Factorize 2x²+y²+8z²-2√2xy+4√2yz-8xz.
Find the value when x=1,y=1,z=-1
Please solve both parts. Will mark brainliest asap!
Answers
Step by step explanation:
(i) 2x2+y2+8z2–2√2xy+4√2yz–8xz
) 2x2+y2+8z2–2√2xy+4√2yz–8xzUsing identity, (x +y+z)2 = x2+y2+z2+2xy+2yz+2zx
) 2x2+y2+8z2–2√2xy+4√2yz–8xzUsing identity, (x +y+z)2 = x2+y2+z2+2xy+2yz+2zxWe can say that, x2+y2+z2+2xy+2yz+2zx = (x+y+z)2
) 2x2+y2+8z2–2√2xy+4√2yz–8xzUsing identity, (x +y+z)2 = x2+y2+z2+2xy+2yz+2zxWe can say that, x2+y2+z2+2xy+2yz+2zx = (x+y+z)22x2+y2+8z2–2√2xy+4√2yz–8xz
) 2x2+y2+8z2–2√2xy+4√2yz–8xzUsing identity, (x +y+z)2 = x2+y2+z2+2xy+2yz+2zxWe can say that, x2+y2+z2+2xy+2yz+2zx = (x+y+z)22x2+y2+8z2–2√2xy+4√2yz–8xz= (-√2x)2+(y)2+(2√2z)2+(2×-√2x×y)+(2×y×2√2z)+(2×2√2×−√2x)
) 2x2+y2+8z2–2√2xy+4√2yz–8xzUsing identity, (x +y+z)2 = x2+y2+z2+2xy+2yz+2zxWe can say that, x2+y2+z2+2xy+2yz+2zx = (x+y+z)22x2+y2+8z2–2√2xy+4√2yz–8xz= (-√2x)2+(y)2+(2√2z)2+(2×-√2x×y)+(2×y×2√2z)+(2×2√2×−√2x)= (−√2x+y+2√2z)2
) 2x2+y2+8z2–2√2xy+4√2yz–8xzUsing identity, (x +y+z)2 = x2+y2+z2+2xy+2yz+2zxWe can say that, x2+y2+z2+2xy+2yz+2zx = (x+y+z)22x2+y2+8z2–2√2xy+4√2yz–8xz= (-√2x)2+(y)2+(2√2z)2+(2×-√2x×y)+(2×y×2√2z)+(2×2√2×−√2x)= (−√2x+y+2√2z)2= (−√2x+y+2√2z)(−√2x+y+2√2z)