Sjmarf@sh.itjust.works to Math Memes@lemmy.blahaj.zoneEnglish · 29 days agoSimplifysh.itjust.worksimagemessage-square75fedilinkarrow-up1330
arrow-up1330imageSimplifysh.itjust.worksSjmarf@sh.itjust.works to Math Memes@lemmy.blahaj.zoneEnglish · 29 days agomessage-square75fedilink
minus-squarethreelonmusketeers@sh.itjust.workslinkfedilinkEnglisharrow-up8·28 days agoCouldn’t you combine a lot of like terms as you went along, though? A polynomial of the order x26 would only have 27 terms.
minus-squareLostXOR@fedia.iolinkfedilinkarrow-up2·28 days agoNo, because each coefficient is its own variable; they’re not constants.
minus-squarethreelonmusketeers@sh.itjust.workslinkfedilinkEnglisharrow-up1·edit-227 days agoHuh, I’m so used to polynomials being in the form ax^2 + bx + c that I never considered that every letter might be a variable.
minus-square💡𝚂𝗆𝖺𝗋𝗍𝗆𝖺𝗇 𝙰𝗉𝗉𝗌📱@programming.devlinkfedilinkEnglisharrow-up1·18 days ago because each coefficient There’s only 1 coefficient - in this case it’s (a-x) - the rest are just factors. they’re not constants They could be - we haven’t been given that information.
Couldn’t you combine a lot of like terms as you went along, though? A polynomial of the order x26 would only have 27 terms.
No, because each coefficient is its own variable; they’re not constants.
Huh, I’m so used to polynomials being in the form ax^2 + bx + c that I never considered that every letter might be a variable.
There’s only 1 coefficient - in this case it’s (a-x) - the rest are just factors.
They could be - we haven’t been given that information.