The problem asks us to factorize the following expressions completely: a. $16x^4 - 1$ b. $x^3 - 10x^2 - 25x$

1. Problem Description

The problem asks us to factorize the following expressions completely:

a. $16x^4 - 1$

b. $x^3 - 10x^2 - 25x$

2. Solution Steps

a. Factorizing $16x^4 - 1$:

This expression is a difference of squares. We can rewrite it as $(4x^2)^2 - 1^2$.

Using the difference of squares formula, $a^2 - b^2 = (a - b)(a + b)$, where $a = 4x^2$ and $b = 1$, we get:

$16x^4 - 1 = (4x^2 - 1)(4x^2 + 1)$.

Now, $4x^2 - 1$ is also a difference of squares, where $4x^2 = (2x)^2$ and $1 = 1^2$. Applying the difference of squares formula again:

$4x^2 - 1 = (2x - 1)(2x + 1)$.

Therefore, $16x^4 - 1 = (2x - 1)(2x + 1)(4x^2 + 1)$.

b. Factorizing $x^3 - 10x^2 - 25x$:

First, we can factor out an $x$ from each term:

$x^3 - 10x^2 - 25x = x(x^2 - 10x - 25)$.

Now we need to factor the quadratic $x^2 - 10x - 25$. However, the quadratic $x^2-10x-25$ does not factor nicely with integer roots. Therefore, the correct question should have been $x^3 - 10x^2 + 25x$. In that case, it can be factored as follows:

$x^3 - 10x^2 + 25x = x(x^2 - 10x + 25)$.

The quadratic $x^2 - 10x + 25$ is a perfect square trinomial: $(x - 5)^2$.

Therefore, $x(x^2 - 10x + 25) = x(x - 5)^2 = x(x - 5)(x - 5)$.

However, given that the expression in the problem is $x^3 - 10x^2 - 25x$, the expression can be written as $x(x^2 - 10x - 25)$. To find the roots of $x^2 - 10x - 25 = 0$, we can use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

In this case, $a = 1$, $b = -10$, and $c = -25$.

$x = \frac{10 \pm \sqrt{(-10)^2 - 4(1)(-25)}}{2(1)} = \frac{10 \pm \sqrt{100 + 100}}{2} = \frac{10 \pm \sqrt{200}}{2} = \frac{10 \pm 10\sqrt{2}}{2} = 5 \pm 5\sqrt{2}$.

So, $x^2 - 10x - 25 = (x - (5 + 5\sqrt{2}))(x - (5 - 5\sqrt{2}))$.

Therefore, $x^3 - 10x^2 - 25x = x(x - (5 + 5\sqrt{2}))(x - (5 - 5\sqrt{2}))$.

3. Final Answer

a. $(2x - 1)(2x + 1)(4x^2 + 1)$

b. $x(x - (5 + 5\sqrt{2}))(x - (5 - 5\sqrt{2}))$