Problem 74 - Algebra A Combined Function

Question

Solve. $$ x^{3}+x-3 x^{2}-3=0 $$

Step-by-Step Solution

Answer

The solutions are \(x = 3\), \(x = i\), and \(x = -i\).

Step 1: Rearranging the Equation

The given equation is \(x^3 + x - 3x^2 - 3 = 0\). First, we need to rearrange it into a more standard form: \(x^3 - 3x^2 + x - 3 = 0\).

Step 2: Factoring by Grouping

We can factor the equation by grouping terms: \((x^3 - 3x^2) + (x - 3)\). This gives us \(x^2(x-3) + 1(x-3)\).

Step 3: Factoring Out Common Terms

Notice in step 2 that \((x-3)\) is common, so we can factor it out: \((x^2 + 1)(x-3) = 0\).

Step 4: Solving Each Factor

Set each factor equal to zero: \(x^2 + 1 = 0\) and \(x-3 = 0\). The solution for \(x-3 = 0\) is \(x = 3\).

Step 5: Solving the Quadratic Factor

The equation \(x^2 + 1 = 0\) doesn't have real roots because \(x^2 = -1\), but its complex roots are \(x = i\) and \(x = -i\).

Step 6: Final Solutions

The solutions to the equation are \(x = 3\), \(x = i\), and \(x = -i\).

Key Concepts

Factoring by Grouping

Factoring by grouping is a method used to solve polynomial equations, especially when they are not easily factorable by standard methods like the quadratic formula. This technique is a powerful way to find solutions without complicated calculations.

Here's how it works:

• First, we break down the polynomial into groups of terms that have common factors. In this case, our equation is rearranged to look like this: \((x^3 - 3x^2) + (x - 3)\).
• We then factor out the greatest common factor from each group. So, from \(x^3 - 3x^2\), we factor out \(x^2\), which gives us \(x^2(x-3)\). From \(x - 3\), we factor out \(1\), resulting in \(1(x-3)\).
• The key to this method is identifying common terms between the groups. Here, \((x-3)\) becomes the common factor for both groups.

Once the common factor is identified, it can be factored out of the entire polynomial, simplifying it greatly. Thus, \((x^2 + 1)(x-3) = 0\) is the result.

Factoring by grouping is an essential algebraic tool that simplifies complex polynomials, making them manageable and easier to solve.

Complex Numbers

Complex numbers come into play when dealing with values that cannot be represented on the real number line. In algebra, complex numbers are an essential concept, especially when solving equations without real roots.

A complex number is composed of two parts: a real part and an imaginary part, usually written as \(a + bi\), where \(i\) is the imaginary unit with the property \(i^2 = -1\).

In our exercise, after factoring by grouping, we are left with a quadratic equation, \(x^2 + 1 = 0\). This equation does not have real solutions because no real number squared will result in \(-1\).

• To solve this, we turn to complex numbers and set \(x^2 = -1\).
• Taking the square root of both sides, we find \(x = i\) and \(x = -i\), where \(i\) is the square root of \(-1\).

Understanding complex numbers allows us to find solutions in situations where traditional algebra falls short, offering insights into the behavior of quadratic equations that don't intersect the real number axis.

Quadratic Equations

Quadratic equations are a fundamental part of algebra, forming the cornerstone for understanding polynomial behaviors. They typically take the form \(ax^2 + bx + c = 0\).

When solving quadratic equations, several methods are available, such as factoring, completing the square, and using the quadratic formula. However, not all quadratic equations have real roots.

In our problem, after grouping, we encounter \(x^2 + 1 = 0\), a quadratic equation with no real roots since the sum of squares does not result in zero with real numbers.

• This is where the concept of discriminant comes into play; the discriminant \(b^2 - 4ac\) determines the nature of solutions for a quadratic equation.
• For \(x^2 + 1 = 0\), \(b = 0\), \(a = 1\), and \(c = 1\), giving us a negative discriminant \((-4)\).
• A negative discriminant indicates the existence of complex solutions, which we found as \(x = i\) and \(x = -i\).

Recognizing and solving quadratic equations with complex roots is critical in advance. Understanding quadratics expands our mathematical toolset beyond the real numbers to tackle more challenging problems in algebra.