The image shows three separate math problems, labeled a, b, and c. Problem a is $16x^4 - 1$. Problem b is $x^3 - 10x^2 - 25x$. Problem c is $ax^2 - 9a + 2x^2 - 18$. We will factor each of these. The first one is already factored as $4(3x^4 - 1)$.

1. Problem Description

The image shows three separate math problems, labeled a, b, and c. Problem a is $16x^4 - 1$. Problem b is $x^3 - 10x^2 - 25x$. Problem c is $ax^2 - 9a + 2x^2 - 18$. We will factor each of these. The first one is already factored as $4(3x^4 - 1)$.

2. Solution Steps

a. $16x^4 - 1$

This is a difference of squares: $(4x^2)^2 - 1^2$.

$a^2 - b^2 = (a - b)(a + b)$

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

The first term is also a difference of squares: $4x^2 - 1 = (2x)^2 - 1^2$.

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

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

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

First, factor out an $x$:

$x(x^2 - 10x - 25)$

The quadratic is a perfect square trinomial: $x^2 - 10x - 25$ does not factor as easily because $-25$ should be $+25$ for it to be a perfect square trinomial.

The quadratic can be written as $(x - 5)^2$, which expands to $x^2 - 10x + 25$.

It appears that there is an error in the problem. If it's $x^3 - 10x^2 + 25x$ then

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

Assuming the intended expression was $x^3 - 10x^2 + 25x$, the factored form is $x(x-5)^2$.

c. $ax^2 - 9a + 2x^2 - 18$

Group terms: $(ax^2 + 2x^2) + (-9a - 18)$

Factor out $x^2$ from the first group and $-9$ from the second group:

$x^2(a + 2) - 9(a + 2)$

Factor out $(a + 2)$:

$(a + 2)(x^2 - 9)$

$x^2 - 9$ is a difference of squares: $x^2 - 3^2$.

$x^2 - 9 = (x - 3)(x + 3)$

Therefore, $ax^2 - 9a + 2x^2 - 18 = (a + 2)(x - 3)(x + 3)$.

3. Final Answer

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

b. $x(x-5)^2$ assuming the intended expression was $x^3 - 10x^2 + 25x$.

c. $(a + 2)(x - 3)(x + 3)$