Problem 8 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

Zeros are 0, \(\frac{-1 \pm i\sqrt{3}}{2}\). Factored form: \(x(x^2 + x + 1)\).

Step 1: Identify Common Factors

The polynomial is given as \(P(x) = x^3 + x^2 + x\). First, observe if there's a common factor in all terms. Here, \(x\) is a common factor. So, we can factor \(x\) out from each term.

Step 2: Factor the Polynomial

Factoring \(x\) out of the polynomial yields:\[P(x) = x(x^2 + x + 1). \]We now have a factored form involving \(x\) and another quadratic \(x^2 + x + 1\).

Step 3: Find Real Zeros

To find the real zeros of \(P(x)\), set each factor equal to zero:\[x = 0\]This gives us one real zero at \(x = 0\).

Step 4: Solve Quadratic for Complex Zeros

The other factor is the quadratic \(x^2 + x + 1 = 0\). Use the quadratic formula to find its zeros:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \]where \(a = 1\), \(b = 1\), \(c = 1\).

Step 5: Calculate Discriminant

Calculate the discriminant \(b^2 - 4ac = 1^2 - 4(1)(1) = 1 - 4 = -3\). Since the discriminant is negative, the solutions will be complex.

Step 6: Find Complex Zeros

Using the quadratic formula:\[x = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2}.\]This gives us the complex zeros: \(\frac{-1 + i\sqrt{3}}{2}\) and \(\frac{-1 - i\sqrt{3}}{2}\).

Step 7: Write Complete Factored Form

The complete factored form of the polynomial \(P(x)\) including all real and complex zeros is:\[P(x) = x(x - (\frac{-1 + i\sqrt{3}}{2}))(x - (\frac{-1 - i\sqrt{3}}{2})).\]

Key Concepts

Factoring Polynomials

Factoring polynomials is like finding a puzzle for mathematicians. It involves breaking down a polynomial into simpler components, known as factors, that when multiplied back together, give the original polynomial. This is useful because it simplifies expressions and makes solving algebraic equations easier.

In the case of the polynomial given in the exercise, \(P(x) = x^3 + x^2 + x\), we noticed something interesting: all the terms share a common factor, which is \(x\). Factoring \(x\) out is the first step. When we factor out the \(x\), we write the expression as a product: \(P(x) = x(x^2 + x + 1)\). This reveals a second polynomial, \(x^2 + x + 1\), within the original expression. The first step makes the polynomial simpler to handle and crucial for finding zeros.

Another key point is considering all the terms in a given polynomial to determine common factors. Once identified, factoring them out could be crucial for revealing additional structure within the polynomial.

Complex Numbers in Algebra

Complex numbers are an extension of real numbers, introducing a new dimension to solving equations. In algebra, complex numbers often pop up when we are solving quadratics that don't have real solutions. This is especially evident when working with polynomials and their roots.

For example, in the problem, after factoring, we encounter the quadratic \(x^2 + x + 1\). To solve this, we use the quadratic formula. Interestingly, the discriminant of this quadratic equation is negative \((-3)\). This means there are no real solutions; instead, we'll find complex numbers as solutions. The solutions are given by \(x = \frac{-1 \pm i\sqrt{3}}{2}\). "i" is the imaginary unit, defined by the property \(i^2 = -1\). These solutions reflect the nature of complex numbers: consisting of both a real part and an imaginary part.

Having complex roots means that polynomials can have a variety of solutions, and understanding complex numbers helps to see a more comprehensive picture of what the polynomial represents.

Quadratic Formula Applications

The quadratic formula is a reliable tool for finding the roots of quadratic equations of the form \(ax^2 + bx + c = 0\). Given by \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), this formula is pivotal in determining whether the quadratic has real or complex roots.

In our exercise, once we factored the polynomial to isolate \(x^2 + x + 1\), we used the quadratic formula to find its zeros. We calculated the discriminant \(b^2 - 4ac\), which in this scenario was \(-3\), signaling the presence of complex roots. The formula not only tells us whether roots are real or complex but also spells out the precise values—including both real and imaginary parts. This is particularly useful in scenarios where factoring alone might not immediately reveal all solutions.

Applying the quadratic formula equips students with the capability to solve more complex and diverse algebraic problems, especially those involving expressions not easily factored through standard methods.