Problem 35 - Beginning and Intermediate Algebra

Question

Factor completely by first taking out \(-1\) and then by factoring the trinomial, if possible. Check your answer. $$-a^{2}-10 a-16$$

Step-by-Step Solution

Answer

The completely factored form of the given expression \(-a^2 - 10a - 16\) is \(-1(a + 2)(a + 8)\).

Step 1: Factor out the common factor (-1)

Since all the terms in this expression are negative, we can start by taking out the common factor \(-1\): \[ -1(a^2 + 10a + 16) \]

Step 2: Factor the trinomial

Now, we need to factor the trinomial \(a^2 + 10a + 16\). To do this, we will look for two numbers that multiply to the last term (16) and add up to the middle term (10). In this case, those numbers are 2 and 8. Therefore, we can write the trinomial as: \[ a^2 + 10a + 16 = (a + 2)(a + 8) \]

Step 3: Rewrite expression with factored trinomial

Now we'll combine the factors we found in steps 1 and 2 to rewrite the original expression completely factored: \[ -a^2 - 10a - 16 = -1(a + 2)(a + 8) \]

Step 4: Check the answer

To check our answer, we will expand the factored expression and make sure it matches the original expression: \[ -1(a + 2)(a + 8) = -1[a(a + 8) + 2(a + 8)] = -1(a^2 + 8a + 2a + 16) = -a^2 - 10a - 16 \] Since the expanded expression matches the original expression, our factoring is correct. So, the final factored form of the given expression is: \[ -a^2 - 10a - 16 = -1(a + 2)(a + 8) \]

Key Concepts

Understanding Algebraic Expressions

Algebraic expressions are mathematical phrases that can include numbers, variables, and operation symbols. In the expression \(-a^2 - 10a - 16\), each part (\(-a^2\), \(-10a\), and \(-16\)) is called a term. These terms are combined using addition or subtraction. Understanding the components of algebraic expressions is essential in mathematics, especially in operations like factoring.

• **Variables**: Symbols like \(a\) that represent numbers.
• **Coefficients**: Numbers in front of variables, such as \(-10\) in \(-10a\).
• **Constants**: Numbers without variables, such as \(-16\).

To simplify or manipulate these expressions, such as factoring, you need to break them down into easier parts. This can help solve equations or find solutions in word problems.

The Art of Polynomial Factorization

Polynomial factorization involves breaking down a polynomial into simpler pieces called factors. Think of it like breaking a number into prime numbers. In our example, the goal is to factor the trinomial \(a^2 + 10a + 16\).
Firstly, it's essential to identify any common factors. In this exercise, we noticed all terms are negative, so we took out \(-1\) first. This simplifies the work later. Then the trinomial is further dissected:

• Look for two numbers that multiply to the last term (in this case, 16).
• These numbers must also add up to the middle coefficient (10). Here, 2 and 8 work perfectly.

We can rewrite the trinomial as \( (a + 2)(a + 8) \). The process of identifying these factors is crucial for solving quadratic equations and other higher-degree polynomials in algebra.

Solving Quadratic Equations Through Factoring

Quadratic equations are equations of the form \( ax^2 + bx + c = 0 \). Factoring is a method that can simplify solving these equations. Given a quadratic expression like \(-a^2 - 10a - 16\), we start by rearranging it and factoring as shown in the original exercise.
Factoring a quadratic enables us to find the *roots* or solutions of the equation. In our case, once the quadratic trinomial \(a^2 + 10a + 16\) is factored into \((a + 2)(a + 8)\), setting each binomial equal to zero gives the solutions:

• \(a + 2 = 0 \Rightarrow a = -2\)
• \(a + 8 = 0 \Rightarrow a = -8\)

These roots represent the values of \(a\) for which the entire expression equals zero. Understanding how to factor and solve quadratics is foundational in algebra and beyond.