Question

Solve the equation by factoring. $$28 x^{2}-9 x-1=-4 x+2$$

Short answer

The solutions for the equation are \(x = \frac{3}{4}\) and \(x = -\frac{1}{7}\).

Step-by-step solution

Arrange the equation

Begin by arranging the equation to a standard polynomial format. This can be achieved by adding \(4x\) and subtracting \(2\) to both sides of the equation so that we have everything on one side and zero on the other. Doing this gives:\[28x^{2}-9x-1 +4x-2 = 0\]which simplifies to:\[28x^{2}-5x-3 = 0\].

Factor the equation

Now, factor the equation. It's a trinomial equation, hence can be factored to the form \((ax + b)(cx + d) = 0\).Finding the right coefficients a,b,c and d it factors out to be:\[(4x - 3)(7x + 1) = 0\].

Solve for x

From the factorized equation, \((4x - 3)(7x + 1) = 0\), we can now solve for \(x\). For this equation to equal zero, either \(4x - 3 = 0\) or \(7x + 1 = 0\). Solving each of these gives:\(4x - 3 = 0\) yields \(x = \frac{3}{4}\) and \(7x + 1 = 0\) yields \(x = -\frac{1}{7}\).

Key concepts

Factoring Polynomials

Factoring polynomials is an essential algebraic tool used to simplify expressions and to solve polynomial equations. When dealing with a polynomial, factoring involves breaking it down into a product of simpler polynomials that, when multiplied together, give the original polynomial.

Here are the common techniques for factoring polynomials:

• Greatest Common Factor (GCF): Identify the largest polynomial that divides all the terms.
• Grouping: Group terms to create pairs that are easier to factor.
• Trinomial Factoring: Rewriting a trinomial as a product of two binomials if possible.
• Special Products: Recognizing and using patterns such as difference of squares and perfect square trinomials.

For example, in the exercise given, the polynomial is arranged into standard form, then we search for two binomials whose product results in the original trinomial. This process breaks a complex problem into simpler parts that can be more easily solved.

It is important to note that not all polynomials can be factored. When they can't be factored using integer coefficients, they are considered to be ‘prime’ within the set of real numbers.

Trinomial Equations

Trinomial equations are a type of polynomial equation that specifically have three terms, usually written in the form \(ax^2+bx+c=0\). The process of solving a trinomial equation by factoring hinges on finding two binomials that multiply to give the original trinomial.

\textbf{The Process of Factoring Trinomial Equations}\textbf{}:

1. Make sure the equation is equal to zero.2. Look for a pair of numbers whose product equals \(ac\) (where \(a\) and \(c\) are the coefficients of \(x^2\) and the constant term, respectively) and whose sum equals \(b\) (the coefficient of \(x\)).3. Use these numbers to factor the trinomial into two binomials.

Returning to our exercise, once the equation is set to zero and arranged, we look for binomial factors that when multiplied give us -3 (the product of \(28\) and -1) and when added, they result in -5 (the coefficient of \(x\)). This process eventually yields \(4x - 3\) and \(7x + 1\) as the factors of the trinomial.

Quadratic Equations

Quadratic equations are second-degree polynomial equations, typically written in the format \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, with \(a \eq 0\). These equations are one of the most common and fundamental types encountered in algebra.

\textbf{Methods to Solve Quadratic Equations}:

• \textbf{Factoring:} If the quadratic can be factored into the product of two binomials, setting each factor equal to zero and solving for \(x\) will yield the solutions.
• \textbf{Quadratic Formula:} When factoring is difficult or impossible, the quadratic formula can always be applied.
• \textbf{Completing the Square:} This method involves creating a perfect square trinomial from the quadratic and then solving for \(x\).
• \textbf{Graphing:} Visualize the solutions as the points at which the graph of the quadratic equation intersects the x-axis.

For the exercise provided, we solve the quadratic by first factoring it into two binomials. This turned a complex quadratic equation into two simple linear equations, with the solutions being the x-intercepts of the original quadratic equation graph. When \(4x - 3 = 0\), \(x = \frac{3}{4}\), and when \(7x + 1 = 0\), \(x = - \frac{1}{7}\). These indicate the points where the graph of the corresponding quadratic function touches or crosses the x-axis.