Factor Polynomial: x^3-27, Find A, B, C

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SUMMARY

The polynomial $$x^{3}-27$$ can be factored using the difference of cubes formula. By recognizing it as $$x^3 - 3^3$$, the factorization is established as $$(x - 3)(x^{2} + 3x + 9)$$. The values of A, B, and C are definitively A = 3, B = 3, and C = 9. This method simplifies the process of polynomial factorization significantly.

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cchacon92
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Hello, for homework I was given $$x^{3}-27$$ to factor and my answer can be written this way:

$$(x - A)(x^{2} + Bx + C)$$

I am being asked to find the value of A, B and C.
I am sure this is simpler than I think but it's perhaps the Bx that is confusing me as I haven't practiced a lot of these examples yet.
 
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Hint: Write the polynomial as the difference of cubes $x^3-3^3$ then use the difference of cubes formula $a^3-b^3=(a-b)(a^2+ab+b^2)$.
 
MarkFL said:
Hint: Write the polynomial as the difference of cubes $x^3-3^3$ then use the difference of cubes formula $a^3-b^3=(a-b)(a^2+ab+b^2)$.

Alright, so I obtained $$(x -3)(x^{2} + 3x + C)$$

I still can't find C
 
In the formula I gave for the difference of cubes, you want to substitute $x$ for $a$, and $3$ for $b$. So what does this mean $C$ is equal to?
 
9 ! Great
 
Yes, good work! (Yes)
 

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