Help factoring this polynomial please? :)?

Question

x^3 +3x^2 -x -3

I tried this and got x(x^2 + 3x -1) -3

which isn't exactly factored..and now I'm stuck.

Answer 1

x^3 +3x^2 -x -3=

x^2(x-1)+4x(x-1)+3(x-1)=

(x-1)(x^2+4x+3)=

(x-1)[x(x+1)+3(x+1)]=

(x-1)(x+1)(x+3)

Answer 2

instead of factoring out an x from all terms containing an x, make it so you have to binomials like so.

(x^3 + 3x^2) (-x-3)

then you can take an x^2 out of the first parenthesis and a -1 out of the second, leaving you with

x^2(x + 3) -1(x + 3)

then you combine like terms... kind of like terms... (sorry, i dont know how to explain this step :( sorry)

(x^2 - 1)(x + 3)

you do that because, they both have an x + 3 attached so you can combine them together like i showed.

next thing you should see is that you have a factorable term - the (x^2 - 1) it is able to be factored like so...

(x - 1)(x +1)(x +3)

and that would be your final answer :)

Hope this helps!!!

Answer 3

You can factor this further...

I started by guessing that (x + 3) is a factor since the first value is x^3 (for the 1) and the last is -3 (for the 3).

Using polynomial long division....

........x^2............-1

......._______________

x+3 | x^3 + 3x^2 - x - 3

........x^3 + 3x^2

........---------------

..........0......0....-x - 3

.......................-x - 3

.......................--------

........................0...0

So

(x + 3)(x^2 - 1)

Now, you can factor the difference of squares (you must memorize this formula)

(x + 3)(x - 1)(x + 1)

Answer 4

sure that is factored, maybe just not completely...you are on the right track. now try to factor the part in the parentheses further. i you can do it, great, if not, then it is already factored.

Answer 5

Regroup it:

x^3-x+3x^2-3

x(x^2-1)+3(x^2-1)

(x+3)(x^2-1)

Answer 6

x^3 +3x^2 -x -3 is

x(x^2-1)+3(x^2-1)

(x+3)(x^2-1)

(x+3)(x+1)(x-1)