Let-t-be-a-root-of-x-3-3x-1-0-if-t-2-pt-1-t-2-t-1-can-be-written-as-t-c-for-some-p-c-Z-then-p-c-equals-

Question Number 110419 by Aina Samuel Temidayo last updated on 28/Aug/20

Let t be a root of x^{3} - 3 x + 1 = 0, if

\frac{t^{2} + pt + 1}{t^{2} - t + 1} can be written as t + c for

some p, c \in Z, then p - c equals ?

Commented by Her_Majesty last updated on 28/Aug/20

p, z \in Z o r p, c \in Z ?

Commented by Aina Samuel Temidayo last updated on 28/Aug/20

Oh . Sorry for that . It has been

corrected .

Answered by Her_Majesty last updated on 28/Aug/20

t^{2} + p t + 1 = (t + c) (t^{2} - t + 1)

\Leftrightarrow

t^{3} + (c - 2) t^{2} - (c + p - 1) t + c - 1 = 0

b u t a l s o x = t i n 1^{s t} e q u a t i o n

t^{3} - 3 t + 1 = 0

\Rightarrow

(1) c - 2 = 0

(2) c + p - 1 = 3

(3) c - 1 = 1

\Rightarrow c = 2 \land p = 2

Commented by Aina Samuel Temidayo last updated on 28/Aug/20

Implies p - c = 0 right ?

Commented by Aina Samuel Temidayo last updated on 28/Aug/20

Why are we dealing with the

numerator only ? Could you please

shed more light on it ? And why did

you equate the numerator to 0 ?

Commented by Her_Majesty last updated on 28/Aug/20

“ \frac{t^{2} + pt + 1}{t^{2} - t + 1} can be written as t + c^{″}

m e a n s

\frac{t^{2} + pt + 1}{t^{2} - t + 1} = t + c \Leftrightarrow t^{2} + pt + 1 = (t + c) (t^{2} - t + 1)

t h i s i s w h a t I d i d, t h e r e s t f o l l o w s

Commented by Aina Samuel Temidayo last updated on 28/Aug/20

Thanks, I really appreciate that . You

can also look up other questions I

posted .