Problem 60 - Beginning and Intermediate Algebra

Question

The following equations are not quadratic but can be solved by factoring and applying the zero product rule. Solve each equation. $$-13 b(12 b+7)(b-11)=0$$

Step-by-Step Solution

Answer

The possible values for b that make the given equation true are \(b_1 = 0\), \(b_2 = -\frac{7}{12}\), and \(b_3 = 11\).

Step 1: Identify the factors

First, identify the factors of the given equation. The given equation is already factored in the form: $$-13b (12b + 7)(b - 11) = 0$$ Here, the factors are -13b, 12b + 7, and b - 11.

Step 2: Apply the zero-product rule

Using the zero-product rule, set each factor equal to zero individually, and solve for b. $$-13b = 0,$$ $$(12b + 7) = 0,$$ $$(b - 11) = 0.$$

Step 3: Solve for b

Now, we just need to solve the three individual equations for b. For the first factor -13b, $$-13b = 0 \Rightarrow b = 0 / -13$$ Thus, $$b_1 = 0$$ (since any number divided by itself equals to 0) For the second factor 12b + 7, $$12b + 7 = 0 \Rightarrow 12b = -7 \Rightarrow b = -7/12$$ Thus, $$b_2 = -\frac{7}{12}$$ For the third factor b - 11, $$b - 11 = 0 \Rightarrow b = 11$$ Thus, $$b_3 = 11$$

Step 4: Write the solution

The possible values for b that make the given equation true are $$b_1 = 0$$, $$b_2 = -\frac{7}{12}$$, and $$b_3 = 11$$. These are the solutions to the equation.

Key Concepts

Factoring

Factoring involves breaking down an expression into simpler components, called factors, that, when multiplied together, return to the original expression. Think of it like splitting a number into its prime numbers, but here it’s with algebraic expressions. In the given equation, \(-13b (12b + 7)(b - 11) = 0\), it’s already provided in factored form.

• The factor \(-13b\) accounts for the variable \(b\) being multiplied by \(-13\).
• Then we have \((12b + 7)\), a binomial expression that is one of the factors.
• The final factor, \(b - 11\), is another binomial expression.

Being able to identify these factors is crucial, as it greatly simplifies solving the equation using the zero-product property, as each factor represents a potential solution to the equation when set to zero. If an equation is not factored to start with, you’d first rearrange and rewrite it in this form.

Solving Equations

Solving equations involves finding all possible values that make the equation true. Here, we use the zero-product property, which states that if the product of multiple terms is zero, then at least one of the terms must be zero. Applying this property involves setting each factor of the equation equal to zero and solving:

• -13b = 0: Even though \(-13\) cannot be zero, \(b\) can indeed be zero, leading us to \(b = 0\). This is straightforward because any number times zero gives zero.
• 12b + 7 = 0: Rearranging gives \(12b = -7\), and dividing by 12 gives \(b = -\frac{7}{12}\). This is the value of \(b\) that resolves this specific factor to zero.
• b - 11 = 0: Solving gives \(b = 11\), the value that zeroes out this factor.

Each solution of \(b\) is valid and independent, meaning substituting back into the original equation results in zero when multiplied by all other factors. This ensures the equation holds true.

Algebraic Expressions

An algebraic expression is a combination of numbers, variables, and operations. They represent mathematical phrases, and understanding how they work is essential to solving equations effectively. The equation \(-13b (12b + 7)(b - 11) = 0\) showcases how multiple algebraic expressions are multiplied together.

• Each component in our factored equation is an algebraic expression.
• Expressions like \(12b + 7\) and \(b - 11\) are linear, containing variables raised to the first power.
• The simplest form is \(-13b\), which is a product of a constant and a variable.

Recognizing these components helps students apply operations like factoring and using properties like the zero-product property effectively. Algebraic expressions are versatile and form the backbone of many more complex mathematical concepts and solutions. By simplifying and analyzing these expressions, we make seemingly complex problems much more manageable.