- #1

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ya title is pretty much all i need (x^3 - 8)

if you could factor that it would be of great help

if you could factor that it would be of great help

- Thread starter whatdofisheat
- Start date

- #1

- 24

- 0

ya title is pretty much all i need (x^3 - 8)

if you could factor that it would be of great help

if you could factor that it would be of great help

- #2

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- 552

HINT:[tex]8=2^{3} [/tex]

Daniel.

Daniel.

- #3

- 789

- 4

I assume you know the formulas for the sum / difference of cubes, since your homework is asking a question that pertains to this method of factoring.

Rewrite your problem as [tex](x^3-2^3)[/tex]

Can you see it now?

Jameson

EDIT: In case you don't have the formula, I'll be nice... here you go.

[tex](x^3-y^3) = (x-y)(x^2+xy+y^2)[/tex]

Rewrite your problem as [tex](x^3-2^3)[/tex]

Can you see it now?

Jameson

EDIT: In case you don't have the formula, I'll be nice... here you go.

[tex](x^3-y^3) = (x-y)(x^2+xy+y^2)[/tex]

Last edited:

- #4

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but another methode we are trying to use is synthetic division

if any one can do it that way it would also help

thanks

fish

- #5

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x^{3} x^{2} x^{1} x^{0}

coeff. 1 0 0 -8

2 1 2 4 0

coeff. 1 0 0 -8

2 1 2 4 0

Solution

[tex] x^{3}-8=(x-2)\left(x^{2}+2x+4\right) [/tex]

Daniel.

Last edited:

- #6

uart

Science Advisor

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Yes well if you don't know the formula for factorization ofwhatdofisheat said:

but another methode we are trying to use is synthetic division

if any one can do it that way it would also help

thanks

fish

In order to use the division method you must first obtain one factor by some means, possibly guess. With the difference of two cubes,

Now just do the polynoimial division

Last edited:

- #7

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thanks for all your help

i got it now

i got it now

- #8

mathwonk

Science Advisor

Homework Helper

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For some reason this fact seems to be unknown to most first year college calculus students today.

- #9

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[tex] a^{3}-b^{3}=(a-b)\left(a^{2}+ab+b^{2}\right) [/tex]

Daniel.

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