Why Do We "Factor Out" When Solving the Gambler's Ruin Recurrence?

Question

While solving the Gambler’s Ruin recurrence

$P_i = pP_{i+1} + qP_{i-1}$

we guess a solution of the form

$P_i = x^i$

and substitute it into the recurrence to get

$x^i = px^{i+1} + qx^{i-1}$

The instructor then says to “factor out” and eventually obtains

$px^2 - x + q = 0$

What exactly does “factor out” mean here?

Factoring out means identifying a common factor that appears in every term and rewriting the expression using that common factor.

Starting with

$x^i = px^{i+1} + qx^{i-1}$

notice that every term contains .

Using the exponent rule

$x^a = x^b x^{a-b}$

we can rewrite each term as

$x^i = x^{i-1}x$ $x^{i+1} = x^{i-1}x^2$

Substituting these forms gives

$x^{i-1}x = px^{i-1}x^2 + qx^{i-1}$

Now is a common factor in every term, so factor it out:

$x^{i-1}(x) = x^{i-1}(px^2 + q)$

Assuming

$x \neq 0$

we can divide both sides by

$x^{i-1}$

to obtain

$x = px^2 + q$

Moving everything to one side yields the characteristic equation

$px^2 - x + q = 0$

The guess

$P_i = x^i$

is useful because shifting the index only changes the exponent:

$P_{i+1} = x^{i+1}$ $P_{i-1} = x^{i-1}$

This allows the recurrence relation to be converted into an algebraic quadratic equation, which is much easier to solve.