Answer:
[tex]=\left(x-4\right)^2[/tex]
Step-by-step explanation:
[tex]x^2-8x+16\\\mathrm{Rewrite\:}x^2-8x+16\mathrm{\:as\:}x^2-2x\cdot \:4+4^2\\x^2-8x+16\\\mathrm{Rewrite\:}16\mathrm{\:as\:}4^2\\=x^2-8x+4^2\\\mathrm{Rewrite\:}8x\mathrm{\:as\:}2x\cdot \:4\\=x^2-2x\cdot \:4+4^2\\\mathrm{Apply\:Perfect\:Square\:Formula}:\quad \left(a-b\right)^2=a^2-2ab+b^2\\a=x,\:b=4\\=\left(x-4\right)^2[/tex]
Answer:
0
Step-by-step explanation:
Since it is asking you to factor out the quadratic, you need to make it equal zero first. -4 times -4 is both 16 and -8. So that means (x-4)^2
