Question 350349
{{{P(x) = x^3 - 2x^2 + 9x - 18}}}
When factoring always start with the Greatest Common Factor (GCF). The GCF of all four terms is 1. Since we rarely factor out a 1, we will move on to other factoring techniques. The most commonly used patterns all have 2 or 3 terms. Since P(x) has 4 terms it looks like patterns won't help. There is also trimomial factoring but, as the name implies, it is for three term expressions.<br>
So we are left with Factoring by grouping or by trial and error of the possible rational roots. For factoring by grouping you look for groups of terms that have a GCF. In P(x) the first two terms have a GCF (other then 1) and the last two terms also have a GCF (other than 1) so this looks promising. Factoring out the GCF of each group we get:
{{{P(x) = x^2*red((x - 2)) + 9*red((x - 2))}}}
As you can see, in red, the two groups have a common factor: (x-2). Factoring this common factor from each group we get:
{{{P(x) = red((x-2))(x^2 + 9)}}}
Since {{{(x^2 + 9)}}} is not linear we must continue trying to factor. This will not be simple. And I'm not sure how you've been taught to do something like this. But here's one way:
Rewrite as a subtraction:
{{{x^2 - (-9))}}}
Think of this as a difference of squares. It's easy to think of {{{x^2}}} as a perfect square. But what to so square to get -9?? If you know about Real and Imaginary numbers you might realize that if yous square 3i (or -3i) you get -9! (If this is not clear, try squaring 3i and see what you get.)
Now that we know what to square to get -9 we and look at {{{x^2 - (-9))}}} as a difference of squares and use the pattern of the same name, {{{a^2 - b^2 = (a + b)(a - b)}}}, to factor {{{x^2 - (-9))}}} into {{{(x +3i)(x-3i)}}}. So now we have:
{{{P(x) = (x-2)(x+3i)(x-3i)}}}
We have have P(x) factored into linear factors.