- Factor out the common variable \(y\) from the expression: \(y(x^4 - 4x^2 - 5)\).
- Recognize \(x^4 - 4x^2 - 5\) as a quadratic in \(x^2\) and factor the trinomial by identifying numbers that multiply to -5 and add to -4.
- Express the quadratic as \((x^2 - 5)(x^2 + 1)\).
- Conclude that the completely factored form is \(\boxed{y(x^2 - 5)(x^2 + 1)}\).
Explanation
- Identifying the Greatest Common Factor
The expression to be factored is \(x^4 y - 4x^2 y - 5y\). First, identify the greatest common factor (GCF) of all the terms. Observing the expression, we see that every term contains the variable \(y\). We can factor out \(y\) from the entire expression:
\[x^4 y - 4x^2 y - 5y = y(x^4 - 4x^2 - 5)\]
- Factoring the Quadratic Expression
Now we look at the expression inside the parentheses, which is \(x^4 - 4x^2 - 5\). We can treat this as a quadratic expression by performing a substitution. Let \(u = x^2\). Then \(u^2 = (x^2)^2 = x^4\). Substituting this into our expression gives us:
\[u^2 - 4u - 5\]
We need to factor this quadratic trinomial. To do so, we search for two numbers that multiply to the constant term \(-5\) and add up to the coefficient of the middle term \(-4\). These two numbers are \(-5\) and \(1\), since \((-5) \times 1 = -5\) and \(-5 + 1 = -4\). Thus, the quadratic factors as \((u - 5)(u + 1)\).
- Combining the Factors
Now that we have factored the quadratic expression in terms of \(u\), we substitute \(u = x^2\) back into our expression. Replacing \(u\) with \(x^2\), we get:
\[(x^2 - 5)(x^2 + 1)\]
Combining this with the common factor \(y\) that we factored out in the first step, the fully factored expression is:
\[y(x^2 - 5)(x^2 + 1)\]
- Final Conclusion
We compare our result with the provided choices:
- \(y(x^2 - 5)(x^2 + 1)\)
- \(y(x^2 + 5)(x^2 - 1)\)
- \((x^2 y - 5)(x^2 + 1)\)
- \((x^2 y + 5)(x^2 - 1)\)
Our result matches the first choice. Therefore, the completely factored form is \(y(x^2 - 5)(x^2 + 1)\).
Examples
Factoring polynomials is a fundamental skill in algebra with applications in fields like engineering and physics. For example, if you are calculating the trajectory of a projectile and obtain a complex expression for its displacement, factoring the expression can simplify the equation, allowing you to easily identify the time at which the object hits the ground (by setting the factors equal to zero). This process, often referred to as finding the zeros of a function, is essential for predicting the behavior of moving objects.