Problem 30 - Algebra A Combined Function

Question

Factor each trinomial by grouping. Exercises 9 through 12 are broken into parts to help you get started. $$ 4 y^{2}-2 y-12 $$

Step-by-Step Solution

Answer

The trinomial factors to \( 2(2y + 3)(y - 2) \).

Step 1: Multiply and Find Factors

To factor by grouping, identify two numbers whose product is equal to the product of the coefficient of the first term and the constant term, and whose sum is the coefficient of the middle term. Here, multiply the coefficient of \(y^2\) (which is 4) by the constant term (-12): \[ 4 \times -12 = -48 \] Now, we need two numbers that multiply to -48 and add to -2 (the coefficient of the middle term). These numbers are -8 and 6, since \[ -8 \times 6 = -48 \] and \[ -8 + 6 = -2 \].

Step 2: Split the Middle Term

Rewrite the trinomial by splitting the middle term using the numbers found in the previous step. Replace \[ 4y^2 - 2y - 12 \] with \[ 4y^2 - 8y + 6y - 12 \].

Step 3: Group Terms

Group the terms in pairs and factor out the greatest common factor (GCF) from each group. Group the terms as follows: \[ (4y^2 - 8y) + (6y - 12) \].

Step 4: Factor Each Group

Factor out the GCF from each group.For the first group \[ 4y^2 - 8y \], the GCF is 4y:\[ 4y(y - 2) \].For the second group \[ 6y - 12 \], the GCF is 6:\[ 6(y - 2) \].

Step 5: Combine Groups

Now, notice that both groups contain the binomial \( y - 2 \). Factor \( y - 2 \) out, resulting in \[ (4y + 6)(y - 2) \].

Step 6: Simplify Factors

Simplify the expression \( (4y + 6)(y - 2) \) by factoring out any common factor in the components. For \( 4y + 6 \), the GCF is 2:\[ 2(2y + 3)(y - 2) \].This gives the fully factored form of the original polynomial.

Key Concepts

Algebraic Expressions

Algebraic expressions are combinations of variables, constants, and mathematical operations such as addition, subtraction, multiplication, and division. Understanding how to manipulate these expressions is an essential part of algebra.

These expressions can be as simple as a single constant or variable, like 5 or \(y\), or they can be more complex, with multiple terms and operations, such as \(4y^2 - 2y - 12\).
When dealing with algebraic expressions, the following components are typical:

• Terms: These are the building blocks of expressions. In the expression \(4y^2 - 2y - 12\), there are three terms: \(4y^2\), \(-2y\), and \(-12\).
• Coefficients: These are the numbers that multiply the variables. For \(4y^2\), 4 is the coefficient.
• Variables: These are symbols that represent numbers. In the expression above, \(y\) is a variable.
• Constants: These are the numbers on their own, like \(-12\).

Mastery of these components aids in understanding more complex operations, such as factoring polynomials, which is crucial for solving many algebra problems.

Polynomial Factoring

Polynomial factoring is a process where we express a polynomial as a product of its factors.
This is crucial in solving equations, simplifying algebraic expressions, and finding roots of functions.

Imagine you have a trinomial like \(4y^2 - 2y - 12\). To factor it, you're looking for two binomials that multiply to give you the original trinomial.

• Factoring by grouping: This is one way to factor polynomials by rewriting them into groups that can be easily factored.
• Common factors: Before grouping, it's beneficial to look for any common factors in all the terms of the polynomial.
• Product and sum: Identifying numbers whose product equals a part of the polynomial (like a multiplied constant term here) and whose sum equals the middle term's coefficient is a common strategy.

By factoring, you effectively "break down" the polynomial into simpler components, which can then be solved or simplified further.
It's like finding the prime factors of a number, which gives you more insight into its properties.

Grouping Method

The grouping method is a systematic approach for factoring trinomials and other polynomials by separating the polynomial into two binomial groups.
It involves several orderly steps:

• Step 1: Multiply and Find Factors: Multiply the coefficient of the first term by the constant term and find two numbers that multiply to this result and sum to the middle term.
• Step 2: Split the Middle Term: Use the two numbers to rewrite or "split" the middle term of the trinomial.
• Step 3: Group Terms: Arrange the polynomial in pairs, allowing you to focus on smaller, easily factorable parts.
• Step 4: Factor Each Group: Factor out the greatest common factor from each group.
• Step 5: Combine Groups: If the groups have a common binomial factor, this binomial can be factored out.
• Step 6: Simplify: Simplify the expression further if possible, ensuring it is in its simplest form.

This method is especially useful when the trinomial cannot be easily factored by traditional methods, providing a clear path to the solution.
It's like piecing together a puzzle by first breaking it down into manageable parts and then reconstructing it into something that makes sense.
Understanding this technique enhances problem-solving skills and deepens comprehension of algebraic structure.