Question

unit factoring
progress:
the movement of the progress bar may be uneven because questions can be worth more or less (including zero) depending on your answer.
question 10 bim2
which of the following factorizations is correct?
○ $x^3 - 27 = (x - 3)(x^2 + 9)$
○ $x^3 + 27 = (x - 3)(x^2 + 3x - 9)$
○ $x^3 + 27 = (x - 3)(x^2 - 9)$
○ $x^3 - 27 = (x - 3)(x^2 + 3x + 9)$

Explanation:

Step1: Recall the formula for difference of cubes

The formula for factoring a difference of cubes is \(a^3 - b^3=(a - b)(a^2+ab + b^2)\).
Here, \(a = x\) and \(b = 3\) since \(3^3=27\). So \(x^3-27=x^3 - 3^3\).

Step2: Apply the formula

Using the difference of cubes formula \(a^3 - b^3=(a - b)(a^2+ab + b^2)\) with \(a=x\) and \(b = 3\), we get:
\(x^3-3^3=(x - 3)(x^2+(x)(3)+3^2)=(x - 3)(x^2 + 3x+9)\)
Let's check the other options:

• For \(x^3 - 27=(x - 3)(x^2 + 9)\): Expanding \((x - 3)(x^2 + 9)=x^3+9x-3x^2 - 27

eq x^3-27\)

• For \(x^3 + 27=(x - 3)(x^2+3x - 9)\): \(x^3 + 27\) is a sum of cubes (\(x^3+3^3\)) and the formula for sum of cubes is \(a^3 + b^3=(a + b)(a^2 - ab + b^2)\), also expanding \((x - 3)(x^2+3x - 9)=x^3+3x^2-9x-3x^2-9x + 27=x^3-18x + 27

eq x^3 + 27\)

• For \(x^3 + 27=(x - 3)(x^2 - 9)\): Expanding \((x - 3)(x^2 - 9)=x^3-9x-3x^2 + 27

eq x^3 + 27\)

Answer:

\(x^3 - 27=(x - 3)(x^2 + 3x + 9)\) (the last option in the given choices)