y=x^5 +ax^4 +bx^3 +cx^2 +dx+e If we let x=t+h can we find h in terms of a,b,c,d,e such that y=(t+R)(t^2 +pt+q)(t^2 +s) this means two roots are of opposite sign, of course its possible by shifting the curve along x, but can we find the shift h ?


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Question Number 67819 by ajfour last updated on 31/Aug/19

$y = x^{5} + {a x}^{4} + {b x}^{3} + {c x}^{2} + d x + e$ $I f w e l e t x = t + h$ $c a n w e f i n d h i n t e r m s o f a, b, c, d, e$ $s u c h t h a t$ $y = (t + R) (t^{2} + p t + q) (t^{2} + s)$ $t h i s m e a n s t w o r o o t s a r e o f$ $o p p o s i t e s i g n, o f c o u r s e i t s$ $p o s s i b l e b y s h i f t i n g t h e c u r v e$ $a l o n g x, b u t c a n w e f i n d t h e$ $s h i f t h ?$
Commented by TawaTawa last updated on 31/Aug/19

$Weldone sir on the evetyday trial . It will be fruitful sir .$ $God bless you sir$
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