Problem 2 - A Graphical Approach to College Algebra

Question

Find all real solutions. $$x^{4}-x^{3}-6 x^{2}=0$$

Step-by-Step Solution

Answer

The real solutions are \(x = 0, 3, -2\).

Step 1: Set the Equation to Zero

The given equation is already set to zero: \[ x^{4}-x^{3}-6x^{2}=0 \] This is a polynomial equation that we need to factor to find its real solutions.

Step 2: Factor Out the Greatest Common Factor

Identify the greatest common factor (GCF) in the equation. Here, all terms share a common factor of \(x^2\). Factor \(x^2\) out of the equation:\[ x^2(x^2 - x - 6) = 0 \]

Step 3: Solve the Simple Factors

The equation \(x^2(x^2 - x - 6) = 0\) can be split into two distinct factors. The first factor is \(x^2\).Solve \(x^2 = 0\) for \(x\):\[ x = 0 \] Thus, \(x = 0\) is a real solution, with multiplicity 2.

Step 4: Factor and Solve the Quadratic

Now solve the quadratic factor \(x^2 - x - 6 = 0\). We will factor this by finding two numbers that multiply to -6 and add to -1. These numbers are -3 and 2.Thus, it factors as:\[ (x - 3)(x + 2) = 0 \] Set each factor equal to zero and solve for \(x\):\[ x - 3 = 0 \Rightarrow x = 3 \]\[ x + 2 = 0 \Rightarrow x = -2 \]

Step 5: List All Real Solutions

The complete set of real solutions from the factors are:- \(x = 0\) with multiplicity 2- \(x = 3\)- \(x = -2\)Hence, the real solutions to the original equation \(x^4 - x^3 - 6x^2 = 0\) are \(x = 0, 3, -2\).

Key Concepts

Factoring

Factoring is an essential technique when dealing with polynomial equations. It involves breaking down a complex expression into simpler, more manageable parts, known as factors.
In the given exercise, the goal is to find the real solutions by factoring the polynomial equation. The equation is initially set up as \[x^{4}-x^{3}-6x^{2}=0\] To make the expression simpler, start by identifying any common factors across all terms. In this case, all terms share a common factor of \(x^2\). By factoring \(x^2\) out of the equation, you simplify the equation to \[x^2(x^2 - x - 6) = 0\] This first step of factoring makes the equation more straightforward and manageable for finding real solutions. Essentially, factoring reduces the degrees of the polynomial, making it easier to solve.

Real Solutions

Real solutions are the actual values of \(x\) that satisfy the given polynomial equation, making the entire equation equal to zero. In the process of solving the given equation \[x^{4}-x^{3}-6x^{2}=0\] through factoring, we determined that one factor is \(x^2\). Upon solving \(x^2 = 0\), we find that one real solution is \[x = 0\] This solution, however, has a multiplicity of 2, meaning it appears twice in the solution set.
After factoring the remaining quadratic part \(x^2 - x - 6 = 0\), it further breaks down into two factors: \((x - 3)\) and \((x + 2)\). Solving those gives two additional real solutions:

• \(x = 3\)
• \(x = -2\)

Together, the complete list of real solutions for the equation is \(x = 0, 3, -2\). Each solution discussed is a real number because they can be plotted on a number line and verified within the original equation.

Quadratic Equations

Quadratic Equations are polynomial equations of the form \[ax^2 + bx + c = 0\] where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). Solving a quadratic equation involves finding the values of \(x\) that make the equation equal zero.
In our exercise, the quadratic equation to solve is extracted after factoring out \(x^2\): \[x^2 - x - 6 = 0\]This specific equation is handled by factoring, where you need two numbers that multiply to the constant term \(-6\) and add up to the middle coefficient \(-1\). The numbers \(-3\) and \(2\) perfectly suit this need, hence the quadratic equation factors into two linear expressions: \[(x - 3)(x + 2) = 0\] To find the values of \(x\), each factor is set to zero separately:

• \(x - 3 = 0\) implies \(x = 3\)
• \(x + 2 = 0\) implies \(x = -2\)

Quadratic equations can often be factored into simpler, linear terms, as demonstrated, making them easier to solve.