FACTOR COMPLETELY s^2 + 16.?
回答 (7)
2009-11-18 1:30 pm
✔ 最佳答案
s^2+16 is prime, you can not factor any further. This is a question meant to trap you into a common mistake - you can easily factor a difference of squares, students commonly try to do the same thing with a sum of squares.
2009-11-18 9:28 pm
s^2 + 16 = s^2 + 4^2
does that help?
does that help?
2009-11-18 9:38 pm
s^2 + 16 is already factored over the integers.
If complex numbers are allowed we would have (s+4i)(s-4i).
If complex numbers are allowed we would have (s+4i)(s-4i).
2009-11-18 9:35 pm
s^2 + 16
= s^2 - (-16)
= s^2 - (-1 * 16)
= s^2 - (i^2 * 4^2)
= s^2 - (4i)^2
= (s + 4i)(s - 4i)
= s^2 - (-16)
= s^2 - (-1 * 16)
= s^2 - (i^2 * 4^2)
= s^2 - (4i)^2
= (s + 4i)(s - 4i)
2009-11-18 9:31 pm
s^2+16=(s+4i)(s-4i)
2009-11-18 9:31 pm
s² + 16 = 0
Use the quadratic formula to factor
   as²+bs+c=0
where
   a=1
   b=0
   c=16
s = [-b ± â(b²-4ac)] / (2a)
   = [-(0) ± â((0)² - 4(1)(16))] / (2·1)
   = [0 ± â(-64)] / 2
   = [0 ± â64â(-1)] / 2
   = [0 ± 8i] / 2
   â 0 ± 4i
   = -4i, 4i
s² + 16 = (s + 4i)(s - 4i)
Use the quadratic formula to factor
   as²+bs+c=0
where
   a=1
   b=0
   c=16
s = [-b ± â(b²-4ac)] / (2a)
   = [-(0) ± â((0)² - 4(1)(16))] / (2·1)
   = [0 ± â(-64)] / 2
   = [0 ± â64â(-1)] / 2
   = [0 ± 8i] / 2
   â 0 ± 4i
   = -4i, 4i
s² + 16 = (s + 4i)(s - 4i)
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