## Solving 4th level polynomial

This problem was set in

\(x^4 - 2x^3 + x - 132 = 0\) .... (1)

Note \((x^2 - x)^2 - (x^2 - x) = x^4 - 2x^3 + x^2 - x^2 + x\)

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

Let \( y = (x^2 - x) \)

So \( y^2 - y - 132 = 0 \)

\( (y - 12)(y + 11) = 0 \)

*Pyrgic Puzzles*in the*Guardian 29-04-2017*. I have left out preamble etc.\(x^4 - 2x^3 + x - 132 = 0\) .... (1)

**My approach.**- Note: factors of 132 are 2,2,3,11
- Using the
*Remainder Theorem*it is found that \( (x - 4 )\) and \( (x + 3) \) are factors. - Divide equation by 1 by these factors leaves \( x^2 - x + 1 = 0 \) .... (2)
- Using
*Quadratic Equation*formula the last two factors are \( \Big(x \pm \frac{\sqrt{-43}}{2}\Big) \)

**Sugested approach.**Note \((x^2 - x)^2 - (x^2 - x) = x^4 - 2x^3 + x^2 - x^2 + x\)

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

Let \( y = (x^2 - x) \)

So \( y^2 - y - 132 = 0 \)

\( (y - 12)(y + 11) = 0 \)