Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

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?

$$\mathrm{Let}\:\mathrm{t}\:\mathrm{be}\:\mathrm{a}\:\mathrm{root}\:\mathrm{of}\:\mathrm{x}^{\mathrm{3}} −\mathrm{3x}+\mathrm{1}=\mathrm{0},\:\mathrm{if}\: \\ $$$$\frac{\mathrm{t}^{\mathrm{2}} +\mathrm{pt}+\mathrm{1}}{\mathrm{t}^{\mathrm{2}} −\mathrm{t}+\mathrm{1}}\:\mathrm{can}\:\mathrm{be}\:\mathrm{written}\:\mathrm{as}\:\mathrm{t}+\mathrm{c}\:\mathrm{for} \\ $$$$\mathrm{some}\:\mathrm{p},\mathrm{c}\:\in\:\mathbb{Z},\:\mathrm{then}\:\mathrm{p}−\mathrm{c}\:\mathrm{equals}? \\ $$

Commented by Her_Majesty last updated on 28/Aug/20

p, z ∈Z or p, c ∈Z?

$${p},\:{z}\:\in\mathbb{Z}\:{or}\:{p},\:{c}\:\in\mathbb{Z}? \\ $$

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

Oh. Sorry for that. It has been corrected.

$$\mathrm{Oh}.\:\mathrm{Sorry}\:\mathrm{for}\:\mathrm{that}.\:\mathrm{It}\:\mathrm{has}\:\mathrm{been} \\ $$$$\mathrm{corrected}. \\ $$

Answered by Her_Majesty last updated on 28/Aug/20

t^2 +pt+1=(t+c)(t^2 −t+1) ⇔ t^3 +(c−2)t^2 −(c+p−1)t+c−1=0 but also x=t in 1^(st) equation t^3 −3t+1=0 ⇒ (1) c−2=0 (2) c+p−1=3 (3) c−1=1 ⇒ c=2∧p=2

$${t}^{\mathrm{2}} +{pt}+\mathrm{1}=\left({t}+{c}\right)\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right) \\ $$$$\Leftrightarrow \\ $$$${t}^{\mathrm{3}} +\left({c}−\mathrm{2}\right){t}^{\mathrm{2}} −\left({c}+{p}−\mathrm{1}\right){t}+{c}−\mathrm{1}=\mathrm{0} \\ $$$${but}\:{also}\:{x}={t}\:{in}\:\mathrm{1}^{{st}} \:{equation} \\ $$$${t}^{\mathrm{3}} −\mathrm{3}{t}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{1}\right)\:{c}−\mathrm{2}=\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:{c}+{p}−\mathrm{1}=\mathrm{3} \\ $$$$\left(\mathrm{3}\right)\:{c}−\mathrm{1}=\mathrm{1} \\ $$$$\Rightarrow\:{c}=\mathrm{2}\wedge{p}=\mathrm{2} \\ $$

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

Implies p−c=0 right?

$$\mathrm{Implies}\:\mathrm{p}−\mathrm{c}=\mathrm{0}\:\mathrm{right}? \\ $$

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?

$$\mathrm{Why}\:\mathrm{are}\:\mathrm{we}\:\mathrm{dealing}\:\mathrm{with}\:\mathrm{the} \\ $$$$\mathrm{numerator}\:\mathrm{only}?\:\mathrm{Could}\:\mathrm{you}\:\mathrm{please} \\ $$$$\mathrm{shed}\:\mathrm{more}\:\mathrm{light}\:\mathrm{on}\:\mathrm{it}?\:\mathrm{And}\:\mathrm{why}\:\mathrm{did} \\ $$$$\mathrm{you}\:\mathrm{equate}\:\mathrm{the}\:\mathrm{numerator}\:\mathrm{to}\:\mathrm{0}? \\ $$

Commented by Her_Majesty last updated on 28/Aug/20

“((t^2 +pt+1)/(t^2 −t+1)) can be written as t+c” means ((t^2 +pt+1)/(t^2 −t+1))=t+c ⇔ t^2 +pt+1=(t+c)(t^2 −t+1) this is what I did, the rest follows

$$``\frac{\mathrm{t}^{\mathrm{2}} +\mathrm{pt}+\mathrm{1}}{\mathrm{t}^{\mathrm{2}} −\mathrm{t}+\mathrm{1}}\:\mathrm{can}\:\mathrm{be}\:\mathrm{written}\:\mathrm{as}\:\mathrm{t}+\mathrm{c}'' \\ $$$${means} \\ $$$$\frac{\mathrm{t}^{\mathrm{2}} +\mathrm{pt}+\mathrm{1}}{\mathrm{t}^{\mathrm{2}} −\mathrm{t}+\mathrm{1}}={t}+{c}\:\Leftrightarrow\:\mathrm{t}^{\mathrm{2}} +\mathrm{pt}+\mathrm{1}=\left({t}+{c}\right)\left(\mathrm{t}^{\mathrm{2}} −\mathrm{t}+\mathrm{1}\right) \\ $$$${this}\:{is}\:{what}\:{I}\:{did},\:{the}\:{rest}\:{follows} \\ $$

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.

$$\mathrm{Thanks},\:\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{that}.\:\mathrm{You} \\ $$$$\mathrm{can}\:\mathrm{also}\:\mathrm{look}\:\mathrm{up}\:\mathrm{other}\:\mathrm{questions}\:\mathrm{I} \\ $$$$\mathrm{posted}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com