Problem 10 - Precalculus Mathematics for Calculus

Question

A polynomial \(P\) is given. (a) Find all zeros of \(P\), real and complex. (b) Factor \(P\) completely. $$P(x)=x^{4}-x^{2}-2$$

Step-by-Step Solution

Answer

Zeros: \( \sqrt{2}, -\sqrt{2}, i, -i \). Factorization: \( (x - \sqrt{2})(x + \sqrt{2})(x - i)(x + i) \).

Step 1: Set the Polynomial Equal to Zero

To find the zeros of the polynomial, we first set the polynomial equal to zero:\[ x^4 - x^2 - 2 = 0 \]

Step 2: Substitute and Simplify

Let's substitute \( y = x^2 \), so we replace each \( x^2 \) with \( y \).This gives us a quadratic equation in \( y \):\[ y^2 - y - 2 = 0 \]

Step 3: Factor the Quadratic Equation

Now, we factor the quadratic equation:\[ y^2 - y - 2 = (y - 2)(y + 1) = 0 \]

Step 4: Solve for \( y \)

Next, we solve for \( y \) using the factors:- \( y - 2 = 0 \) gives \( y = 2 \)- \( y + 1 = 0 \) gives \( y = -1 \)

Step 5: Resubstitute \( y \) and Solve for \( x \)

We substitute back \( x^2 = y \) to find \( x \):For \( y = 2 \):\( x^2 = 2 \) implies \( x = \sqrt{2} \) or \( x = -\sqrt{2} \).For \( y = -1 \):\( x^2 = -1 \) implies \( x = i \) or \( x = -i \), where \( i \) is the imaginary unit.

Step 6: List the Zeros of the Polynomial

The zeros of the polynomial are \( \sqrt{2} \), \( -\sqrt{2} \), \( i \), and \( -i \).

Step 7: Write the Complete Factorization of \( P(x) \)

Using the zeros found, the complete factorization of the polynomial is:\[ P(x) = (x - \sqrt{2})(x + \sqrt{2})(x - i)(x + i) \]

Key Concepts

Quadratic Equation

A quadratic equation is a type of polynomial that is of degree two. It generally takes the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable. In the context of the problem we’re solving, the quadratic equation emerged when we made a substitution,\( y = x^2 \), to simplify the polynomial \( P(x) \).
By converting the polynomial to a quadratic one, \( y^2 - y - 2 = 0 \), it becomes easier to find solutions. Generally, these equations can be solved using several methods:

• Factoring: Expressing the quadratic as a product of two binomials.
• Completing the square: Adjusting the equation into a perfect square.
• Quadratic formula: Applying the classic formula \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \).

The step-by-step conversion of a complex polynomial to a quadratic equation via substitution showcases efficiency in solving polynomial equations.

Factorization

Factorization is a key method for solving polynomial equations. It involves expressing a polynomial as a product of its linear factors or simpler polynomials. In our example, the quadratic \( y^2 - y - 2 = 0 \) is factorized as \((y - 2)(y + 1) = 0\).

When you factor, you're identifying values of \( y \) that make the equation zero—these are critical because they directly give us the zeros of the polynomial. In simple terms, to factor means to break down a complex expression into multiples of simpler ones.

• This process is particularly powerful as it can easily reveal the roots or zeros of the polynomial.
• Through factorization, we identified two solutions, \( y = 2 \) and \( y = -1 \), simplifying our task of finding possible values for \( x \).

This approach underpins many algebraic solutions and always leads us to possible roots, which in their turn allow us to express the polynomial in its complete factorized form.

Complex Numbers

Complex numbers expand the concept of solutions beyond real numbers. They take the form \( a + bi \), where \( a \) and \( b \) are real numbers and \( i \) is the imaginary unit. Whenever you encounter a polynomial with a solution like \( x^2 = -1 \), you know complex numbers come into play.

In our solution, after factorization, we found a case where \( x^2 = -1 \). This naturally resulted in complex solutions. These are solutions that can’t be represented on the number line traditionally, as they incorporate both real and imaginary components:

• The solutions \( x = i \) and \( x = -i \) illustrate terms where no real numbers suffice. This is where the beauty of complex numbers lies—they generalize numbers to a higher realm.
• Complex numbers allow for polynomials to have complete sets of roots, demonstrating a comprehensive solution encapsulating all possible cases.

Understanding them is pivotal as they form the basis for advanced mathematical interpretation across real and theoretical constructs.

Imaginary Unit

The imaginary unit \( i \) is a fundamental concept when dealing with complex numbers. By definition, it’s the square root of \(-1\), making it an indispensable tool in finding solutions outside the realm of real numbers.

In the problem, when \( y = -1 \) led to \( x^2 = -1 \), the use of the imaginary unit was highlighted. With \( x = i \) and \( x = -i \), \( i \) facilitated the solution beyond natural and real numbers:

• It is used to bridge the gap for quantities that square to negative numbers, which real numbers cannot address.
• Imaginary numbers, like those formed with \( i \), allow us to factorize complex equations into completely solvable parts, ensuring no solutions are overlooked.

The imaginary unit expands algebra’s reach, giving us tools to solve thoroughly, even within polynomials, and it proofs rest intricately within the exploration of polynomial zeros.