Problem 88 - Algebra A Combined Function

Question

Factor each trinomial completely. $$ y^{2}(x+1)-2 y(x+1)-15(x+1) $$

Step-by-Step Solution

Answer

The trinomial factors to \((x+1)(y-5)(y+3)\).

Step 1: Identify Common Factor

Observe that the trinomial has a common factor of \((x+1)\). Factoring out \((x+1)\), we rewrite the expression as \((x+1)(y^2 - 2y - 15)\).

Step 2: Factor the Quadratic Expression

The expression \(y^2 - 2y - 15\) is a quadratic trinomial. Find two numbers that multiply to \(-15\) and add to \(-2\). These numbers are \(-5\) and \(3\).

Step 3: Write the Quadratic in Factored Form

Using the numbers found, the quadratic \(y^2 - 2y - 15\) can be factored as \((y - 5)(y + 3)\).

Step 4: Write the Complete Factored Form

Combine the results from Steps 1 and 3 to express the original trinomial in factored form as \((x+1)(y-5)(y+3)\).

Key Concepts

Common Factor

Identifying a common factor is a crucial step in simplifying expressions. A common factor is an element that is shared among the terms of an expression. In the given problem,

• all terms include the factor \((x+1)\).
• This means we can factor it out from each term.

Taking out a common factor simplifies the process of further factoring the expression. Imagine whether you are working with numbers or variables, always check if they share any common elements. When you pull out this common factor, you end up with a simpler expression that is easier to work with. Here, by factoring out \((x+1)\), we reduce the problem to \((x+1)(y^2 - 2y - 15)\). This step is like unwrapping the outer layer before diving deeper.

Quadratic Expressions

A quadratic expression is typically formulated as \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants. Our expression after factoring out the common factor is \(y^2 - 2y - 15\). You might have noticed:

• The highest power of the variable \(y\) is 2, indicating it's quadratic.
• Each quadratic expression takes the shape of a parabola when graphically represented.

The \(b\) term in the expression partly determines how the parabola shifts on a graph, while the \(c\) handles how high or low it crosses the \(y\)-axis. Understanding the nature and layout of a quadratic can dramatically ease the complexity of factors within it. In simple words, it sets the stage for the next part of the process.

Factoring Quadratics

Factoring quadratics means breaking down a trinomial into two binomials. With \(y^2 - 2y - 15\), we need to find two numbers that multiply together to \(-15\) (the constant term \(c\)) and add up to \(-2\) (the coefficient on \(b\)).

• The numbers \(-5\) and \(3\) fit perfectly—when multiplied, \(-5 \times 3 = -15\), and added together, \(-5 + 3 = -2\).
• This allows us to split the middle term and factor faster.

Rewriting the trinomial as \(y^2 - 5y + 3y - 15\) and then grouping these terms gives \((y - 5)(y + 3)\). The initial problem expression thus becomes \((x+1)(y-5)(y+3)\), accomplishing our factorization goal.