Problem 30 - College Algebra and Calculus: An Applied Approach

Question

Factor the trinomial.\(z^{2}-4 z-21\)

Step-by-Step Solution

Answer

The factored form of the trinomial \(z^{2}-4 z-21\) is \( (z - 7)(z + 3)

Step 1: Identifying the Coefficients

Firstly, identify the coefficients 'a', 'b', and 'c' in the given trinomial. In the trinomial \(z^{2}-4 z-21\), the coefficients are a=1, b=-4, and c=-21.

Step 2: Finding the Pair of Numbers

Find two numbers that add up to the coefficient of the middle term (-4 in this case), and multiply to the constant term (-21 in this case). The numbers that satisfy these conditions are -7 and 3. (-7) + 3 = -4 and (-7) * 3 = -21.

Step 3: Factoring the Trinomial

Use the two numbers found in the previous step to factor the trinomial into the product of two binomials. In this case, the trinomial \(z^{2}-4 z-21\) factors as \( (z - 7)(z + 3)\)

Key Concepts

Algebraic Expressions

Understanding algebraic expressions is crucial in the world of algebra. These expressions are combinations of numbers, variables (like the 'z' in our example), and arithmetic operations (+, -, *, /). In the case of the exercise, we dealt with a specific type of algebraic expression known as a trinomial, which is a polynomial with three terms. Here, the trinomial \(z^2-4z-21\) comprises a squared term (\(z^2\)), a linear term (\(-4z\)), and a constant term (\(-21\)).

To make the concept of algebraic expressions more tangible, let's visualize them as building blocks of algebra. Each term in an expression can interact with others, following the rules of algebra to simplify or transform the expression into a different form. This is exactly what we do when factoring trinomials—break them down to their simplest building blocks, which, in this case, are binomials.

Binomial Products

When we talk about binomial products, we're referring to the result of multiplying two binomials together. A standard formula many students learn is \( (x + m)(x + n) = x^2 + (m+n)x + mn \), where \( m \) and \( n \) are numbers or expressions. If you look closely, this formula resembles the original trinomial we had in our exercise, \( z^2 - 4z - 21 \).

In the step-by-step solution, we found that this trinomial could be factored into binomials as \( (z - 7)(z + 3) \). The process we used highlights a key relationship: factoring a trinomial is essentially reverse-engineering the product of two binomials. We look for the binomials whose product gives us back the original trinomial, completing a puzzle that reveals the fundamental structure of the expression. The ability to identify and produce binomial products is a foundational skill in algebra that can simplify complex expressions and solve equations.

Quadratic Equations

The trinomial we factored earlier is actually related to what we call a quadratic equation. A quadratic equation is an equation of the form \( ax^2 + bx + c = 0 \), where \( a \) is not zero. The equation represents a parabola when plotted on a graph. The exercise we tackled, if set equal to zero (\( z^2 - 4z - 21 = 0 \)), becomes a quadratic equation.

Factoring is one method of solving these equations. By finding the values of \( z \) that make each binomial equal to zero, we can determine the solutions to the equation (in this case, \( z = 7 \) and \( z = -3 \)). Factoring is particularly handy when the coefficients are manageable numbers. However, when factoring is too complex or impossible with real numbers, other methods like completing the square or using the quadratic formula may be employed. Quadratic equations are a mainstay in algebra and are foundational for further studies in calculus and other areas of advanced mathematics.