- Extract the common factor \(y\) to simplify the expression to \(y(x^4 - 4x^2 - 5)\).
- Recognize that \(x^4 - 4x^2 - 5\) is a quadratic form in terms of \(x^2\), which factors as \((x^2 - 5)(x^2 + 1)\).
- Combine the factors to obtain the complete factorization \(y(x^2 - 5)(x^2 + 1)\).
- Identify that this matches option (c), resulting in \(\boxed{y(x^2 - 5)(x^2 + 1)}\).
Explanation
- Analyzing and Extracting the Common Factor
The objective is to find the completely factored form of the polynomial expression \(x^4 y - 4 x^2 y - 5 y\). First, we identify the common factor in all terms of the expression, which is \(y\). By factoring out \(y\), the expression simplifies to \(y(x^4 - 4x^2 - 5)\).
- Factoring the Quadratic-like Expression
Now we focus on factoring the quadratic-like expression inside the parentheses: \(x^4 - 4x^2 - 5\). To make this easier, we can perform a substitution by letting \(u = x^2\). This transforms the expression into a standard quadratic: \(u^2 - 4u - 5\). We seek two numbers that multiply to \(-5\) and add to \(-4\). These numbers are \(-5\) and \(1\). Thus, we can factor the quadratic as \((u - 5)(u + 1)\).
- Finalizing the Factored Form
Substituting back \(x^2\) for \(u\), the expression \(u^2 - 4u - 5\) becomes \((x^2 - 5)(x^2 + 1)\). Combining this with the factor \(y\) that we extracted initially, the fully factored form of the original polynomial is \(y(x^2 - 5)(x^2 + 1)\).
- Conclusion and Selection
Comparing our result \(y(x^2 - 5)(x^2 + 1)\) with the given choices:
A\(y(x^2 + 5)(x^2 - 1)\)
B\((x^2 y + 5)(x^2 - 1)\)
C\(y(x^2 - 5)(x^2 + 1)\)
D\((x^2 y - 5)(x^2 + 1)\)
We can clearly see that our result matches option c. Therefore, the completely factored form is \(\boxed{y(x^2 - 5)(x^2 + 1)}\).
Examples
Factoring polynomials is a fundamental tool in algebra used to simplify complex equations, solve for roots, and understand the behavior of functions. For instance, if you are designing a product and the cost \(C\) is modeled by a polynomial like \(C(x) = x^4 y - 4 x^2 y - 5 y\), factoring it into \(C(x) = y(x^2 - 5)(x^2 + 1)\) makes it much easier to determine at what values of \(x\) the cost becomes zero or to analyze the sensitivity of the cost with respect to changes in the variables \(x\) and \(y\). Similar techniques are used in fields ranging from engineering to economics to break down complex systems into more manageable, understandable components.