# Trouble with factoring polonomial to the 3rd degree

I am having trouble factoring this problem: $\displaystyle{-x^{3} + 6x^{2} - 11x + 6}$

I know the answer but i can't figure out how it is done with this. I have tried by grouping and is doesn't seem to work. Can someone show me how to do this.

• That's not a "problem," it is a polynomial. The problem is factoring the polynomial. That might seem pedantic, but it is important to name things correctly in mathematics. – Thomas Andrews Nov 28 '13 at 23:18

\begin{align} & -x^3+6x^2-11x+6 \\ =& -x^3+1+6x^2-11x+5 \\ =& (1-x^3)+6x^2-6x-5x+5 \\ =& (1-x)(1+x+x^2)-6x(1-x)+5(1-x) \\ =& (1-x)(1+x+x^2-6x+5) \\ =& (1-x)(x^2-5x+6) \\ =& (1-x)(x^2-2x-3x+6) \\ =& (1-x)(x(x-2)-3(x-2)) \\ =& (1-x)(x-2)(x-3) \end{align}

• you have an error somewhere – wolfcall Nov 28 '13 at 23:26
• You may want to edit your answer after reading about the align environment for MathJax. – Lord_Farin Nov 28 '13 at 23:28
• There we go. That's a lot more readable. – Lord_Farin Nov 28 '13 at 23:40
• @Lord_Farin. Thank you I agree with with you – Adi Dani Nov 28 '13 at 23:42
• You're welcome. Be sure to check the source, so that you can typeset your answers like this yourself in the future! – Lord_Farin Nov 28 '13 at 23:44

Hint: $$-1 + 6 - 11 + 6 = 0$$

This method may be suggested by examining the rational roots theorem, as well.

Big hint:

In this case, just look for zeros of the cubic polynomial, since each zero tells you a linear factor (by the factor theorem).

Then try looking at factors of 6 (the constant term) - there are not many integer factors of 6 ...