# How do you factor completely #P(x)=x^3-6x^2+11x-6#?

##### 1 Answer

#### Explanation:

Given:

#P(x) = x^3-6x^2+11x-6#

Note that linear factors correspond to zeros. That is,

**Rational Roots Theorem**

By the Rational Roots Theorem, any rational zeros of

That means that the only possible *rational* zeros of

#+-1, +-2, +-3, +-6#

**Descartes' Rule of Signs**

Note that the pattern of the signs of the coefficients of

The pattern of the signs of the coefficients of

So, in combination with the Rational Roots Theorem, we can deduce that the only possible rational zeros of

#1, 2, 3, 6#

We could try each of these values in turn and find that:

#P(1) = P(2) = P(3) = 0#

Hence

but here's another way...

**Sum of coefficients shortcut**

Note that the sum of the coefficients of

#1-6+11-6 = 0#

Hence we can deduce that

#x^3-6x^2+11x-6 = (x-1)(x^2-5x+6)#

Then to factor the remaining quadratic note that

#x^2-5x+6 = (x-2)(x-3)#

Either way we arrive at our result:

#x^3-6x^2+11x-6 = (x-1)(x-2)(x-3)#