Question

Factor each polynomial completely. If the polynomial cannot be factored, say it is prime. $$ x^{6}-2 x^{3}+1 $$

Short answer

The polynomial \(x^{6}-2x^{3}+1\) factors to \( (x-1)^2 (x^2+x+1)^2 \).

Step-by-step solution

Recognize the polynomial structure

Observe that the given polynomial is in the form of a quadratic in terms of \(x^3\): \(x^{6} - 2x^{3} + 1\). This can be rewritten as \( (x^3)^2 - 2(x^3) + 1 \).

Set a substitution

Let \(y = x^3\). Thus, the polynomial becomes \(y^2 - 2y + 1\).

Factor the quadratic expression

Factor the quadratic expression: \(y^2 - 2y + 1\). This can be written as \((y - 1)^2\).

Substitute back the original variable

Replace \(y\) with \(x^3\) to get \((x^3 - 1)^2\).

Factor the expression \(x^3-1\)

Recognize that \(x^3 - 1\) is a difference of cubes and can be factored as \((x-1)(x^2+x+1)\).

Write the complete factorization

Combine all steps to write the complete factorization of the original polynomial: \( (x-1)^2 (x^2+x+1)^2 \).

Key concepts

quadratic substitution

Quadratic substitution is a helpful technique for simplifying more complicated polynomial expressions. In our exercise, we start by recognizing the structure of the given polynomial: \( x^6 - 2x^3 + 1 \). It can be observed that we can treat this polynomial as a quadratic in terms of \( x^3 \).

To make the polynomial simpler, we set \( y = x^3 \). Our polynomial then transforms from \( (x^3)^2 - 2(x^3) + 1 \) to \( y^2 - 2y + 1 \). This substitution now makes it easier to factor because we are dealing with a quadratic equation.

difference of cubes

The difference of cubes is a special factoring formula that helps in breaking down cubic terms. The formula states that \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \). This is very useful in our exercise.

After substituting back the value of \( y \) with \( x^3 \), we get \( (x^3 - 1)^2 \). We can now recognize that \( x^3 - 1 \) is a difference of cubes since it can be written as \( x^3 - 1^3 \).

Applying the difference of cubes formula, we get \( x^3 - 1 = (x - 1)(x^2 + x + 1) \). Rewriting the expression, we have \( (x - 1)^2 (x^2 + x + 1)^2 \).

polynomial factorization

Polynomial factorization is the process of breaking down a polynomial into a product of simpler polynomials. This helps in solving polynomial equations and simplifying algebraic expressions. In the exercise, after the quadratic substitution, we reached the polynomial \( y^2 - 2y + 1 \).

This is a quadratic expression that can be factored using methods such as factoring by grouping or using the formula for perfect square trinomials. Here, we recognize that \( y^2 - 2y + 1 \) can be rewritten as \( (y - 1)^2 \).

After substituting back \( y = x^3 \), the expression becomes \( (x^3 - 1)^2 \), which we further break down using the difference of cubes formula.

algebra

Algebra is a branch of mathematics dealing with symbols and the rules for manipulating those symbols. It is all about finding the unknowns. Techniques such as substitution, factorization, and special formulas play a crucial role in solving algebraic expressions.

In our exercise, we used several core principles of algebra. First, we used quadratic substitution to transform a complex polynomial into a simpler form. Then, we used our understanding of the difference of cubes to further factorize the polynomial.

Through these steps, we successfully factored the complex polynomial \( x^6 - 2x^3 + 1 \) into \( (x - 1)^2 (x^2 + x + 1)^2 \).