# How do you determine the binomial factors of #x^3-6x^2+11x-6#?

##### 1 Answer

Test possible factors using synthetic division; retain those which are true factors. Repeat with the factored polynomial until only binomial factors remain.

#### Explanation:

In order for *(We'll see why in a bit).* So if we're looking for integer-based factors, this reduces our options considerably.

For our polynomial

#x+-1# ,#x+-2# ,#x+-3# , and#x+- 6# .

Let's start with the factors

If

#{:("-"1,|,1, "-"6, 11, "-"6),(,|,ul(" "),ul("-"1),ul7,ul("-"18)),(,,1,"-"7,18,|ul("-"24)):}#

The -24 is our remainder after division. Since this is not zero,

What about

#{:(1,|,1, "-"6, 11, "-"6),(,|,ul(" "),ul(1),ul("-"5),ul(6)),(,,1,"-"5,6,|ul(0)):}#

Since the remainder is zero, we know *coefficients* of the factored polynomial. Meaning:

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

We can find the remaining factors by using synthetic division on other possible factors, or, since the remaining polynomial is only a quadratic, we can use the old classic "find two numbers that add to -5 and multiply to 6" method.

This is fairly easily done. The two numbers are -2 and -3:

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

So we can replace the

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

## Bonus:

Remember how we said the binomial factors needed to have *needs* to be our

*(A similar argument follows for the #a#-values needing to be factors of 1.)*