Algebra Practice Test 2025 – Complete Exam Preparation

Question: 1 / 400

What is the factorization of the polynomial x^3 - 27?

(x - 3)(x² + 3x + 9)

To understand how to factor the polynomial \(x^3 - 27\), it is helpful to recognize that this expression is a difference of cubes. The general form for factoring a difference of cubes is:

\[

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

\]

In this case, we can identify \(a\) as \(x\) and \(b\) as \(3\) because \(27\) can be expressed as \(3^3\). Therefore, we have:

\[

x^3 - 3^3 = (x - 3)(x^2 + x \cdot 3 + 3^2)

\]

Now, let's break it down:

1. The first term is \(x - 3\).

2. Next, for the second term \(x^2 + x \cdot 3 + 3^2\), we compute it:

- The middle term becomes \(3x\).

- The last term is \(9\).

As a result, we arrive at:

\[

x^3 - 27 = (x - 3)(x^2 + 3x

Get further explanation with Examzify DeepDiveBeta

(x + 3)(x² - 3x + 9)

(x - 3)(x² - 3x - 9)

(x + 3)(x² + 3x + 9)

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