$$ \begin{array}{l} 5\left(x^{5}-5 x^{3}-6 x ight) \ 5 \cdot x\left(x^{4}-5 x^{2}-6 ight) \ 5 x\left(x^{2}+1 ight)\left(x^{2}-6 ight) \ 5 x\left(x^{2}+1 ight)\left(x^{2}-6 ight) \end{array} $$

Question

$$ \begin{array}{l} 5\left(x^{5}-5 x^{3}-6 xight) \ 5 \cdot x\left(x^{4}-5 x^{2}-6ight) \ 5 x\left(x^{2}+1ight)\left(x^{2}-6ight) \ 5 x\left(x^{2}+1ight)\left(x^{2}-6ight) \end{array} $$

UpStudy AI · Accepted Answer

Let's analyze the expressions provided step by step.

1. **First Expression:**
   $$
   5\left(x^{5}-5 x^{3}-6 x\right)
   $$
   This expression can be factored out by taking $$5$$ as a common factor. We can focus on factoring the polynomial inside the parentheses.

2. **Second Expression:**
   $$
   5 \cdot x\left(x^{4}-5 x^{2}-6\right)
   $$
   Here, we can also factor out $$5x$$ and then focus on the polynomial $$x^{4}-5 x^{2}-6$$.

3. **Third Expression:**
   $$
   5 x\left(x^{2}+1\right)\left(x^{2}-6\right)
   $$
   This expression is already factored, showing $$5x$$ multiplied by two other factors.

4. **Fourth Expression:**
   $$
   5 x\left(x^{2}+1\right)\left(x^{2}-6\right)
   $$
   This is identical to the third expression.

Now, let's focus on factoring the polynomial $$x^{4}-5 x^{2}-6$$ from the second expression. We can use substitution to make it easier. Let $$y = x^{2}$$, then we have:

$$
y^{2} - 5y - 6
$$

Next, we will factor this quadratic expression.

The factors of $$-6$$ that add up to $$-5$$ are $$-6$$ and $$1$$. Thus, we can write:

$$
y^{2} - 5y - 6 = (y - 6)(y + 1)
$$

Substituting back $$y = x^{2}$$, we get:

$$
x^{4} - 5x^{2} - 6 = (x^{2} - 6)(x^{2} + 1)
$$

Now, substituting this back into the second expression:

$$
5 \cdot x\left((x^{2} - 6)(x^{2} + 1)\right)
$$

This matches the third and fourth expressions.

Thus, we can conclude that all expressions are equivalent and can be factored as:

$$
5x(x^{2}+1)(x^{2}-6)
$$

In summary, the factored form of all the expressions is:

$$
5x(x^{2}+1)(x^{2}-6)
$$
All expressions can be factored as $$5x(x^{2}+1)(x^{2}-6)$$.

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