Problem 84 - Elementary Algebra

Question

Use any of the factoring methods to factor. Identify any prime polynomials. $$ 4 p^{2}+4 p-3 $$

Step-by-Step Solution

Answer

The factored form is \((2p - 1)(2p + 3)\). The polynomial is not prime.

Step 1: Identify the Quadratic Expression

The given expression is a quadratic polynomial: \[ 4p^2 + 4p - 3 \]

Step 2: Apply the Quadratic Formula for Factoring

To factor the quadratic expression, use the quadratic formula to find the roots, which is given by: \[ p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 4 \), \( b = 4 \), and \( c = -3 \).

Step 3: Calculate the Discriminant

Calculate the discriminant \( b^2 - 4ac \): \[ b^2 - 4ac = 4^2 - 4 \cdot 4 \cdot (-3) = 16 + 48 = 64 \]

Step 4: Find the Roots Using the Quadratic Formula

Substitute the discriminant back into the quadratic formula to find the roots: \[ p = \frac{-4 \pm \sqrt{64}}{2 \cdot 4} = \frac{-4 \pm 8}{8} \] So, the roots are: \[ p = \frac{4}{8} = \frac{1}{2} \quad \text{and} \quad p = \frac{-12}{8} = \frac{-3}{2} \]

Step 5: Write the Factors

Use the roots to write the polynomial as a product of its factors: \[ 4p^2 + 4p - 3 = 4(p - \frac{1}{2})(p + \frac{3}{2}) \]

Step 6: Adjust to Standard Factored Form

Combine and rewrite the factors in standard forms: \[ 4(p - \frac{1}{2})(p + \frac{3}{2}) = (2p - 1)(2p + 3) \]

Step 7: Identify Prime Polynomials

Since the given polynomial has been successfully factored into two binomials, it is not a prime polynomial.

Key Concepts

quadratic formula

The quadratic formula is essential for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). It helps us find the roots, or solutions, of these equations. The formula is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. Here, \( a \), \( b \), and \( c \) are constants from the quadratic equation. To apply this formula, follow these steps:

• Identify the values of \( a \), \( b \), and \( c \) from the equation.
• Calculate the discriminant, \( b^2 - 4ac \).
• Substitute these values into the quadratic formula.

The solutions, or roots, are the \( x \)-values that make the equation true. Understanding this formula will help you solve any quadratic equation.

discriminant

The discriminant is a key part of the quadratic formula and is given by \( b^2 - 4ac \). It tells us about the nature of the roots of the quadratic equation. Based on its value, we can determine:

• If the discriminant is positive: There are two distinct real roots.
• If the discriminant is zero: There is one real double root.
• If the discriminant is negative: There are two complex roots.

In our exercise, the discriminant is \( 64 \), which is positive. This means there are two distinct real roots. Calculating the discriminant is an essential step in using the quadratic formula, as it directly impacts the value and nature of the solutions.

polynomial roots

Polynomial roots, also known as zeros, are the values of the variable that make the polynomial equal to zero. For a quadratic polynomial \( 4p^2 + 4p - 3 \), the roots are found using the quadratic formula. In the step-by-step solution, we found the roots by calculating: \[ p = \frac{-4 \pm \sqrt{64}}{8} = \frac{{4}}{8} \quad \text{and} \quad \frac{{-12}}{8} \].Simplifying, we get: \( p = \frac{1}{2} \) and \( p = \frac{-3}{2} \).Roots are vital for factoring the polynomial as they inform us how it can be split into simpler expressions. By setting each root in a factor form, we reassemble the original polynomial in a factored form.

factoring methods

Factoring quadratics involves rewriting the polynomial as a product of its factors. This makes solving equations easier. There are several techniques for factoring quadratics, such as:

• Factoring by grouping: Useful when a polynomial has four terms.
• Using special patterns: Identifies forms like \( a^2 - b^2 = (a - b)(a + b) \).
• Quadratic formula: Especially helpful when other methods don't work easily.

In our solution, we used the quadratic formula. After finding the roots \( \frac{1}{2} \) and \( \frac{-3}{2} \), we wrote the polynomial as: \[ 4(p - \frac{1}{2})(p + \frac{3}{2}) \].Then, simplifying further, we get: \[ (2p - 1)(2p + 3) \].This is the factored form. Factors help break down complex polynomials into simpler expressions, making them easier to work with.