#1
+118733 How do you factorise x3+5.x2+8.x+4=0
I know the solution is (x+1).(x+2)²=0, but how do you get there?
x 3 + 5x 2 + 8x + 4 = 0
It is important to note that all the coefficients are integers - otherwise this wouldn't work.
Rational root theorem tell me that all possible rational roots will be +1, -1, +2, -2, +4, -4 because these are the factors of the constant. (If the coefficient of x 3 was not 1 it would be a bit different)
Now I am going to use Remainder theorem to find one that works: I look at the + and - numbers more or less at the same time.
+-1 is the smallest, easiest one so I'll start there.
I'm substituting +-1 into your equation and seeing if it will make it true.
x 3 + 5x 2 + 8x + 4 = +-1 +5 +-8 +4 = 9 +-1 +-8 This looks promising if x=-1 then it equals 9-1-8 = 0 which is what I want so x=-1 is a solution so (x+1) is a factor.
Now I would divide x 3 + 5x 2 + 8x + 4 by (x+1) You can do this with algebra long division or with synthetic divison (which is much easier)
There are a number of you tube clips on synthetic division, here is one of them.
http://www.youtube.com/watch?v=_zl63HL7Npw
After you divide you get another factor which is (x 2 + 4x + 4) which, hopefully you can see, can be factorised to (x+2)(x+2)
so
x 3 + 5x 2 + 8x + 4 = (x+1)(x+2) 2
and that is how it is done.
This was a great question, I had never used synthetic division before so I learned something too.
Thanks.
