Problem 46 - Fundamentals of Mathematics

Question

Find the value of each expression. $$6 a^{2}+2 a-15, \text { if } a=-2$$

Step-by-Step Solution

Answer

The value of the expression is 5.

Step 1: Substitute the Value of 'a'

First, we need to substitute the given value of \(a = -2\) into the expression \(6a^2 + 2a - 15\). This means wherever you see 'a', replace it with -2. This gives us \(6(-2)^2 + 2(-2) - 15\).

Step 2: Simplify the Exponents

Next, calculate \((-2)^2\). We have \((a^2) = (-2)\times(-2) = 4\). Substitute this back into the expression: \(6(4) + 2(-2) - 15\).

Step 3: Perform Multiplication

Now, multiply the coefficients by their respective numbers. Calculate \(6 \times 4 = 24\) and \(2 \times (-2) = -4\). The expression now looks like this: \(24 - 4 - 15\).

Step 4: Simplify the Expression

Finally, perform the arithmetic operations left. Subtract 4 from 24 to get 20, and then subtract 15 from 20. This results in \(20 - 15 = 5\).

Key Concepts

Expression Evaluation

Expression evaluation is a crucial concept in algebra, where we find the numerical value of an algebraic expression once the values of its variables are provided.
Understanding this can help simplify and find precise answers across mathematical problems.When given an algebraic expression like \(6a^2 + 2a - 15\), and a specific value for \(a\), such as \(a = -2\), our first task is to insert this value wherever \(a\) appears in the expression.
This substitution turns the variable-based expression into a purely numerical one.
By replacing \(a\) with \(-2\), our expression transforms: \(6(-2)^2 + 2(-2) - 15\).
With this step, we can further simplify by following typical mathematical operations: exponents, multiplication, addition, and subtraction. The initial substitution step is vital as it initiates the transformation process from variable to number output.

Substitution Method

The substitution method is an essential technique in algebra that entails replacing variables with specific values.
This method is particularly beneficial when dealing with algebraic expressions containing unknowns.When you encounter a problem such as evaluating \(6a^2 + 2a - 15\) for \(a = -2\), substitution becomes your primary tool.
Here's how it works:

• Identify the variable in the expression, which is \(a\) in this context.
• Replace \(a\) with \(-2\) everywhere in the expression.
• Simplify the new numerical expression obtained.

Executing these steps accurately ensures that you transform every appearance of the variable into a calculable number, which helps you move further into simplifying the overall expression.

Polynomial Simplification

Simplification of polynomials involves reducing the expression into its simplest form by performing calculations and combining like terms.
It takes a multi-step approach that, once mastered, makes working with complex expressions much easier.For example, when evaluating \(6(-2)^2 + 2(-2) - 15\), we:

• Calculate exponents: \((-2)^2 = 4\). The expression turns into \(6(4) + 2(-2) - 15\).
• Perform multiplication: \(6 \times 4 = 24\) and \(2 \times (-2) = -4\). This results in \(24 - 4 - 15\).
• Complete any remaining arithmetic: Subtract \(4\) from \(24\) to get \(20\), then subtract \(15\) to finalize the answer as \(5\).

By simplifying the expression, you uncover the simplest form of the polynomial and solve it effectively, enhancing both problem-solving skills and overall algebraic understanding.