theory of equations finding roots from given polynomial

If the equation $x^4-4x^3+ax^2+bx+1=0$ has four positive roots then $a=\,?$ and $b=\,?$

$\textbf{A.}\,6,-4$

$\textbf{B.}\,-6,4$

$\textbf{C.}\,6,4$

$\textbf{D.}\,-6,-4$

we can replace options and check answers .. are there any other shortcuts we can use

• Try the binomial theorem. – copper.hat Dec 20 '13 at 18:33
• @copper.hat But what if we don't know there is only one solution? – Jean-Claude Arbaut Dec 20 '13 at 18:48
• @arbautjc: Well, it is a 4th order monic polynomial. It is completely determined by its roots. – copper.hat Dec 20 '13 at 18:55
• @copper.hat Understood! :-) – Jean-Claude Arbaut Dec 20 '13 at 18:57
• The problem is badly formulated. I would prefer it if the condition "and $|a|=6, |b|=4$" were added. – Hagen von Eitzen Dec 20 '13 at 19:27

$$(x-x_1)(x-x_2)(x-x_3)(x-x_4)$$ $$=x^4-x^3\left(\sum_{4\ge i\ge1} x_i\right)+x^2\left(\sum_{4\ge i>j\ge1}x_ix_j\right)-x\left(\sum_{4\ge i>j>k\ge1}x_ix_jx_k\right)+x_1x_2x_3x_4$$
If $x_i>0,1\le i\le4$ the coefficient of $x^2$ must be $>0$ and that of $x$ must be $<0$
$$x^4-4x^3+6x^2-4x+1=(x-1)^4$$ $$a=6,b=-4$$