Factoring The Polynomial 3r^2 - 8r + 5

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Factoring polynomials is a fundamental skill in algebra. It involves breaking down a polynomial expression into a product of simpler expressions. This process is crucial for solving equations, simplifying expressions, and understanding the behavior of functions. In this article, we will focus on factoring the quadratic polynomial 3r2−8r+53r^2 - 8r + 5. We will explore the steps involved in factoring this polynomial and discuss the different techniques that can be applied.

Understanding Quadratic Polynomials

Before we dive into factoring the specific polynomial 3r2−8r+53r^2 - 8r + 5, let's first understand the general form of a quadratic polynomial. A quadratic polynomial is a polynomial of degree two, meaning the highest power of the variable is two. The general form of a quadratic polynomial is:

ax2+bx+cax^2 + bx + c

Where:

  • a, b, and c are constants.
  • x is the variable.

In our case, the polynomial 3r2−8r+53r^2 - 8r + 5 is a quadratic polynomial where a = 3, b = -8, and c = 5. The goal of factoring a quadratic polynomial is to express it as a product of two binomials, if possible. A binomial is a polynomial with two terms.

Methods for Factoring Quadratic Polynomials

There are several methods for factoring quadratic polynomials, including:

  1. Trial and Error: This method involves guessing and checking different combinations of binomials until you find the correct factors.
  2. Factoring by Grouping: This method involves rewriting the middle term (bx) as the sum of two terms and then grouping the terms to factor.
  3. Using the Quadratic Formula: The quadratic formula can be used to find the roots of the polynomial, which can then be used to determine the factors.

For the polynomial 3r2−8r+53r^2 - 8r + 5, we will primarily use the factoring by grouping method, as it is a systematic approach that works well for many quadratic polynomials. However, it's worth noting that the trial and error method can also be effective, especially with practice.

Factoring 3r2−8r+53r^2 - 8r + 5 using Factoring by Grouping

The factoring by grouping method involves the following steps:

Step 1: Find two numbers that multiply to ac and add up to b

In our polynomial, 3r2−8r+53r^2 - 8r + 5, a = 3, b = -8, and c = 5. Therefore, we need to find two numbers that multiply to ac = 3 * 5 = 15 and add up to b = -8. Let's list the pairs of factors of 15:

  • 1 and 15
  • 3 and 5
  • -1 and -15
  • -3 and -5

Among these pairs, -3 and -5 satisfy our condition because (-3) * (-5) = 15 and (-3) + (-5) = -8.

Step 2: Rewrite the middle term (-8r) using the two numbers found in step 1

We rewrite the middle term -8r as the sum of -3r and -5r:

3r2−8r+5=3r2−3r−5r+53r^2 - 8r + 5 = 3r^2 - 3r - 5r + 5

This step is crucial because it sets up the polynomial for grouping in the next step.

Step 3: Group the terms and factor out the greatest common factor (GCF) from each group

We group the first two terms and the last two terms:

(3r2−3r)+(−5r+5)(3r^2 - 3r) + (-5r + 5)

Now, we factor out the GCF from each group. The GCF of 3r23r^2 and -3r is 3r, and the GCF of -5r and 5 is -5:

3r(r−1)−5(r−1)3r(r - 1) - 5(r - 1)

Notice that both terms now have a common factor of (r - 1).

Step 4: Factor out the common binomial factor

We factor out the common binomial factor (r - 1) from the entire expression:

(r−1)(3r−5)(r - 1)(3r - 5)

This is the factored form of the polynomial 3r2−8r+53r^2 - 8r + 5.

Verification

To verify our factoring, we can multiply the two binomials together to see if we get back the original polynomial:

(r−1)(3r−5)=r(3r−5)−1(3r−5)(r - 1)(3r - 5) = r(3r - 5) - 1(3r - 5)

=3r2−5r−3r+5= 3r^2 - 5r - 3r + 5

=3r2−8r+5= 3r^2 - 8r + 5

This confirms that our factored form is correct.

Alternative Method: Trial and Error

As mentioned earlier, trial and error can also be used to factor this polynomial. This method involves considering the possible factors of the leading coefficient (3) and the constant term (5) and trying different combinations of binomials. Since 3 is a prime number, its only factors are 1 and 3. Similarly, the factors of 5 are 1 and 5. We need to consider the signs as well.

We can start by trying different combinations such as:

  • (3r+1)(r+5)(3r + 1)(r + 5)
  • (3r−1)(r−5)(3r - 1)(r - 5)
  • (3r+5)(r+1)(3r + 5)(r + 1)
  • (3r−5)(r−1)(3r - 5)(r - 1)

Expanding each of these, we find that:

(3r−5)(r−1)=3r2−3r−5r+5=3r2−8r+5(3r - 5)(r - 1) = 3r^2 - 3r - 5r + 5 = 3r^2 - 8r + 5

Thus, the trial and error method also leads us to the factored form (3r−5)(r−1)(3r - 5)(r - 1).

Conclusion

Factoring the polynomial 3r2−8r+53r^2 - 8r + 5 using the factoring by grouping method involves finding two numbers that multiply to 15 and add up to -8, rewriting the middle term, grouping the terms, and factoring out the common factors. The result is the factored form (3r−5)(r−1)(3r - 5)(r - 1). We verified this result by multiplying the binomials and confirming that we obtained the original polynomial. The trial and error method also provides an alternative approach to factoring this polynomial, reinforcing the importance of understanding different factoring techniques.

Understanding how to factor quadratic polynomials is essential for solving a wide range of algebraic problems. Whether you use factoring by grouping, trial and error, or other methods, the ability to break down complex expressions into simpler factors is a valuable skill in mathematics.

Final Answer: The factored form of the polynomial 3r2−8r+53r^2 - 8r + 5 is (3r−5)(r−1)(3r - 5)(r - 1).