Solving 4th level polynomial
This problem was set in Pyrgic Puzzles in the Guardian 29-04-2017. I have left out preamble etc.
\(x^4 - 2x^3 + x - 132 = 0\) .... (1)
My approach.
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 \( y^2 - y - 132 = 0 \)
\( (y - 12)(y + 11) = 0 \)
\(x^4 - 2x^3 + x - 132 = 0\) .... (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 \( y^2 - y - 132 = 0 \)
\( (y - 12)(y + 11) = 0 \)