Problem 36 - Algebra and Trigonometry

Question

Factor the polynomial and use the factored form to find the zeros. Then sketch the graph. $$ P(x)=x^{6}-2 x^{3}+1 $$

Step-by-Step Solution

Answer

Factored form: \((x - 1)^2(x^2 + x + 1)^2\) Zero: \(x = 1\)

Step 1: Recognize the Structure of the Polynomial

The polynomial \(P(x) = x^6 - 2x^3 + 1\) can be seen as a quadratic in form by letting \(y = x^3\). This transforms the polynomial into \(y^2 - 2y + 1\).

Step 2: Factor the Quadratic Expression

The expression \(y^2 - 2y + 1\) is a perfect square trinomial. It factors as \((y - 1)^2\). Substituting back \(y = x^3\), we have \((x^3 - 1)^2\).

Step 3: Factor the Expression Completely

Notice that \(x^3 - 1\) is a difference of cubes, which can be factored using the identity \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\). Thus, \(x^3 - 1 = (x - 1)(x^2 + x + 1)\).

Step 4: Combine the Factored Parts

Substitute back into \((x^3 - 1)^2\) to get the fully factored form: \[(x - 1)^2(x^2 + x + 1)^2\].

Step 5: Find the Zeros of the Polynomial

To find the zeros, set \((x - 1)^2(x^2 + x + 1)^2 = 0\). The zero at \(x - 1 = 0\) gives \(x = 1\). The expression \(x^2 + x + 1 = 0\) has no real zeros, as its discriminant \(b^2 - 4ac = 1^2 - 4 \times 1 \times 1 = -3\) is negative.

Step 6: Sketch the Graph

The graph of \(P(x) = x^6 - 2x^3 + 1\) touches the x-axis at \(x=1\) and does not cross it, since it has a zero with even multiplicity. The function increases rapidly as \(x\) moves away from zero in either direction, consistent with the leading term \(x^6\).

Key Concepts

Polynomial Zeros

Finding the zeros of a polynomial is crucial when solving an equation. Zeros, also known as roots or solutions, refer to the values of the variable that make the polynomial equal to zero.
In this exercise, we are tasked to find the zeros of the polynomial after factoring. Here, the polynomial given, \(P(x) = x^6 - 2x^3 + 1\), has been transformed and factored. The trick was to see it as a quadratic \((y^2 - 2y + 1)\) by temporarily setting \(y = x^3\).
Once factored completely, the expression becomes \((x - 1)^2(x^2 + x + 1)^2\).

• The factor \((x - 1)^2\) yields a zero at \(x = 1\).
• The other factor, \(x^2 + x + 1\), does not give real zeros since its discriminant is negative \(-3\).

This means the only real solution is \(x = 1\), despite there being other complex roots from \(x^2 + x + 1 = 0\). A zero of even multiplicity, like \(x = 1\) here, indicates that the graph touches the x-axis but doesn't cross it.

Graphing Polynomials

Understanding how to graph polynomials allows us to visualize their behavior and determine important characteristics, like intercepts and end behaviors. After we find the zeros, sketching the graph becomes more straightforward.
For the polynomial \(P(x) = x^6 - 2x^3 + 1\), it can help us see its shape and behavior. Here are some key points for graphing this particular polynomial:

• The zero at \(x = 1\), derived from the factor \((x - 1)^2\), signals the graph will touch the x-axis at this point but won't cross it.
• The polynomial \(x^6\) gives an idea of the end behavior. As \(x\) becomes very large or very small, the graph leans upwards (ends up) on both sides.
• This even-degree polynomial line will symmetrically reflect around the y-axis, though it’s based on the detailed polynomial factorization.
• Since \(x^2 + x + 1\) contributes no real zeros, the graph stays above the x-axis for other parts of the curve.

The graph, hence, lightly touches at \(x=1\) and displays a consistent rise with steep curves on each end.

Difference of Cubes

The difference of cubes is a special factorization identity useful in solving polynomial equations. It appears when rewriting expressions that take the form \(a^3 - b^3\).
The general factorization for a difference of cubes is:

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

In our problem, \(x^3 - 1\) fits as a difference of cubes where \(a = x\) and \(b = 1\), factoring into:

• \((x - 1)(x^2 + x + 1)\)

This step was crucial to reach the fully factored form \((x^3 - 1)^2\) and understand both the polynomial's real and complex roots.
Factoring in this way allows the identification of both linear and quadratically involved roots, providing a clearer understanding of how the polynomial behaves. Such factorization pieces, especially differences of cubes, offer insights into solutions and graphing beyond simple expressions.