Problem 18 - Elementary Algebra

Question

Use the guess and check method to factor. Identify any prime polynomials. $$ 2 f^{2}-13 f-7 $$

Step-by-Step Solution

Answer

(2f + 1)(f - 7)

Step 1 - Setup the Polynomial for Factoring

The given polynomial is \(2f^2 - 13f - 7\). Our goal is to factor it into the form \((af + b)(cf + d)\), where \(a, b, c,\) and \(d\) are constants.

Step 2 - Identify Possible Factor Pairs

We need factor pairs of the quadratic coefficient (2) and the constant term (-7). Possible pairs are: - For 2: (2, 1) - For -7: (7, -1) and (-7, 1)

Step 3 - Apply the Guess and Check Method

Using the pairs (2, 1) and (-7, 1), try different combinations to see which product gives the middle term \(-13f\). Testing \((2f + 1)(f - 7)\): \((2f)(f) = 2f^2\) (2f)(-7) + (1)(f) = -14f + 1f = -13f (1)(-7) = -7 Thus, \((2f + 1)(f - 7)\) is the correct factorization.

Step 4 - Identify if it's a Prime Polynomial

Since we successfully factored \(2f^2 - 13f - 7\) into \( (2f + 1)(f - 7)\), it is not a prime polynomial.

Key Concepts

Factoring Polynomials

Factoring polynomials is a method used to simplify expressions and solve equations. It's like breaking down numbers into their prime factors. For example, the number 12 is broken down into 2 x 2 x 3. Similarly, when we factor a polynomial, we break it down into simpler polynomials that, when multiplied together, give the original polynomial.
Polynomials can often be factored using several methods:

• Guess and check method
• Grouping
• Special products (like difference of squares, perfect square trinomials)

In our problem, we used the guess and check method to factor the polynomial. This involves identifying pairs of factors of the coefficients and the constant term, then testing combinations to see which pair gives the correct middle term.
Let's revisit the factor pairs:

• For 2: (2, 1)
• For -7: (7, -1) and (-7, 1)

We tested (2f + 1)(f - 7) and checked if the middle term matches -13f. This method can be time-consuming, but it's straightforward and can be quite effective for simpler polynomials.

Quadratic Equations

Quadratic equations are polynomial equations of degree 2. They have the general form ax^2 + bx + c = 0, where a, b, and c are constants and x is the variable.
There are several methods to solve quadratic equations:

• Factoring
• Completing the square
• Quadratic formula

In our problem, we focused on factoring. To factor a quadratic equation, we write it in the form (px + q)(rx + s). Then, by expanding this factored form, we ensure it matches the original polynomial. When the equation is factored correctly, we can use the zero-product property, which states if ab = 0, then either a = 0 or b = 0. This property helps us find the solutions to the equation.
From our exercise, we factored 2f^2 - 13f - 7 into (2f + 1)(f - 7). This form can then be used to find the values of f that satisfy the equation 2f^2 - 13f - 7 = 0.

Prime Polynomials

Prime polynomials are polynomials that cannot be factored into simpler polynomials with integer coefficients. Think of them like prime numbers, which can't be broken down into other numbers except by multiplying by 1 and themselves.
To determine if a polynomial is prime, we often attempt to factor it using known methods. In our case, we wanted to check if 2f^2 - 13f - 7 is prime. After using the guess and check method, we found it could be factored into (2f + 1)(f - 7). This means that 2f^2 - 13f - 7 is not a prime polynomial because it can be expressed as the product of two simpler polynomials.
This process is essential not only in simplifying polynomials but also in understanding their nature. If a polynomial is prime, it tells us about its intrinsic properties and helps us during algebraic manipulations. It’s crucial in higher math, where such properties are foundational to more advanced theories.