Problem 3 - Algebra and Trigonometry

Question

\(1-12\) . A polynomial \(P\) is given. (a) Find all zeros of \(P\) , real and complex. (b) Factor \(P\) completely. $$ P(x)=x^{3}-2 x^{2}+2 x $$

Step-by-Step Solution

Answer

Zeros: 0, 1 + i, 1 - i; Factored form: \( x(x - (1+i))(x - (1-i)) \).

Step 1: Identify the Polynomial and Set It to Zero

We start with the polynomial \( P(x) = x^3 - 2x^2 + 2x \). To find the zeros of the polynomial, we set \( P(x) = 0 \). Thus, we need to solve \( x^3 - 2x^2 + 2x = 0 \).

Step 2: Factor the Polynomial

Notice that all terms in the polynomial share a common factor of \( x \). Factor \( x \) out of the polynomial: \[ x(x^2 - 2x + 2) = 0 \]. This gives us two factors to consider: \( x = 0 \) and \( x^2 - 2x + 2 = 0 \).

Step 3: Solve the Simple Linear Equation

The factor \( x = 0 \) gives us one zero of the polynomial directly: \( x = 0 \).

Step 4: Solve the Quadratic Equation

Now, solve the quadratic equation \( x^2 - 2x + 2 = 0 \). Use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \], where \( a = 1 \), \( b = -2 \), and \( c = 2 \).

Step 5: Calculate Using the Quadratic Formula

Substitute \( a = 1 \), \( b = -2 \), and \( c = 2 \) into the quadratic formula: \[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} \]. Simplify inside the square root: \[ x = \frac{2 \pm \sqrt{4 - 8}}{2} \], which becomes \( x = \frac{2 \pm \sqrt{-4}}{2} \).

Step 6: Simplify the Expression

Simplify further: since \( \sqrt{-4} = 2i \), we have \[ x = \frac{2 \pm 2i}{2} \]. This results in two complex solutions: \( x = 1 + i \) and \( x = 1 - i \).

Step 7: Factor the Polynomial Completely

Rewrite the polynomial using its zeros: \[ P(x) = x(x - (1 + i))(x - (1 - i)) \].To confirm the correctness, expand these factors back into polynomial form to check whether we get the original polynomial. However, this form represents the polynomial factored over the complex numbers.

Key Concepts

Complex Numbers

Complex numbers extend the idea of one-dimensional numbers on a number line to a two-dimensional plane. This is achieved by incorporating an imaginary unit, denoted as \(i\). The fundamental characteristic of \(i\) is that \(i^2 = -1\). This allows the expression of complex numbers in the standard form \(a + bi\), where:

• \(a\) is the real part
• \(b\) is the imaginary part

Complex numbers are crucial in solving equations that do not have real number solutions, such as \(x^2 = -1\). In our specific problem, solving the quadratic equation involved finding solutions for \(x^2 - 2x + 2 = 0\). On simplification, this step led to the appearance of \(\sqrt{-4}\), resulting in complex roots. Hence, the solutions were \(x = 1 + i\) and \(x = 1 - i\). These solutions demonstrate how complex numbers help extend the solutions beyond real numbers.

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\). No matter what values \(a\), \(b\), or \(c\) take (as long as \(a eq 0\)), you can find the solutions for \(x\) given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]In this solution process, you first calculate the discriminant \(D = b^2 - 4ac\), which decides the nature of the roots:

• If \(D > 0\), we have two distinct real solutions.
• If \(D = 0\), there's exactly one real solution (a repeated root).
• If \(D < 0\), two complex solutions exist.

In our example, the discriminant became \(4 - 8 = -4\), indicating complex roots. Substituting back into the formula yielded the complex solutions \(x = 1 + i\) and \(x = 1 - i\), illustrating the utility of the quadratic formula in scenarios involving non-real solutions.

Factoring Polynomials

Factoring polynomials is the process of expressing a polynomial as a product of its factors. This technique is useful for solving polynomial equations and simplifying expressions. The goal is to break down a complex polynomial into simpler factors, which reveal the polynomial's roots.When factoring the polynomial \(P(x) = x^3 - 2x^2 + 2x\), the first observation is a common factor among all terms, which is \(x\). Factoring this out leads to \(x(x^2 - 2x + 2) = 0\). This gives us one immediate solution at \(x = 0\). The remaining quadratic \(x^2 - 2x + 2\), although not easily factorable over the reals, was tackled using the quadratic formula to identify complex roots.After finding all roots, the polynomial is completely factored using its roots as \(P(x) = x(x - (1 + i))(x - (1 - i))\). Confirming the factorization by expanding the expression should yield the original polynomial, thereby verifying the correctness of finding both real and complex factors.