Problem 5 - College Algebra Essentials

Question

By factoring and then using the zero-product principle. $$ 2 x-3=8 x^{3}-12 x^{2} $$

Step-by-Step Solution

Answer

The solutions are roughly \(x=0\), \(x=0.78\) and \(x=-0.28\).

Step 1: Arrange all terms

Rewrite the equation with all terms arranged on one side of the equation: \[8x^3 - 12x^2 - 2x + 3 = 0 \]

Step 2: Factor out the common factor

Factor out the common factor of each term. In this case, the common factor in each term of the equation is 2x, this yields: \[2x(4x^2 - 6x - 1) + 3 = 0 \] or simplified:\[x(4x^2 - 6x - 1) + \frac{3}{2} = 0 \]

Step 3: Apply the zero-product principle

Through the zero-product principle, if \(4x^2 - 6x - 1 = 0 \), then \(x = 0\) or if \(\frac{3}{2} = 0 \), but \(\frac{3}{2} = 0 \) is not possible, so we should solve \(4x^2 - 6x - 1 = 0 \) quadratic equation via the quadratic formula to find the solution, i.e., \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Plug the coefficients of the quadratic into the formula\[a = 4, b = -6, c = -1 \]We get:\[x_1 = \frac{3 + \sqrt{25}}{8} \approx 0.78 \]\[x_2 = \frac{3 - \sqrt{25}}{8} \approx -0.28 \]Therefore, the solutions of the initial equation are \(x_1, x_2, and 0\)

Key Concepts

Zero-Product Principle

Understanding the zero-product principle is essential for solving equations efficiently. This principle states that if the product of two factors equals zero, then at least one of the factors must be zero. In terms of an equation, if you have two expressions multiplied together to give zero, you can set each expression to zero to find the possible solutions. For example, in the equation

• \(4x^2 - 6x - 1 = 0\)
• \(0 = 0\)

formations like these arise after factoring an equation or polynomial. Once you apply the zero-product principle, you can derive potential solutions efficiently by setting each factor of the product equal to zero and solving for the variable.

Quadratic Equations

Quadratic equations are a fundamental part of algebra. They come in the standard form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. Solving these equations often involves methods like factoring, completing the square, or using the quadratic formula. The quadratic formula is very useful, especially when the quadratic does not easily factor.

For example, we used it in the form

• \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
• to solve for \(x\) when \(4x^2 - 6x - 1 = 0\)
• with \(a = 4\), \(b = -6\), and \(c = -1\).

In this case, substituting the values of \(a\), \(b\), and \(c\) into the formula allows us to calculate the roots of the quadratic. The discriminant, \(b^2 - 4ac\), determines the nature of the roots. If it is positive, there are two real solutions; if it is zero, there's one real solution; and if it is negative, there are no real solutions, but instead two complex solutions.

Polynomial Factoring

Polynomial factoring is like breaking down a complex equation into simpler, more manageable parts. When you factor a polynomial, you are expressing it as a product of its factors. For example, in the polynomial

• \(8x^3 - 12x^2 - 2x + 3\),

the goal is to identify any common factors first.

This usually starts with factoring out the greatest common factor (GCF) of all terms. In this example, the GCF is 2x. After factoring, the polynomial simplifies into a product form, making it easier to apply methods like the zero-product principle.

Keep in mind, not every polynomial can be factored into real-number factors; this is where techniques like synthetic division, long division, or algebraic identities can be useful. By factoring, you transform complex equations into simpler tools you can use to solve algebraic problems effectively.