Problem 39 - Algebra for College Students

Question

Solve each equation. You will need to use the factoring techniques that we discussed throughout this chapter. $$-3 x^{2}-19 x+14=0$$

Step-by-Step Solution

Answer

The solutions are \(x = -7\) and \(x = \frac{2}{3}\).

Step 1: Identify and write down the equation

The given quadratic equation is \[-3x^2 -19x + 14 = 0.\] Our goal is to solve for \(x\) using factoring techniques.

Step 2: Move to a standard form

The equation is already in standard quadratic form, \(ax^2 + bx + c = 0\). Here, \(a = -3\), \(b = -19\), and \(c = 14\).

Step 3: Factor by grouping (Split the Middle Term)

We need to factor the quadratic expression. We first find two numbers that multiply to give \(a \times c = (-3) \times 14 = -42\), and add to give \(b = -19\). These numbers are \(-21\) and \(2\).Rewrite \(-19x\) as \(-21x + 2x\) and then factor by grouping:\[-3x^2 - 21x + 2x + 14 = 0.\]

Step 4: Apply the grouping method

Group the terms to factor by common factors:\[(-3x^2 - 21x) + (2x + 14) = 0.\]Factor out the common factors:\[-3x(x + 7) + 2(x + 7) = 0.\] The terms \((x + 7)\) are common factors.

Step 5: Factor out the common binomial

Factor out \((x + 7)\) from both terms:\[(x + 7)(-3x + 2) = 0.\]

Step 6: Solve each factor for x

Set each factor equal to zero to find the solutions for \(x\):1. \((x + 7) = 0\) gives \(x = -7\).2. \((-3x + 2) = 0\) gives\[-3x = -2 \x = \frac{2}{3}\]

Key Concepts

Quadratic Equation

A quadratic equation is a special type of polynomial equation of degree 2. It typically takes the form:

• \(ax^2 + bx + c = 0\)

where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). The solution to a quadratic equation involves finding the values of \(x\) that satisfy the equation. Usually, there are at most two solutions because the equation represents a parabola, which can intersect the x-axis at two points.

The method of factoring converts the quadratic into a product of two binomials set to zero. This allows us to solve each binomial separately, ultimately finding the values of \(x\) that solve the original equation. Understanding the structure of a quadratic equation helps in choosing the appropriate methods to solve it.

Polynomial Expression

Polynomials are expressions that can have multiple terms consisting of variables and coefficients. Each term is made up of the product of a constant and a variable raised to an exponent. In a quadratic polynomial expression like \(-3x^2 - 19x + 14\), the degree is 2.

Polynomials are usually easier to handle when expressed in standard form, which arranges terms in descending order of their exponents. For quadratics, this is \(ax^2 + bx + c\). This standard form helps in identifying the coefficients that you will need to solve the equation, such as when finding numbers that multiply to \(a \times c\) and add to \(b\).

Dealing with polynomial expressions correctly is crucial because it sets the ground for translating word problems into mathematical solutions. Knowing how to manipulate these expressions allows you to simplify and solve equations effectively.

Factor by Grouping

Factoring by grouping is a method used for factoring polynomials that have four terms. It's an algebraic technique that simplifies complex expressions by reorganizing and factoring in parts.

To use this method for the quadratic \(-3x^2 - 19x + 14\), you start by looking for two numbers that multiply to the product of \(a\) and \(c\) (here, \(-42\)) and add to \(b\) (here, \(-19\)). Once these numbers are identified, they are used to split the middle term, transforming the expression into four terms:

• \(-3x^2 -21x + 2x + 14\)

Group these terms:

• \((-3x^2 - 21x) + (2x + 14)\)

Next, factor each group by their common factor:

• \(-3x(x + 7) + 2(x + 7)\)

Finally, the expression can be factored into the product of two binomials:

• \((x + 7)(-3x + 2) = 0\)

This technique is especially useful for expressions that aren't straightforward quadratics or don't have immediate, obvious factors. Factoring by grouping breaks down the problem into simpler parts, making it more manageable to solve for \(x\).