Problem 8 - Precalculus Mathematics for Calculus

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

The zeros of \(P(x)\) are \(x = 0, \frac{-1 \pm i\sqrt{3}}{2}\). The factorization is \(x(x^2 + x + 1)\).

Step 1: Identify Common Factor

First, observe that each term of the polynomial \(P(x) = x^3 + x^2 + x\) has a factor of \(x\). Factor out \(x\) from each term to simplify the polynomial.

Step 2: Factor Out the Common Term

Factor \(x\) out of \(P(x) = x^3 + x^2 + x\) to get:\[P(x) = x(x^2 + x + 1)\]

Step 3: Identify Roots from Linear Factor

Since we have factored out \(x\) in Step 2, one zero of \(P(x)\) is \(x = 0\). We now need to find the zeros of the quadratic \(x^2 + x + 1\).

Step 4: Solve the Quadratic Equation

Find the remaining zeros by solving the quadratic equation \(x^2 + x + 1 = 0\) using the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, \(a = 1\), \(b = 1\), and \(c = 1\).

Step 5: Calculate Discriminant

The discriminant of the quadratic equation \(x^2 + x + 1 = 0\) is given by \(b^2 - 4ac\). Calculate this:\[b^2 - 4ac = 1^2 - 4 \times 1 \times 1 = 1 - 4 = -3\]

Step 6: Calculate Complex Roots

Since the discriminant is negative, the quadratic has complex roots. Using the quadratic formula:\[x = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2}\]Therefore, the roots are \(x = \frac{-1 + i\sqrt{3}}{2}\) and \(x = \frac{-1 - i\sqrt{3}}{2}\).

Step 7: Write the Complete Factorization

Using the roots found, we can write the complete factorization of the polynomial as:\[P(x) = x \left( x - \frac{-1 + i\sqrt{3}}{2} \right) \left( x - \frac{-1 - i\sqrt{3}}{2} \right)\]

Key Concepts

Complex Zeros

Complex zeros are the solutions of a polynomial equation that are not real numbers. These are answers that involve imaginary numbers. An imaginary number is when you have a square root of a negative number, involving the unit 'i', where

• i = \(\sqrt{-1}\)

When we solve a quadratic equation and the discriminant (we'll get to that later) is negative, we find complex zeros.
In the polynomial \(x^2 + x + 1\), when we apply the quadratic formula and the discriminant turns out to be negative, the resulting solutions involve imaginary numbers.
Specifically, these can be expressed as:

• \(\frac{-1 + i\sqrt{3}}{2}\)
• \(\frac{-1 - i\sqrt{3}}{2}\)

These are the conjugate pairs, meaning they are mirror images over the real axis in the complex plane. In polynomial factorization, it is common to encounter complex zeros when the discriminant isn't positive.

Quadratic Formula

The quadratic formula is a crucial tool in solving quadratic equations of the form \(ax^2 + bx + c = 0\).
This formula provides the solutions (or 'zeros') of the quadratic equation. It is given by:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

The components of the formula are the coefficients of the quadratic equation.

• 'a' is the coefficient in front of \(x^2\).
• 'b' is the coefficient in front of \(x\).
• 'c' is the constant term.

Using this formula helps you find both real and complex solutions to the quadratic equation. In our original problem with the quadratic \(x^2 + x + 1\), the quadratic formula was used to find the solutions because direct factorization wasn't straightforward. The result was two complex solutions due to the negative discriminant.

Discriminant in Quadratics

The discriminant is a specific part of the quadratic formula and plays a major role in determining the nature of the roots of a quadratic equation. It is given by:

• \(b^2 - 4ac\)

The discriminant allows us to evaluate:

• If it is positive, the quadratic equation has two distinct real roots.
• If it equals zero, there is exactly one real root (a repeated root).
• If it is negative, the quadratic equation has two complex roots.

In our case, the quadratic \(x^2 + x + 1\) had a discriminant of \(-3\). This negative value indicated that the roots would be complex, not real. The magnitude of the discriminant helps in sketching or understanding the nature of the roots without actually solving the entire equation. Therefore, examining the discriminant is a helpful step before moving onto solving quadratic equations fully.