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The polynomial [tex]\( x^3 + 8 \)[/tex] is equal to:

A. [tex]\((x + 2)(x^2 - 2x + 4)\)[/tex]

B. [tex]\((x - 2)(x^2 + 2x + 4)\)[/tex]

C. [tex]\((x + 2)(x^2 - 2x + 8)\)[/tex]

D. [tex]\((x - 2)(x^2 + 2x + 8)\)[/tex]

Sagot :

GinnyAnswer
To factor the polynomial [tex]\( x^3 + 8 \)[/tex], we need to recognize it as a sum of cubes. The sum of cubes formula is:

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

In this problem, [tex]\( a = x \)[/tex] and [tex]\( b = 2 \)[/tex], because [tex]\( x^3 + 8 \)[/tex] can be rewritten as [tex]\( x^3 + 2^3 \)[/tex].

Using the formula:
- [tex]\( a = x \)[/tex]
- [tex]\( b = 2 \)[/tex]

Plug these values into the sum of cubes formula:

[tex]\[ x^3 + 2^3 = (x + 2)\left(x^2 - x \cdot 2 + 2^2\right) \][/tex]

Simplify inside the parentheses:

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

So, the factored form of the polynomial [tex]\( x^3 + 8 \)[/tex] is:

[tex]\[ (x + 2)\left(x^2 - 2x + 4\right) \][/tex]

Therefore, the correct answer is:

[tex]\[ (x+2)\left(x^2-2x+4\right) \][/tex]
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