Problem 20 - Algebra A Combined Function

Question

Factor each trinomial completely. If a polynomial can't be factored, write "prime." $$ 6 q-27+q^{2} $$

Step-by-Step Solution

Answer

The trinomial factors to \((q + 9)(q - 3)\).

Step 1: Rearrange the trinomial in standard form.

To make it easier to factor, rearrange the given trinomial in descending order of the variable. The trinomial is presented as \( 6q - 27 + q^2 \). We can rewrite this as \( q^2 + 6q - 27 \).

Step 2: Identify the coefficients for factoring.

In the trinomial \( q^2 + 6q - 27 \), identify \( a = 1 \), \( b = 6 \), and \( c = -27 \). Use these values to look for two numbers that multiply to \( a imes c = 1 imes (-27) = -27 \) and add up to \( b = 6 \).

Step 3: Find the numbers for factoring.

To factor the quadratic, find two numbers whose product is \(-27\) and sum is \(6\). The numbers \(9\) and \(-3\) satisfy these conditions because \(9 imes (-3) = -27\) and \(9 + (-3) = 6\).

Step 4: Break down the middle term using the identified numbers.

Using the numbers from Step 3, break the middle term into two terms. This gives \( q^2 + 9q - 3q - 27 \).

Step 5: Factor by grouping.

Group the four terms as two pairs: \((q^2 + 9q) + (-3q - 27)\). Factor out the greatest common factor from each pair: \(q(q + 9) - 3(q + 9)\).

Step 6: Complete the factoring process.

Notice the common factor \((q + 9)\) in both terms. Factor it out to get \((q + 9)(q - 3)\). This is the completely factored form of the trinomial.

Key Concepts

Quadratic Equation

A quadratic equation is an equation that can be expressed in the form of \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). These equations are called "quadratic" because the highest power of the variable \( x \) is 2, which is known as a square.Quadratic equations can have:

• Two distinct real roots
• One double real root
• No real roots (complex or imaginary roots)

The solutions or roots of a quadratic equation can be found by various methods, including:

• Factoring
• Using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
• Completing the square

In specific cases where the quadratic can be factored, like in the exercise with \( q^2 + 6q - 27 \), factoring is often the preferred method due to its straightforward approach and revealing insight into the equation's solutions.

Polynomial Factoring

Polynomial factoring involves expressing a polynomial as a product of its factors. For a trinomial like \( q^2 + 6q - 27 \), factoring it completely means breaking it down into simpler polynomials that, when multiplied, give back the original trinomial.The process of polynomial factoring can be done through various methods, such as:

• Identifying greatest common factors
• Factoring by grouping
• Utilizing special factoring formulas (difference of squares, perfect square trinomials)

In the quadratic trinomial \( q^2 + 6q - 27 \):

• The coefficients are \( a = 1 \), \( b = 6 \), and \( c = -27 \).
• We need two numbers whose product is \( a \cdot c = 1 \times -27 = -27 \) and whose sum is \( b = 6 \).
• These numbers are 9 and -3, leading to factors \((q + 9)\) and \((q - 3)\).

The end result is the neat expression \((q + 9)(q - 3)\), which is the completely factored form of the polynomial.

Algebraic Expressions

Algebraic expressions are combinations of variables, numbers, and operations (such as addition and multiplication). In solving problems like factoring, we manipulate algebraic expressions to simplify or rewrite them in another form.Key components of algebraic expressions include:

• Terms: Building blocks of expressions, which can be a constant, a variable, or both multiplied together. An expression like \( q^2 + 6q - 27 \) has three terms: \( q^2 \), \( 6q \), and \( -27 \).
• Coefficients: The numerical part of terms with variables. In 6q, 6 is the coefficient.
• Variables: Symbols that represent unspecific numbers. Here, 'q' is the variable.

Understanding these components is essential when working with algebraic expressions as it helps in operations like addition, subtraction, multiplication, and particularly in factoring. Mastering the manipulation of algebraic expressions allows for efficient solving of complex problems, making it a critical skill in algebra and beyond.