Problem 65 - Algebra A Combined Function

Question

Factor each trinomial completely. See Examples 1 through 7. \(10 x^{4}+25 x^{3} y-15 x^{2} y^{2}\)

Step-by-Step Solution

Answer

The factored form is \(5x^2(2x - y)(x + 3y)\).

Step 1: Identify the Common Factor

First, look at all the terms in the expression: \(10x^4\), \(25x^3y\), and \(-15x^2y^2\). Identify the greatest common factor (GCF) among the coefficients and the variable parts. Here, the GCF is \(5x^2\), since it is common in all terms.

Step 2: Factor Out the Greatest Common Factor

Factor out the GCF \(5x^2\) from the expression. This means you divide each term by \(5x^2\). \[10x^4 + 25x^3y - 15x^2y^2 = 5x^2(2x^2 + 5xy - 3y^2)\]

Step 3: Simplify the Factored Expression

Now, check if the quadratic trinomial \(2x^2 + 5xy - 3y^2\) can be factored further. Here, look for two numbers that multiply to \(2 \times (-3) = -6\) and add to \(5\). These numbers are \(6\) and \(-1\).

Step 4: Decompose the Middle Term and Factor by Grouping

Use the numbers \(6\) and \(-1\) to decompose the middle term:\[2x^2 + 5xy - 3y^2 = 2x^2 + 6xy - xy - 3y^2\]Now, factor by grouping:\[= (2x^2 + 6xy) + (-xy - 3y^2)\]\[= 2x(x + 3y) - y(x + 3y)\]\[= (2x - y)(x + 3y)\]

Step 5: Write the Fully Factored Expression

Substitute the factors back with the GCF to find the completely factored form of the original expression:\[10x^4 + 25x^3y - 15x^2y^2 = 5x^2(2x - y)(x + 3y)\]

Key Concepts

Greatest Common Factor

Factoring trinomials often begins with finding the Greatest Common Factor (GCF). This is a crucial step because it simplifies expressions and paves the way for easier factoring of more complex expressions.
In a polynomial like the one in the original exercise, each term has both a coefficient (a number) and a variable part (like powers of \(x\) and \(y\)).
To find the GCF:

• Identify the smallest power of each variable that commonly appears in all terms.
• Identify the largest number that divides all coefficients evenly.

In our expression \(10x^4 + 25x^3y - 15x^2y^2\), the GCF is \(5x^2\).
This calculation involves:- Taking the smallest power of \(x\) that appears in all terms (\(x^2\) in this case).- Choosing \(5\) because it's the largest number that divides \(10\), \(25\), and \(-15\) evenly.
By factoring out the GCF, the expression simplifies greatly, making further steps easier to manage.

Factoring Quadratic Expressions

After factoring out the GCF, the next step involves dealing with the quadratic expression you are left with. In this exercise, it is \(2x^2 + 5xy - 3y^2\).
Factoring a quadratic trinomial requires finding two numbers that multiply to the product of the coefficient of the first and the last terms, yet add up to the middle coefficient. This process is often called 'factor by grouping'.
Here's how you can navigate this step:

• Multiply the coefficient of the squared term (\(2\)) by the constant multiplier of the last term (\(-3\)), giving \(-6\).
• Identify two numbers that multiply to \(-6\) and simultaneously add to \(5\) (the middle coefficient), which are \(6\) and \(-1\).
• Decompose the middle term \(5xy\) into \(6xy - xy\).

Now you have \(2x^2 + 6xy - xy - 3y^2\). From here you can group terms and factor these smaller parts, eventually finding \((2x - y)(x + 3y)\).
This method ensures we can fully simplify complex quadratic expressions by breaking them into more manageable pieces.

Algebraic Expressions

Understanding algebraic expressions is key in solving and simplifying polynomial equations. An algebraic expression includes numbers, variables like \(x\) and \(y\), and operations such as addition and multiplication.
When working with expressions such as \(10x^4 + 25x^3y - 15x^2y^2\), it's vital to comprehend how each part fits together:

• The terms are separated by '+' or '-' signs, with each term composed of a coefficient and variable(s).
• Expressions can be simplified or factored to reveal underlying structures.

Factoring is particularly important as it breaks down complicated expressions into simpler multiplications.
This not only assists in solving equations but also makes it easier to see relationships between different components of the equation.
In our exercise, factoring the expression completely makes it simpler, and understanding these transformations gives insight into the behavior of algebraic functions.