Problem 40 - Algebra and Trigonometry

Question

\(27-40\) 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

The factored form is \( (x-1)^2(x^2+x+1)^2 \); the only real zero is \( x=1 \).

Step 1: Recognize the Form

Notice that the given polynomial \( P(x) = x^6 - 2x^3 + 1 \) can be viewed as a quadratic polynomial in terms of \( u = x^3 \). Thus, rewrite the polynomial in terms of \( u \) as \( P(u) = u^2 - 2u + 1 \).

Step 2: Factor the Quadratic Expression

Recognize \( u^2 - 2u + 1 \) is a perfect square trinomial. Use the identity \( a^2 - 2ab + b^2 = (a-b)^2 \) to factor the expression as \( (u-1)^2 \). Hence, \( P(x) = (x^3 - 1)^2 \).

Step 3: Factor Further Using Difference of Cubes

Recognize \( x^3 - 1 \) as a difference of cubes, which can be factored using the formula \( a^3 - b^3 = (a-b)(a^2 + ab + b^2) \). Apply this to factor \( x^3 - 1 \) as \( (x-1)(x^2 + x + 1) \). Thus, \((x^3 - 1)^2 = (x-1)^2(x^2 + x + 1)^2 \).

Step 4: Find the Zeros

Determine the values of \( x \) that make the factors zero. For \((x-1)^2 = 0\), \( x = 1 \). For \( x^2 + x + 1 = 0 \), use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a=1, b=1, c=1 \), leading to no real solutions since the discriminant \( b^2 - 4ac = 1 - 4 = -3 \) is negative.

Step 5: Sketch the Graph Based on Zeros

Since \( x = 1 \) is a root of multiplicity 2, the graph will touch but not cross the x-axis at \( x = 1 \). The factor \( (x^2 + x + 1)^2 \) contributes no real zeros but ensures the polynomial remains positive for other real values. The graph is symmetric around the y-axis since all even powers are present.

Key Concepts

Zeros of Polynomial

The zeros of a polynomial, also known as roots, are the values of the variable that make the polynomial equation equal to zero. In simpler terms, these are the points where the graph of the polynomial will intersect or touch the x-axis. For the polynomial we are dealing with, \[ P(x) = x^6 - 2x^3 + 1 \]factoring helped us find these zeros. By transforming our polynomial into the factorized form \((x-1)^2(x^2 + x + 1)^2\), we identified that the polynomial has a real zero at \(x = 1\), which occurs with a multiplicity of 2.

• The factor \((x-1)^2 = 0\) gives \(x = 1\) as a zero.
• Other factors, \((x^2 + x + 1)\), yield no real zeroes because the quadratic equation results in a negative discriminant.

Understanding zeros, especially their multiplicities, is essential as they influence the behavior of the polynomial graph. In this example, the graph touches but does not cross the x-axis at \(x = 1\), due to the double root.

Sketching Graphs

Sketching the graph of a polynomial involves plotting its zeros and understanding the general shape and behavior the polynomial adopts between and beyond these zeros. For the polynomial in our example, once the zeros were determined, sketching became a task of representing these insights graphically. The polynomial \( P(x) = x^6 - 2x^3 + 1 \) was transformed into the factorized form \((x-1)^2(x^2 + x + 1)^2\), highlighting a single real zero at \(x = 1\). Due to this being a repeated zero, the graph just touches the x-axis at this point without crossing it.

• The graph remains above the x-axis for other real x-values, because the factor \((x^2 + x + 1)^2\) includes no real zeros.
• Given the polynomial has even powers, the graph is symmetric around the y-axis.

As such, sketching involves drawing a curve that hugs the x-axis at \(x = 1\) and arches upwards, hinting at its minimum, and gracefully follows a symmetric path mirrored along the y-axis.

Quadratic Formula

The quadratic formula is a powerful tool that helps find the zeros of any quadratic equation of the form \(ax^2 + bx + c = 0\). The formula given by \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]is particularly helpful when the quadratic does not factor neatly. In our polynomial solving journey, it aided in determining that \[ x^2 + x + 1 = 0 \]lacks real zeros. Here, choosing \(a = 1, b = 1,\) and \(c = 1\) resulted in a discriminant \(b^2 - 4ac = -3\), which is negative, indicating the absence of real roots.

• The discriminant ( \(b^2 - 4ac\)) decides the nature of the roots: positive for two real roots, zero for one real root, and negative for complex roots.
• Since the quadratic part of our polynomial has no real solution, these complex roots do not affect our real-number sketch of the polynomial.

Grasping the quadratic formula and its implications helps learners understand not only how to find roots but also predict the graphical representation of polynomials.