Can someone help me factor 16x^2+64?
回答 (6)
2009-11-15 11:34 am
✔ 最佳答案
16x^2+64=16(x^2+4) answer//
2009-11-15 7:39 pm
As the other answers say, over the reals the only factor you can take out is 16. However, if you're allowed to factorise over the complex numbers as well, then you can use the difference of squares factorisation.
16x^2+64 = 16(x^2 + 4) = 16(x^2 + 2^2) = 16(x^2 - (2i)^2) = 16(x - 2i)(x + 2i)
16x^2+64 = 16(x^2 + 4) = 16(x^2 + 2^2) = 16(x^2 - (2i)^2) = 16(x - 2i)(x + 2i)
2009-11-15 8:39 pm
16x^2 + 64
= 16(x^2) + 16(4)
= 16(x^2 + 4)
If imaginary number is included, then:
16(x^2 + 4)
= 16[x^2 - (-4)]
= 16[x^2 - (2i)^2]
= 16(x + 2i)(x - 2i)
NB:
i = â(-1)
= 16(x^2) + 16(4)
= 16(x^2 + 4)
If imaginary number is included, then:
16(x^2 + 4)
= 16[x^2 - (-4)]
= 16[x^2 - (2i)^2]
= 16(x + 2i)(x - 2i)
NB:
i = â(-1)
2009-11-15 7:28 pm
16x² + 64 = 16(x² + 4)
That's it. It does not contain any smaller factors.
[Oh, OK, 16 factorises to 2^4, so you could say the complete factorisation is
(2^4) (x² + 4) ]
That's it. It does not contain any smaller factors.
[Oh, OK, 16 factorises to 2^4, so you could say the complete factorisation is
(2^4) (x² + 4) ]
2009-11-15 7:28 pm
erm, i have forgotten my algebra lessons but i guess i can have a try:
: 16x^2+64
answer: 16(x^2+4)
to check:
let x be 2,
given is : 16(2)^2+64 =128
our answer is: 16(x^2+4)=128
thus 128=128
: 16x^2+64
answer: 16(x^2+4)
to check:
let x be 2,
given is : 16(2)^2+64 =128
our answer is: 16(x^2+4)=128
thus 128=128
2009-11-15 7:19 pm
Divide both sides by 16.
x^2=4
Square root both sides.
x = +/- 2
x^2=4
Square root both sides.
x = +/- 2
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