Question 583689
<pre>
{{{(16x^5 - 32x^4 - 81x + 162) / (x - 2)}}}

The numerator can be factored by grouping; no synthetic division required:

{{{(16x^5 - 32x^4 - 81x + 162)/(x-2)}}}

Factor 16x<sup>4</sup> out of the first two terms and -81 out of the last two terms:

{{{(16x^4(x-2)-81(x-2))/(x-2)}}}

Now factor (x-2) out of the two terms in the numerator:

{{{((x-2)(16x^4-81))/(x-2)}}}

That's enough to show that the binomial x-2 is a factor of numerator and
denominator.  However let's finish factoring. We can factor {{{16x^4-81}}} as
the difference of squares:

{{{((x-2)(4x^2-9)(4x^2+9))/(x-2)}}}

We can further factor {{{4x^2-9}}} as the difference of squares:

{{{((x-2)(2x-3)(2x+3)(4x^2+9))/(x-2)}}}

And as long as x isn't 2, we can cancel the (x-2)'s and get

{{{(2x-3)(2x+3)(4x^2+9)}}}

Edwin</pre>