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Question Number 215469 by Abdullahrussell last updated on 08/Jan/25
Answered by alephnull last updated on 08/Jan/25
(122+133)=(14+127)=(27108+4108)therootis31108sumofcoeffecients=P(1)letQ(n)=108⋅P(n)P(1)=Q(1)108letQ(n)=a6n6+a5n5...+a1n+a0P(1)=Q(1)/108didyouprovideaspecificpolynomial?ifnotthesumdependsonQ(n)
Commented by Abdullahrussell last updated on 08/Jan/25
212+313=2+33
Answered by mr W last updated on 08/Jan/25
sayP(x)=a0+a1x+a2x2+...+a6x6witha6=1P(212+313)=∑6n=0an(212+313)n=0(212+313)1=212+313(212+313)2=2+2×212×313+323(212+313)3=3+2×212+6×313+3×212×323(212+313)4=4+8×212×313+12×323+12×212+3×313(212+313)5=60+4×212+20×313+20×212×323+15×212×313+3×323(212+313)6=17+24×212×313+60×323+120×212+90×313+18×212×323n∖1212313323212×313212×323(...)01a0(...)111a1(...)2212a2(...)33263a3(...)44123128a4(...)56042031520a5(...)61712090602418a0+2a2+3a3+4a4+60a5+17=0a1+2a3+12a4+4a5+120=0a1+6a3+3a4+20a5+90=0a2+12a4+3a5+60=02a2+8a4+15a5+24=03a3+20a5+18=0solvingthisequationsystemwegeta0=1,a1=−36,a2=12,a3=−6,a4=−6,a5=0,a6=1(given)sumofallcoefficients∑6n=0an=−34P(x)=x6−6x4−6x3+12x2−36x+1
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