Question-144061

Question Number 144061 by Khalmohmmad last updated on 21/Jun/21

Answered by TheHoneyCat last updated on 21/Jun/21

I am quite sorry to tell you this, but your question is either incomplete (some hypothesis is missing) or unsolvable. Consider: P_1 =3X+8 =0×(X−1)^2 +(3X+8) =3×(X−2) +14 and P_2 =(X−1)^2 +3X+8 =1×(X−1)^2 +(3X+8) =X^2 −2X+1+3X+8 =X×(X−2)+3X+9 =(X+3)×(X−3) + 15 15≠14 The value you are looking for depends on the explicit value of your polynomial if you, at least, gave the value of Q such that P=Q×(X−1)^2 +3X+8 I could tell you that: P=Q×(X^2 −2X+1)+3X+8 =Q×X×(X−2)+Q+3X+8 =(Q×X + 1)×(X−2)+Q+14 so the value you would be looking for would be (Q mod (X−2)) +14 _(deg14=0<deg(X−2)=1) Sorry :∙(

$$\mathrm{I}\:\mathrm{am}\:\mathrm{quite}\:\mathrm{sorry}\:\mathrm{to}\:\mathrm{tell}\:\mathrm{you}\:\mathrm{this},\:\mathrm{but}\:\mathrm{your}\:\mathrm{question}\:\mathrm{is}\:\mathrm{either} \\ $$$$\mathrm{incomplete}\:\left(\mathrm{some}\:\mathrm{hypothesis}\:\mathrm{is}\:\mathrm{missing}\right) \\ $$$$\mathrm{or}\:\mathrm{unsolvable}. \\ $$$$ \\ $$$$\mathrm{Consider}: \\ $$$${P}_{\mathrm{1}} =\mathrm{3}{X}+\mathrm{8} \\ $$$$=\mathrm{0}×\left({X}−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{3}{X}+\mathrm{8}\right) \\ $$$$=\mathrm{3}×\left({X}−\mathrm{2}\right)\:+\mathrm{14} \\ $$$$\mathrm{and} \\ $$$${P}_{\mathrm{2}} =\left({X}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}{X}+\mathrm{8} \\ $$$$=\mathrm{1}×\left({X}−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{3}{X}+\mathrm{8}\right) \\ $$$$={X}^{\mathrm{2}} −\mathrm{2}{X}+\mathrm{1}+\mathrm{3}{X}+\mathrm{8} \\ $$$$={X}×\left({X}−\mathrm{2}\right)+\mathrm{3}{X}+\mathrm{9} \\ $$$$=\left({X}+\mathrm{3}\right)×\left({X}−\mathrm{3}\right)\:+\:\mathrm{15} \\ $$$$ \\ $$$$\mathrm{15}\neq\mathrm{14} \\ $$$$\mathrm{The}\:\mathrm{value}\:\mathrm{you}\:\mathrm{are}\:\mathrm{looking}\:\mathrm{for}\:\mathrm{depends}\:\mathrm{on}\:\mathrm{the}\:\mathrm{explicit}\:\mathrm{value}\:\mathrm{of}\:\mathrm{your}\:\mathrm{polynomial} \\ $$$$ \\ $$$$\mathrm{if}\:\mathrm{you},\:\mathrm{at}\:\mathrm{least},\:\mathrm{gave}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{Q}\:\mathrm{such}\:\mathrm{that} \\ $$$${P}={Q}×\left({X}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}{X}+\mathrm{8} \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{could}\:\mathrm{tell}\:\mathrm{you}\:\mathrm{that}: \\ $$$${P}={Q}×\left({X}^{\mathrm{2}} −\mathrm{2}{X}+\mathrm{1}\right)+\mathrm{3}{X}+\mathrm{8} \\ $$$$={Q}×{X}×\left({X}−\mathrm{2}\right)+{Q}+\mathrm{3}{X}+\mathrm{8} \\ $$$$=\left({Q}×{X}\:+\:\mathrm{1}\right)×\left({X}−\mathrm{2}\right)+{Q}+\mathrm{14} \\ $$$$ \\ $$$$\mathrm{so}\:\mathrm{the}\:\mathrm{value}\:\mathrm{you}\:\mathrm{would}\:\mathrm{be}\:\mathrm{looking}\:\mathrm{for}\:\mathrm{would}\:\mathrm{be} \\ $$$$\left({Q}\:\mathrm{mod}\:\left({X}−\mathrm{2}\right)\right)\:+\mathrm{14}\:_{\mathrm{deg14}=\mathrm{0}<\mathrm{deg}\left({X}−\mathrm{2}\right)=\mathrm{1}} \\ $$$$ \\ $$$${Sorry}\: \\ $$$$:\centerdot\left(\right. \\ $$

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