Problem 52 - Elementary Algebra

Question

For the following problems, factor the polynomials, if possible. $$ a^{2}+6 a+5 $$

Step-by-Step Solution

Answer

Question: Factor the quadratic polynomial \(a^2 + 6a + 5\). Answer: The factored form of the quadratic polynomial is \((a+1)(a+5)\).

Step 1: Identify the terms in the polynomial

The given quadratic polynomial is: $$ a^2 + 6a + 5 $$ Here, the terms are: - Leading term: \(a^2\) - Middle term: \(6a\) - Constant term: \(5\)

Step 2: Find the possible factors of the leading coefficient

The leading coefficient is 1 since it is the coefficient of \(a^2\). The only factor pair for 1 is (1, 1). So, the factors must have the form (a ± something) and (a ± something) since $$ (a ± something)(a ± something) = a^2 $$ To get positive \(a^2\), we can use either both plus signs or both minus signs.

Step 3: Find the possible factors of the constant term

The constant term is 5, and its possible factor pairs are (1, 5) and (-1, -5). Now, we need to find a combination of factors that will also give us the middle term's coefficient (6) when multiplied and added.

Step 4: Try different combinations of factor pairs

Let's try different combinations of the factor pairs for the constant term and check if the middle term's coefficient (6) is obtained: - The combination (1, 5) and both plus signs: \((a+1)(a+5)=a^2+6a+5\). This combination works, as we obtain our given polynomial. Now, we don't need to try any other combination since we've got the solution.

Step 5: Write down the factored form

Since the (1, 5) combination with plus signs works, the factored form of the given polynomial is: $$ a^2 + 6a + 5 = (a+1)(a+5) $$

Key Concepts

Quadratic Polynomial

Polynomials are algebraic expressions that consist of terms added together. A quadratic polynomial is a specific type of polynomial where the highest exponent of the variable is 2. This makes them special and distinguishes them from linear polynomials, for instance. It's called `quadratic` because the term "quad" implies two, referencing the degree of the polynomial, not four as one might assume from "quad". In our exercise, the quadratic polynomial is \(a^2 + 6a + 5\). Here:

• The variable is \(a\), and its highest power is 2, making it a quadratic.
• The solution involves manipulating this expression to put it in a simpler form.

Understanding quadratic polynomials is crucial since they frequently appear in various mathematical problems and real-world applications, such as physics and engineering.

Leading Coefficient

The leading coefficient is the coefficient of the term with the highest power in the polynomial. For quadratic polynomials, this is the coefficient before the squared term. In the expression \(a^2 + 6a + 5\), the leading coefficient is 1 because it's the coefficient of \(a^2\).

• It plays a crucial role in determining the nature of the roots of the polynomial.
• Since there's no number directly written before \(a^2\), it implies a coefficient of 1.

Factoring starts by recognizing this coefficient because it influences the factorization method. With a leading coefficient of 1, we have simpler factor pairs, making the process less complex compared to polynomials with higher leading coefficients.

Constant Term

The constant term is the term in the polynomial that does not contain any variable. It stands alone without any \(a\) attached to it. In our quadratic polynomial, \(a^2 + 6a + 5\), the constant term is 5.

• The value of the constant term is essential for determining the factor pairs, as it dictates the possible integers for factorization.
• The factor pairs of the constant term need to be combined with those of the leading term to obtain a product that matches the middle term.

Understanding the constant term's role aids in the process of breaking down the quadratic polynomial into its factored form, leading to a solution like \((a+1)(a+5)\).

Factor Pairs

Factor pairs are combinations of numbers that multiply together to yield a specific number. In the context of polynomial factoring, finding suitable factor pairs is crucial for breaking down a quadratic polynomial into two or more binomial expressions.

• Regarding \(a^2 + 6a + 5\), we consider factor pairs of the constant term, which is 5.

These pairs are (1, 5) and (-1, -5). However, we also think about how these pairs impact the middle term, 6 in this case.

• The factor pairs need to sum up or subtract correctly to match the middle coefficient for a valid breakdown, leading to an expression like \((a + 1)(a + 5)\).

The accurate combination of factor pairs ensures the expression is correctly factored, paving the way for solving equations or further simplifying the polynomial.