. The second equation should be given without solving this equation

Example

second coefficient of this equation

$\ = b/a = -Σa = -3$

Since we are adding two to every root so second coefficient of the second equation will be $\ =b/a = -Σ(a +2) = -(Σa +Σ2) = -(-3+8) = -5$

thus the equation should have$\ -5x³$

However the problem is arising for the next coefficient

$\ =Σ(a+2)(b+2)$

$\ =Σab+2a+2b+4$

$\ =Σab+ 4Σa +16 = -9$

Correct coefficient however is $\ -7$

Answer 1

The actual formula for the second coefficient is: $$a_{n-2}=\sum_{i=1}^{n-1} \sum_{j=i+1}^n x_ix_j$$

So that $$a'_{n-2}=\sum_{i=1}^{3} \sum_{j=i+1}^4 (x_i+2)(x_j+2)=\sum_{i=1}^{3} \sum_{j=i+1}^4x_ix_j+\sum_{i=1}^{3} \sum_{j=i+1}^42(x_i+x_j)+\sum_{i=1}^{3} \sum_{j=i+1}^44$$ $$=a_{n-2}+3\times2\sum_{i=1}^4 x_i+4\times(4\times 3 -6)$$ Since I don't know how good you are at manipulating sums, I'll write things clearly: $$\sum_{i=1}^{3} \sum_{j=i+1}^42x_i=2\times(x_1+x_1+x_1+x_2+x_2+x_3)$$ $$\sum_{i=1}^{3} \sum_{j=i+1}^42x_j=2\times(x_2+x_3+x_4+x_3+x_4+x_4)$$

What do you get when you add them up??

Answer 2

Perhaps easier to write: $$\sum_{i<j}(r_i+2)(r_j+2)=\sum_{i<j} (r_i r_j + 2(r_i + r_j) + 4) = a_{n-2} - 6 a_{n-1} + 6 \times 4 = -13 -18 + 24 = -7$$ where I used: $$ \sum_{i<j} 2(r_i+r_j) = \sum_{i\neq j} (r_i+r_j) = 2 \sum_i r_i \sum_{j\neq i} 1 = 2 \times 3 \times \sum_i r_i = -6 a_{n-1}$$

But the next equations get a bit complicated...

If you really have to do it you may want to check the mess of combinatoric numbers by comparing to the expansion of: $(X-2)^4 + \sum_{0 \leq k <4} a_k (X-2)^k$. At least this matches with $a_{n-2} - 6 a_{n-1} + 24$ for the second.