How I factor $(1-p) - x + px^2$?

Question

I came across this equation in the solution to problem 35 of "50 Challenging Problems in Probability" (page 37), which deals with a variant of the gambler's ruin problem. Specifically:

A man is 1 step away from falling off a cliff. He takes random steps either towards or away from the cliff. If the probability of his takinng a step away is always 2/3, and of taking a step towards always 1/3, what is his chance of escaping the cliff?

The book gives the answer as $(x-1)(px + p - 1) = 0$, but I can't figure out how to arrive at that answer using standard factoring methods or the quadratic equation. How do I factor this polynomial?

Answer 1

Since $$x^2-1=(x-1)(x+1),$$ we obtain: $$px^2-x+1-p=p(x^2-1)-(x-1)=(x-1)(p(x+1)-1)=(x-1)(px+p-1)$$

Answer 2

.Do you know the factor theorem for polynomials? In particular, since for the polynomial $q(x)=px^2−x+(p−1)$ we have $q(1)=p−1+(p−1)=0$, the theorem tells us that $(x−1)$ is a factor of $q(x)$. Now polynomial division tells us what the other factor is.

Also, as mentioned in another answer, you can use the quadratic formula to get the roots of the polynomial if the roots are not found using these guesses.

Answer 3

Worst case scenario, if you didn't see any of the tricks in the other answer, you could always use the quadratic formula to get the roots $r_1, r_2$. Doing $(x-r_1)(x-r_2)$ won't quite give you the factorization of the polynomial, because as you can see, if you expand the expression above you get a $1$ in front of $x^2$. However, our polynomial is equal to $a(x-r_1)(x-r_2)$ with $a=p$.