Problem 29 - Algebra 2

Question

Factor each expression. $$ 5 t^{2}+28 t+32 $$

Step-by-Step Solution

Answer

(t + 4)(5t + 8)

Step 1: Determine the Coefficients

Identify the coefficients of the quadratic expression. The coefficient of the quadratic term (\(t^2\text{ term}\) is 5, the linear term (\(t\text{ term}\) coefficient is 28, and the constant term is 32.

Step 2: Find Two Numbers

Find two numbers that multiply to give the product of the quadratic term coefficient (5) and the constant term (32) which is 160, and add up to the linear term coefficient (28). After evaluating possible pairs of factors of 160, we find that 20 and 8 satisfy both conditions.

Step 3: Rewrite the Middle Term

Express the middle term (28t) as the sum of two terms using the numbers found in Step 2. Thus, we rewrite 28t as 20t + 8t.

Step 4: Factor by Grouping

Group the terms into two pairs and factor out the common factor from each pair. We have (5t^2 + 20t) + (8t + 32). Factoring out the common factors gives us 5t(t + 4) + 8(t + 4).

Step 5: Factor Out the Common Binomial

Now, take out the common binomial factor, which is (t + 4), to get the final factored form: (t + 4)(5t + 8).

Key Concepts

Coefficients in Quadratic Expressions

Understanding the coefficients in a quadratic expression is critical for various factoring methods. A quadratic expression is generally presented in the form \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are the coefficients of the terms respectively, and \(x\) is the variable.

In the given exercise, the quadratic expression is \(5t^2 + 28t + 32\). Here, \(5\) is the coefficient of the quadratic term \(t^2\), \(28\) is the coefficient of the linear term \(t\), and \(32\) is the constant term. The coefficients play a significant role in determining the method and ease of factoring the expression.

When learning to factor quadratics, one should pay close attention to these coefficients, as they are used to find possible pairs of numbers that can help rewrite and eventually factor the expression. Recognizing and correctly identifying these coefficients is the first vital step in the factoring process.

Factor by Grouping

Factor by grouping is a technique used when a quadratic expression does not readily factor into a simple binomial product. Once the middle term is split using suitable numbers as we have in our example, grouping becomes a valuable strategy.

Let's consider our example \(5t^2 + 28t + 32\), where the middle term \(28t\) is rewritten as \(20t + 8t\) based on the numbers found in the previous step. These terms are strategically chosen so that they still sum up to \(28t\) and their coefficients have common factors with the coefficients in other terms. The expression then becomes \(5t^2 + 20t + 8t + 32\).

The grouping method involves dividing the expression into groups, each of which contains a common factor. In the expression above, we can group the terms into two pairs: \(5t^2 + 20t\) and \(8t + 32\). Each pair has a common factor that can be factored out, which we identify as \(5t\) from the first pair and \(8\) from the second pair. This yields the intermediate factored form \(5t(t + 4) + 8(t + 4)\).

The presence of the common binomial \(t + 4\) in both groups is the key for the final step of this method, leading to the factored form of the original quadratic expression.

Finding Factors of a Quadratic

The search for factors of a quadratic expression typically involves finding two numbers that meet specific criteria based on the coefficients. In factoring the expression \(5t^{2}+28t+32\), we look for two numbers that multiply to \(ac\), the product of the leading coefficient \(a\) and the constant term \(c\), and at the same time, add up to \(b\), the linear term's coefficient.

For our exercise, the product of \(a\) and \(c\) is \(5 \times 32 = 160\), and the linear coefficient \(b\) is \(28\). We search for pairs of factors of \(160\) that add up to \(28\). After some consideration, we recognize that \(20\) and \(8\) fulfill these conditions since \(20 \times 8 = 160\) and \(20 + 8 = 28\).

These numbers are crucial for rewriting the middle term, which is the next step in factoring the expression. By successfully identifying these numbers, we create a pathway to factor the quadratic expression completely, as was demonstrated in our step by step solution. Such understanding is key in not only solving textbook exercises but also in applying the factorization technique to a variety of mathematical problems involving quadratic expressions.