For the polynomial function P(x)=x^4-3x^3-3x^2+7x+6 solve P(x)=0
The Attempt at a Solution
I am not really sure how to break this down and factor it should I break it down into a trinomial and binomial?
yes your approach is very logical and so the answer appears to be x=0 with a multiplicity of 2 and x=-1 with a multiplicity of 3 ? but i checked the function using a graphing calculator and the zeros are -1 2 and 3?
[tex] x = 0 [/tex] isn't a solution. Look at
x\left(x^3 + 1 \right) - 3x^2 \left(x+1\right) + 6\left(x+1\right) = 0
When you substitute 0 into this the left side equals 6.
x^3 + 1
(sum of two cubes) and see what this does to the form of the equation. I believe this is the path the previous poster intended you to take.