Question

Solve the equation. $$(3 a-8)(a+5)=0$$

Short answer

The solutions to the equation are \(a = \frac{8}{3}\) and \(a = -5\).

Step-by-step solution

Set the first factor equal to zero

Set the first factor \(3a - 8 = 0\). Solve this simple linear equation for 'a' by adding 8 to both sides and then dividing by 3.

Set the second factor equal to zero

Set the second factor \(a + 5 = 0\). Solve this simple linear equation for 'a' by subtracting 5 from both sides.

Summarize the solutions

The solutions to equation \((3a - 8)(a + 5) = 0\) are the values of 'a' for which either \(3a - 8 = 0\) or \(a + 5 = 0\).

Key concepts

Factoring Polynomials

When it comes to solving quadratic equations, factoring polynomials is a fundamental skill. A polynomial is an algebraic expression made up of variables and coefficients, with powers only in whole numbers and no variables in the denominator. For instance, in the equation \(3a-8)(a+5)=0\), we are dealing with a polynomial set to zero that has already been factored into two binomials.

Factoring is essentially the process of breaking down a complex expression into simpler parts—typically into a product of smaller polynomials. It’s akin to breaking a number down into its prime factors; just as the number 10 can be broken into the product of primes 2 and 5, a polynomial can be broken down into a product of binomials or trinomials.

• The first step in factoring is finding the greatest common factor (GCF) of the terms if one exists.
• Then, use factoring techniques such as grouping, difference of squares, or trinomial factoring to break down the polynomial further.
• Finally, write the polynomial as a product of its factors, which can often be used to find the roots of the polynomial when set to zero.

If students can master the skill of factoring polynomials, they'll find solving quadratic equations much more straightforward.

Zero Product Property

Another key concept in solving quadratic equations is the zero product property. This property states that if the product of two numbers is zero, then at least one of the numbers must be zero. It's a quite intuitive idea—nothing multiplied by something is nothing, after all. But when applied to algebra, it becomes a powerful tool for finding the roots of quadratic equations.

In the context of the equation from our example, \(3a-8)(a+5)=0\), the zero product property tells us that there are two solutions: one where \(3a-8=0\) and another where \(a+5=0\).

• This property is extremely useful because it simplifies a complex equation into simpler parts that can be solved individually.
• These simpler equations are usually linear equations, which are easier to solve.
• The solutions to these individual equations are the roots of the original quadratic equation.

By applying the zero product property, students can systematically approach solving quadratic equations through factoring.

Linear Equations

At its core, a linear equation represents a straight line and is the simplest form of algebraic equations. They are typically presented in the form \(ax + b = c\), where \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable we aim to solve for. In our example, after applying the zero product property, we ended up with two linear equations: \(3a - 8 = 0\) and \(a + 5 = 0\), which are the products of factoring the original quadratic equation.

To solve a linear equation:

• Isolate the variable on one side of the equation by using inverse operations, like addition or subtraction, to cancel out constants on the variable's side.
• If the variable is multiplied by a coefficient, divide both sides of the equation by that coefficient to solve for the variable.
• The outcome is the value of the variable that makes the equation true.

Linear equations allow us to solve for the unknown in an equation, and in the context of quadratic equations, they help us find the values that satisfy the zero product property. By understanding how to solve linear equations, students can tackle more complex algebraic equations with confidence.