We are given that $x^2 + 2x - 8$ is a factor of the polynomial $f(x) = ax^3 - 4x^2 - 28x - 16$. We need to find the value of $a$.

1. Problem Description

We are given that $x^2 + 2x - 8$ is a factor of the polynomial $f(x) = ax^3 - 4x^2 - 28x - 16$. We need to find the value of $a$.

2. Solution Steps

Since $x^2 + 2x - 8$ is a factor of $f(x)$, we can write $f(x)$ as a product of $x^2 + 2x - 8$ and a linear factor $(cx + d)$, where $c$ and $d$ are constants.

$f(x) = (x^2 + 2x - 8)(cx + d) = ax^3 - 4x^2 - 28x - 16$

Expanding the left side:

$(x^2 + 2x - 8)(cx + d) = cx^3 + dx^2 + 2cx^2 + 2dx - 8cx - 8d = cx^3 + (d + 2c)x^2 + (2d - 8c)x - 8d$

Now, we can equate the coefficients of the polynomial to the given polynomial $ax^3 - 4x^2 - 28x - 16$:

$cx^3 + (d + 2c)x^2 + (2d - 8c)x - 8d = ax^3 - 4x^2 - 28x - 16$

Equating coefficients:

$c = a$

$d + 2c = -4$

$2d - 8c = -28$

$-8d = -16$

From $-8d = -16$, we get $d = 2$.

Substituting $d = 2$ into $d + 2c = -4$, we get $2 + 2c = -4$, so $2c = -6$, and $c = -3$.

Since $c = a$, we have $a = -3$.

We can also check this with the third equation: $2d - 8c = 2(2) - 8(-3) = 4 + 24 = 28 \neq -28$. This means there's an error in the original polynomial. We should have $2d - 8c = -28$, or $d - 4c = -14$. With $d = 2$, we have $2 - 4c = -14$, $-4c = -16$, so $c = 4$. Then, $d + 2c = -4$, so $2 + 2c = -4$, $2c = -6$, $c = -3$.

We have two different values for $c$, which implies that either there is no solution or there's a typo in the question. However, let's assume the given information is correct. We can divide the cubic polynomial by the quadratic.

Perform polynomial long division:

Divide $ax^3 - 4x^2 - 28x - 16$ by $x^2 + 2x - 8$. The quotient is $ax-2a-4$, and the remainder is $(-28+8a+8+4a+16)x + (8a+16-16) = (12a-4)x+8a$

$(ax^3 - 4x^2 - 28x - 16) = (x^2 + 2x - 8)(ax+b)$.

Comparing terms:

$x^3: a = a$.

$x^2: 2a+b = -4$

$x: -8a+2b=-28$

constant term: $-8b=-16$, which means $b=2$.

$2a+2=-4$.

$2a=-6$, $a=-3$.

Let's check:

$(-8a+2b)=-28$.

$(-8(-3)+2(2))=-28$.

$24+4=-28$, $28 = -28$ false.

Consider $x^2 + 2x - 8 = (x+4)(x-2)$.

Therefore, both $x=-4$ and $x=2$ are roots of $ax^3 - 4x^2 - 28x - 16 = 0$.

If $x = -4$:

$a(-4)^3 - 4(-4)^2 - 28(-4) - 16 = 0$

$-64a - 64 + 112 - 16 = 0$

$-64a + 32 = 0$

$64a = 32$

$a = 1/2$

If $x = 2$:

$a(2)^3 - 4(2)^2 - 28(2) - 16 = 0$

$8a - 16 - 56 - 16 = 0$

$8a - 88 = 0$

$8a = 88$

$a = 11$

Since we have two different values for $a$, there must be a mistake in the question or there is no solution. Since $f(x)$ is divisible by the given factor, the remainder must be zero, i.e.,

$(12a-4)=0$ and $8a=0$. This gives $a=1/3$ and $a=0$, impossible. There is something wrong with the problem description. Let's assume the "-E" is "-6".

$x^2 + 2x - 6$.

Going back to the original approach, let's proceed with $a=-3$.

$f(x) = -3x^3 - 4x^2 - 28x - 16 = (x^2 + 2x - 8)(-3x+2)$

$-3x^3 - 6x^2 + 24x + 2x^2 + 4x - 16 = -3x^3 - 4x^2 + 28x - 16$. This is almost correct.

Given the likely typo, let's re-evaluate the problem.

3. Final Answer

There appears to be a typo in the original problem. Assuming the problem is correct as stated, it is likely there's no solution, or infinitely many. If we assume the original polynomial is $-3x^3 - 4x^2 + 28x - 16$ then $a = -3$.

Without correcting the problem, the given information is inconsistent, and we cannot find a unique value for $a$.

Final Answer: Inconsistent.