Sjmarf@sh.itjust.works to Math Memes@lemmy.blahaj.zoneEnglish · 6 个月前Simplifysh.itjust.worksimagemessage-square75linkfedilinkarrow-up1340arrow-down111
arrow-up1329arrow-down1imageSimplifysh.itjust.worksSjmarf@sh.itjust.works to Math Memes@lemmy.blahaj.zoneEnglish · 6 个月前message-square75linkfedilink
minus-squarethreelonmusketeers@sh.itjust.workslinkfedilinkEnglisharrow-up8·6 个月前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.
minus-squareLostXOR@fedia.iolinkfedilinkarrow-up2·6 个月前No, because each coefficient is its own variable; they’re not constants.
minus-square💡𝚂𝗆𝖺𝗋𝗍𝗆𝖺𝗇 𝙰𝗉𝗉𝗌📱@programming.devlinkfedilinkEnglisharrow-up1·5 个月前 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.
minus-squarethreelonmusketeers@sh.itjust.workslinkfedilinkEnglisharrow-up1·edit-26 个月前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.
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.
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.
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.