Question

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

Answer 1

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

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

Explanation:

In order for `(ax+b)` to be a factor of a polynomial `Ax^3+Bx^2+Cx+D`, we need `a` to be a divisor of `A`, and `b` to be a divisor of `D`. (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^3-6x^2+11x-6`, we have `A=1` and `D="-"6`, so the possible factors we will check are

• `x+-1`,
• `x+-2`,
• `x+-3`, and
• `x+- 6`.

Let's start with the factors `(x+1)` and `(x-1)`.
If `(x+1)` is a factor, then dividing the polynomial by `(x+1)` will leave `0` remainder. We divide using synthetic division:

`{:("-"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, `(x+1)` is not a factor of our polynomial.

What about `(x-1)`?

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

Since the remainder is zero, we know `(x-1)` is a factor of our polynomial. The other numbers on our bottom line (1, -5, 6) are the 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^2-5x+6)` in our factorization of the polynomial with `(x-2)(x-3)` to get:

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

Bonus:

Remember how we said the binomial factors needed to have `a`-values that were factors of `A=1` and `b`-values that were factors of `D="-"6`? Take a look at the `b`-values in our factors: -1, -2, and -3. After "FOIL-ing", their product is the only term that has no `x`'s in it, and so that product needs to be our `D`-value of -6. That's why each `b`-value must be a factor of -6: because only factors of -6 can contribute to a product of -6. (Ignoring rational/irrational factors.)

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