Simplify: 2c^5 - 2c^3 - 60c

Understand the Problem
The question requires simplifying the expression 2c^5 - 2c^3 - 60c. This involves factoring out common terms and potentially further factoring the resulting polynomial.
Answer
$2c(c^2 - 6)(c^2 + 5)$
Answer for screen readers
$2c(c^2 - 6)(c^2 + 5)$
Steps to Solve
- Factor out the common term
The terms in the expression $2c^5 - 2c^3 - 60c$ have a common factor of $2c$. Factoring this out, we get: $2c(c^4 - c^2 - 30)$
- Factor the quadratic-like expression
Now, we have a quartic expression inside the parenthesis that resembles a quadratic. We can think of $c^4$ as $(c^2)^2$ and $c^2$ as $(c^2)$. Let $x = c^2$. Then, the expression becomes $x^2 - x - 30$. We need to find two numbers that multiply to $-30$ and add to $-1$. These numbers are $-6$ and $5$. So, we can rewrite the quadratic as $(x - 6)(x + 5)$. Substituting $c^2$ back in for $x$, we get $(c^2 - 6)(c^2 + 5)$.
- Final factored form
Therefore, the completely factored form of the original expression is $2c(c^2 - 6)(c^2 + 5)$.
$2c(c^2 - 6)(c^2 + 5)$
More Information
The polynomial $c^2 - 6$ can be further factored as $(c - \sqrt{6})(c + \sqrt{6})$, but typically when we factor, we are looking for integer or rational coefficients. Similarly, $c^2 + 5$ can be factored using imaginary numbers, but we want real number factors.
Tips
A common mistake is not factoring out the greatest common factor in the beginning. Another mistake may include incorrectly factoring the quadratic-like expression $c^4 - c^2 - 30$.
AI-generated content may contain errors. Please verify critical information