2014-philic Cubic

Since we can see that our polynomial has a lot to do with $2014$ , let's let $a=2014$ .

Our polynomial turns into $x^3-(a+2\sqrt{a})x+2a$ .

Hmm... I don't really like those square roots. So instead, let's let $a=\sqrt{2014}$ .

Our polynomial is now $x^3-(a^2+2a)x+2a^2$ .

Distributing gives $x^3-a^2x-2ax+2a^2$

We can partially factor: $x(x^2-a^2)-2a(x-a)$

We can factor the difference of squares further: $x(x+a)(x-a)-2a(x-a)$

Aha! We have a common $(x-a)$ term so we can complete the factoring to get: $P(x)=(x-a)(x(x+a)-2a)$ or $P(x)=(x-a)(x^2+ax-2a)$

Now let's plug $a=\sqrt{2014}$ back in. We get $P(x)=(x-\sqrt{2014})(x^2+\sqrt{2014}x-2\sqrt{2014})$

We have one factor already: $x=\sqrt{2014}$ . We are left with a quadratic, which we can thankfully solve via the quadratic equation:

$x=\dfrac{-\sqrt{2014}\pm \sqrt{2014+8\sqrt{2014}}}{2}$

We can see that when the $\pm$ sign is negative, the root is negative. When the $\pm$ sign is positive, the root is also positive, because $\sqrt{2014+8\sqrt{2014}} > \sqrt{2014}$ .

Therefore, $r_1=\dfrac{-\sqrt{2014}- \sqrt{2014+8\sqrt{2014}}}{2}$ .

This means we want to find $r_2r_3=\sqrt{2014}\cdot \left(\dfrac{-\sqrt{2014}+\sqrt{2014+8\sqrt{2014}}}{2} \right)$

Distributing, we get $r_2r_3=\dfrac{-2014+\sqrt{2014(2014+8\sqrt{2014})}}{2}$

Simplifying: $r_2r_3=-1007+\sqrt{1007(1007+4\sqrt{2014})}$

Distributing the $1007$ inside the radical: $r_2r_3=-1007+\sqrt{1014049+4028\sqrt{2014}}$

This is the form that we want. We can see that $a=-1007$ $b=1014049$ $c=4028$ $d=2014$

Therefore our final answer is $-1007+1014049+4028+2014\pmod{10000}=\boxed{9084}$

$Hello.\quad My\quad solution\quad was\quad slightly\quad different\quad since\quad I\quad did\\ not\quad notice\quad that\quad one\quad of\quad the\quad roots\quad is\quad \sqrt { 2014 } \quad from\quad the\\ beginning.\quad \\ Let\quad P(x,\alpha )={ x }^{ 3 }-({ \alpha }^{ 2 }+2\alpha )x+2{ \alpha }^{ 2 }={ \alpha }^{ 2 }(2-x)+\alpha (-2x)+{ x }^{ 3 }.\\ After\quad solving\quad for\quad \alpha \quad in\quad P(x,\alpha )=0\quad we\quad obtain\quad \alpha =\frac { { x }^{ 2 } }{ 2-x } \\ or\quad \alpha =x,\quad which\quad means\quad that\quad the\quad three\quad roots\quad of\quad P(x,\alpha )\\ in\quad terms\quad of\quad \alpha \quad are\quad x=\alpha \quad and\quad x=\frac { -{ \alpha }\pm \sqrt { { \alpha }^{ 2 }+8\alpha } }{ 2 } .\\ Since\quad \alpha \quad is\quad non-negative,\quad the\quad smallest\quad root\quad is\\ { r }_{ 1 }=\frac { -{ \alpha }-\sqrt { { \alpha }^{ 2 }+8\alpha } }{ 2 } ,\quad so\quad { r }_{ 2 }{ r }_{ 3 }=\frac { -{ \alpha }^{ 2 }+\alpha \sqrt { { \alpha }^{ 2 }+8\alpha } }{ 2 } .\\ For\quad \alpha =\sqrt { 2014 } ,\quad { r }_{ 2 }{ r }_{ 3 }=-1007+\sqrt { { 1007 }^{ 2 }+4028\sqrt { 2014 } } ,\\ therefore\quad a+b+c+d=1007\times 1012\equiv 9084\quad (mod\quad 10000).$