1. ## Help with factoring

x^4-6x^2-8x-3

I tried using synthetic division but the graphs of the two were no where close

2. (x-3) should be a root. Go from there.

3. Hello, xclo0sive!

What does synthetic division have to do with graphs?

Factor: . $f(x) \:=\:x^4-6x^2-8x-3$
Did you notice that $f(\text{-}1) \:=\:0$ ?
. . This means $(x+1)$ is a factor of $f(x).$

$\text{We have: }\;f(x) \;=\;(x+1)\underbrace{(x^3 - x^2 - 5x - 3)}_{g(x)}$

Note that $g(\text{-}1) \:=\:0$
. . Hence: $(x+1)$ is a factor of $g(x).$

$\text{We have: }\;f(x) \;=\;(x+1)(x+1)\underbrace{(x^2-2x-3)}_{h(x)}$

Finally, we see that $h(x)$ can be factored.

Therefore: . $f(x) \:=\:(x+1)(x+1)(x+1)(x-3) \:=\:(x+1)^3(x-3)$

4. well i assumed that the factored form should have the same graph as the original equation but i didnt get the same graph. I have one question, what did you do to get (x+1)(x^3 - x^2 - 5x - 3)?

5. Since you know one root, you can use synthetic division on the polynomial. Also, you know that the only rational solutions are 1, -1, 3, -3, so you can look at those first, and see which of those are roots.