# Can some one please factor (x^3 - 8)

• whatdofisheat
In summary, the conversation revolved around factoring the polynomial x^3 - 8 and the different methods to do so. The formula for factoring the difference of two cubes was provided, but the conversation also mentioned the use of synthetic division and the root-factor theorem. Ultimately, it was suggested to use polynomial division to obtain the other factor of the polynomial. The conversation also questioned the importance of knowing the root-factor theorem compared to other theorems when graduating high school.
whatdofisheat
ya title is pretty much all i need (x^3 - 8)
if you could factor that it would be of great help

HINT:$$8=2^{3}$$

Daniel.

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 $$(x^3-2^3)$$

Can you see it now?

Jameson

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

$$(x^3-y^3) = (x-y)(x^2+xy+y^2)$$

Last edited by a moderator:
thanks for the formula i have never seen that before
but another methode we are trying to use is synthetic division
if anyone can do it that way it would also help
thanks
fish

x^{3} x^{2} x^{1} x^{0}
coeff. 1 0 0 -8
2 1 2 4 0​

Solution

$$x^{3}-8=(x-2)\left(x^{2}+2x+4\right)$$

Daniel.

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whatdofisheat said:
thanks for the formula i have never seen that before
but another methode we are trying to use is synthetic division
if anyone can do it that way it would also help
thanks
fish

Yes well if you don't know the formula for factorization of difference of two cubes then polynomial division is a good way to proceed.

In order to use the division method you must first obtain one factor by some means, possibly guess. With the difference of two cubes, x^3 - a^3, it's very easy to see that x=a is a zero and hence (x-a) is a factor. So essentially you obtain this first factor by inspection in this case.

Now just do the polynoimial division (x^3 - a^3) / (x-a) to obtain the other less obvious factor.

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i got it now

A basic result once taught early in high school, and called the "root-factor theorem", is that whenever x=a makes a polynomial equal to zero (i.e. if a is a "root"), then x-a is a factor of that polynomial.

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

I wonder which is more important to know when u graduate HS:the root factor theorem (why this theorem and not others) or

$$a^{3}-b^{3}=(a-b)\left(a^{2}+ab+b^{2}\right)$$

Daniel.

## 1. What does it mean to factor a polynomial?

Factoring a polynomial means finding the expressions that can be multiplied together to get the original polynomial. It is essentially the reverse process of multiplying polynomials.

## 2. How do you factor a polynomial with a degree higher than 2?

To factor a polynomial with a degree higher than 2, such as (x^3 - 8), you can use the difference of cubes formula. In this case, the factors would be (x - 2) and (x^2 + 2x + 4).

## 3. Can you factor a polynomial with complex or imaginary roots?

Yes, you can factor a polynomial with complex or imaginary roots by using the quadratic formula or by using the quadratic formula method.

## 4. What is the purpose of factoring a polynomial?

Factoring a polynomial can help us simplify complex expressions and solve equations more easily. It can also help us identify important characteristics of the polynomial, such as its roots and degree.

## 5. Are there any other methods for factoring polynomials?

Yes, there are other methods for factoring polynomials, such as the difference of squares formula and the grouping method. It is important to choose the most efficient method based on the characteristics of the polynomial.

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