Question

For the function \(f(x)=x^{4}+1\) (a) Factor \(f\) into the product of two irreducible quadratics. (b) Find the zeros of \(f\) by finding the zeros of each irreducible quadratic.

Short answer

The function factors into \((x^2 + \sqrt{2}x + 1)(x^2 - \sqrt{2}x + 1)\). The zeros are \( \frac{-\sqrt{2} + i\sqrt{2}}{2}, \frac{-\sqrt{2} - i\sqrt{2}}{2}, \frac{\sqrt{2} + i\sqrt{2}}{2}, \frac{\sqrt{2} - i\sqrt{2}}{2} \).

Step-by-step solution

- Rewrite the Function

Rewrite the function in a form that makes it easier to factor. We start with the function: \[ f(x) = x^4 + 1 \] Notice that this can be rewritten by adding and subtracting a variable squared term: \[ x^4 + 1 = (x^4 + 2x^2 + 1) - 2x^2 = (x^2 + 1)^2 - (\sqrt{2}x)^2 \] Now, it's in the form of a difference of squares.

- Apply the Difference of Squares Formula

Use the difference of squares formula, which states \(a^2 - b^2 = (a + b)(a - b)\), to factor: \[ (x^2 + 1)^2 - (\sqrt{2}x)^2 = (x^2 + 1 + \sqrt{2}x)(x^2 + 1 - \sqrt{2}x) \] Thus, the function factors into two irreducible quadratics.

- Find the Zeros of Each Irreducible Quadratic

To find the zeros of \(f(x)\), solve each quadratic equation separately. First, solve \(x^2 + \sqrt{2}x + 1 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\): \[ a = 1, b = \sqrt{2}, c = 1 \] \[ x = \frac{-\sqrt{2} \pm \sqrt{(\sqrt{2})^2 - 4(1)(1)}}{2(1)} = \frac{-\sqrt{2} \pm \sqrt{2 - 4}}{2} = \frac{-\sqrt{2} \pm \sqrt{-2}}{2} = \frac{-\sqrt{2} \pm i\sqrt{2}}{2} \] Now, solve \(x^2 - \sqrt{2}x + 1 = 0\) using the same method: \[ a = 1, b = -\sqrt{2}, c = 1 \] \[ x = \frac{\sqrt{2} \pm \sqrt{(-\sqrt{2})^2 - 4(1)(1)}}{2(1)} = \frac{\sqrt{2} \pm \sqrt{2 - 4}}{2} = \frac{\sqrt{2} \pm \sqrt{-2}}{2} = \frac{\sqrt{2} \pm i\sqrt{2}}{2} \]

- Summarize the Zeros

Combine the solutions to list all zeros of \(f(x)\): \[ x = \frac{-\sqrt{2} + i\sqrt{2}}{2}, x = \frac{-\sqrt{2} - i\sqrt{2}}{2}, x = \frac{\sqrt{2} + i\sqrt{2}}{2}, x = \frac{\sqrt{2} - i\sqrt{2}}{2} \]

Key concepts

difference of squares

The 'difference of squares' is a powerful technique used in algebra to factor certain types of polynomials. It applies to any expression that can be written in the form \[ a^2 - b^2 \]. The formula to factor this is \[ a^2 - b^2 = (a + b)(a - b) \].

In our given exercise, we start with the polynomial \[ f(x) = x^4 + 1 \], but this is not in the standard form of a difference of squares.

To transform it, we cleverly rewrite it by adding and subtracting \[ 2x^2 \], yielding: \[ x^4 + 1 = (x^4 + 2x^2 + 1) - 2x^2 = (x^2 + 1)^2 - (\sqrt{2}x)^2 \].

Now, we see that \[ (x^2 + 1)^2 \] and \[ (\sqrt{2}x)^2 \] fit the pattern, where \[a = x^2 + 1 \] and \[b = \sqrt{2}x \]. Applying the difference of squares formula, we get: \[ (x^2 + 1 + \sqrt{2}x)(x^2 + 1 - \sqrt{2}x) \].

This factors our original polynomial into two irreducible quadratics. Break down the problem using the difference of squares method and notice how this powerful tool can simplify complex polynomials further.

quadratic formula

The quadratic formula is a fundamental tool in algebra for finding the roots of a quadratic equation. A quadratic equation is any equation that can be written in the form \[ ax^2 + bx + c = 0 \].

The quadratic formula states that the solutions for \[x\] are given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].

In our exercise, we need to solve the quadratics \[ x^2 + \sqrt{2}x + 1 = 0 \] and \[ x^2 - \sqrt{2}x + 1 = 0 \].

For both equations, \[a = 1\text{, } b = \pm\sqrt{2},\text{ and } c = 1\text{ and by plugging these values into the quadratic formula:} \], we find: \[ x = \frac{-\sqrt{2} \pm\sqrt{(\sqrt{2})^2 - 4(1)(1)}}{2(1)} = \frac{-\sqrt{2} \pm \sqrt{2 - 4}}{2} = \frac{-\sqrt{2} \pm \sqrt{-2}}{2} = \frac{-\sqrt{2} \pm i\sqrt{2}}{2} \].

This yields complex solutions due to the negative term inside the square root. Thus, by understanding and applying the quadratic formula, we correctly find the zeros of the factored quadratic polynomials.

complex zeros

In algebra, complex zeros arise when the solutions to a quadratic (or higher-degree) equation are not real numbers. A complex number is of the form \[ a + bi \], where \[ i \] is the imaginary unit, defined such that \[i^2 = -1 \].

In our factorized polynomials, the quadratic equation \[ x^2 + \sqrt{2}x + 1 = 0 \] and \[ x^2 - \sqrt{2}x + 1 = 0 \] both result in complex roots.

Solving these using the quadratic formula, we find the solutions: \[ x = \frac{-\sqrt{2} \pm i\sqrt{2}}{2} \] and \[ x = \frac{\sqrt{2} \pm i\sqrt{2}}{2} \].

Notice these solutions are not real numbers, as they include the imaginary term \[ i\sqrt{2} \]. Instead, they represent complex zeros.

To summarize, the zeros of the polynomial \[ f(x) = x^4 + 1 \] are: \[ x = \frac{-\sqrt{2} + i\sqrt{2}}{2}, x = \frac{-\sqrt{2} - i\sqrt{2}}{2}, x = \frac{\sqrt{2} + i\sqrt{2}}{2} \text{ and } x = \frac{\sqrt{2} - i\sqrt{2}}{2} \].

By understanding complex zeros, we gain a fuller picture of where a polynomial intersects the x-axis, be it in the real or complex plane.