Factor $\frac{x^4 - 9x^2 + 4x + 12}{x - 2}$ using difference of cubes formula

Question

This is a problem I found from Lara Alcock's How to Think About Analysis (page 161) and I've been stumped for a while. So far I have tried to factor by grouping like so: $\displaystyle\frac{x^4 - 9x^2 + 4x + 12}{x - 2}$

First considering the numerator:

$x^4 - 9x^2 + 4x + 12$

$= x^2(x^2 - 3^2) + 4(x + 3)$

$= x^2(x-3)(x+3) + 4(x+3)$

I've also tried rewriting the numerator to get a difference of cubes I could factor but haven't had much success. I would also love any recommendations for books/resources to practice more algebra problems like this. This is my first post, any pointers would be much appreciated!

Answer 1

Since for the numerator $N(2)=0$ an effective way could be starting from

$$N(x) =x^4 - 9x^2 + 4x + 12 =(x^4 - 8x^2+16)-(x^2- 4x + 4)$$

A way to use difference of cubes formula could be as follows

$$x^4 - 9x^2 + 4x + 12 = x^4-8x- 9x^2 +36 + 12x - 24=$$

$$= x(x^3-8)-9(x^2-4)+12(x-2)$$

Answer 2

I've come back to it and, though it doesn't use the difference of cubes formula, I feel this is in spirit of what the text wanted. First, by synthetic division we get the following quotient:

$x^3 + 2x^2 + 5x - 6$

Then by the rational roots theorem we can verify x = -1 is a root. Then we can factor $(x+1)$ out of the quotient:

$x^3 + 2x^2 + 5x - 6 = (x+1)( ... )$

From here I did some brute force guessing-and-checking until I got to the following:

$(x+1)(x^2 + x - 6)$

While this didn't make use of the difference of cubes formula, I believe coming to an expression without relying on long division as much is what the text hoped to guide the reader towards.