Problem 22 - Algebra and Trigonometry

Question

Factor the polynomial completely, and find all its zeros. State the multiplicity of each zero. \(P(x)=x^{3}-x^{2}+x\)

Step-by-Step Solution

Answer

The zeros are \(x = 0\) (multiplicity 1), \(x = \frac{1 + i\sqrt{3}}{2}\) (multiplicity 1), and \(x = \frac{1 - i\sqrt{3}}{2}\) (multiplicity 1).

Step 1: Identify a Common Factor

In the polynomial \(P(x) = x^3 - x^2 + x\), we notice that each term contains the variable \(x\). We can factor \(x\) out of the polynomial:\[ P(x) = x(x^2 - x + 1) \]

Step 2: Factor the Quadratic (Check for Further Factorization)

The quadratic \(x^2 - x + 1\) is part of the expression after extracting the common factor. Check to see if \(x^2 - x + 1\) can be factored further. Since the discriminant \(b^2 - 4ac = (-1)^2 - 4(1)(1) = -3\) is negative, this quadratic doesn't factor further over the real numbers.

Step 3: Find the Zeros of the Function

To find the zeros of \(P(x) = x(x^2 - x + 1)\), set the expression equal to zero:1. \(x = 0\) comes from the factor \(x\).2. 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: \[x = \frac{1 \pm \sqrt{-3}}{2}\] \[x = \frac{1 \pm i\sqrt{3}}{2}\]These are non-real complex zeros.

Step 4: Determine the Multiplicity of Each Zero

The zero \(x = 0\) has a multiplicity of 1 because it appears as a factor only once in \(x(x^2 - x + 1)\).The complex zeros \(x = \frac{1 + i\sqrt{3}}{2}\) and \(x = \frac{1 - i\sqrt{3}}{2}\) each also have a multiplicity of 1 since they arise from the quadratic factor, which is of degree 2 and doesn't simplify further.

Key Concepts

Finding Zeros

Finding the zeros of a polynomial is like locating the points where the graph of the polynomial intersects the x-axis. These are the values of \(x\) for which the polynomial equals zero. For example, the polynomial \(P(x) = x^3 - x^2 + x\) can be rewritten as \(P(x) = x(x^2 - x + 1)\) after factoring out the common factor \(x\).

To find the zeros of the polynomial, set \(P(x) = 0\). This gives us two separate equations to solve:

• \(x = 0\)
• \(x^2 - x + 1 = 0\)

The first equation gives the zero \(x = 0\). For the second equation, we use the quadratic formula since it is a quadratic equation that doesn't factor easily.

Multiplicity of Zeros

The multiplicity of a zero tells us how many times a particular zero appears as a solution of the polynomial equation. It also indicates the behavior of the graph at that zero.

Consider \(P(x) = x(x^2 - x + 1)\):

• The zero \(x = 0\) has a multiplicity of 1 because it appears just once as a factor in the linear term \(x\). The graph touches the x-axis at this point and crosses it.

Next, we focus on the quadratic part, \(x^2 - x + 1\), which gives two complex roots: \(x = \frac{1 + i\sqrt{3}}{2}\) and \(x = \frac{1 - i\sqrt{3}}{2}\).

• Each of these complex zeros also has a multiplicity of 1 because they emerge from a factor of degree 2, which cannot be simplified further in the real number system.

Quadratic Formula

The quadratic formula is a vital tool for finding the zeros of any quadratic equation of the form \(ax^2 + bx + c = 0\). It is given by:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Applying the formula to the equation \(x^2 - x + 1 = 0\), you substitute \(a = 1\), \(b = -1\), and \(c = 1\) to get:

\[ x = \frac{1 \pm \sqrt{-3}}{2} \]

Because the discriminant \(b^2 - 4ac\) is negative, the square root of \(-3\) introduces complex numbers, resulting in the complex zeros \(x = \frac{1 \pm i\sqrt{3}}{2}\). This illustrates the power of the quadratic formula, which not only identifies simple real zeros but also detects complex solutions when the quadratic cannot be factored over the reals.