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2 years ago

(-1)×(-2)×10=? (this is integers)

Mathematics

1 answer:

VLD [36.1K]2 years ago

8 0

Answer:the first one

Step-by-step explanation:

i am teacher lol

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X^2 + X - 12 = 0

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2 years ago

Read 2 more answers

Factor completely. <br> <img src="https://tex.z-dn.net/?f=x%5E%7B8%7D-%5Cfrac%7B1%7D%7B81%7D" id="TexFormula1" title="x^{8}-\fra

Eduardwww [97]

We have 3⁴ = 81, so we can factorize this as a difference of squares twice:

$x^8 - \dfrac1{81} = \left(x^2\right)^4 - \left(\dfrac13\right)^4 \\\\ x^8 - \dfrac1{81} = \left(\left(x^2\right)^2 - \left(\dfrac13\right)^2\right) \left(\left(x^2\right)^2 + \left(\dfrac13\right)^2\right) \\\\ x^8 - \dfrac1{81} = \left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(\left(x^2\right)^2 + \left(\dfrac13\right)^2\right) \\\\ x^8 - \dfrac1{81} = \left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(x^4 + \dfrac19\right)$

Depending on the precise definition of "completely" in this context, you can go a bit further and factorize $x^2-\frac13$ as yet another difference of squares:

$x^2 - \dfrac13 = x^2 - \left(\dfrac1{\sqrt3}\right)^2 = \left(x-\dfrac1{\sqrt3}\right)\left(x+\dfrac1{\sqrt3}\right)$

And if you're working over the field of complex numbers, you can go even further. For instance,

$x^4 + \dfrac19 = \left(x^2\right)^2 - \left(i\dfrac13\right)^2 = \left(x^2 - i\dfrac13\right) \left(x^2 + i\dfrac13\right)$

But I think you'd be fine stopping at the first result,

$x^8 - \dfrac1{81} = \boxed{\left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(x^4 + \dfrac19\right)}$

6 0

3 years ago

(-1)×(-2)×10=? (this is integers)​

(-1)×(-2)×10=? (this is integers)