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

Feb 15, 2017

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

${x}^{3} - 6 {x}^{2} + 11 x - 6 \text{ "=" } \left(x - 1\right) \left(x - 2\right) \left(x - 3\right) .$

#### Explanation:

In order for $\left(a x + b\right)$ to be a factor of a polynomial $A {x}^{3} + B {x}^{2} + C x + D$, 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 ${x}^{3} - 6 {x}^{2} + 11 x - 6$, we have $A = 1$ and $D = \text{-} 6$, so the possible factors we will check are

• $x \pm 1$,
• $x \pm 2$,
• $x \pm 3$, and
• $x \pm 6$.

Let's start with the factors $\left(x + 1\right)$ and $\left(x - 1\right)$.
If $\left(x + 1\right)$ is a factor, then dividing the polynomial by $\left(x + 1\right)$ 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, $\left(x + 1\right)$ is not a factor of our polynomial.

What about $\left(x - 1\right)$?

$\left.\begin{matrix}1 & | & 1 & \text{-"6 & 11 & "-"6 \\ \null & | & ul(" ") & ul(1) & ul("-"5) & ul(6) \\ \null & \null & 1 & "-} 5 & 6 & | \underline{0}\end{matrix}\right.$

Since the remainder is zero, we know $\left(x - 1\right)$ is a factor of our polynomial. The other numbers on our bottom line (1, -5, 6) are the coefficients of the factored polynomial. Meaning:

${x}^{3} - 6 {x}^{2} + 11 x - 6 \text{ "=" } \left(x - 1\right) \left({x}^{2} - 5 x + 6\right) .$

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:

${x}^{2} - 5 x + 6 \text{ "=" } \left(x - 2\right) \left(x - 3\right)$

So we can replace the $\left({x}^{2} - 5 x + 6\right)$ in our factorization of the polynomial with $\left(x - 2\right) \left(x - 3\right)$ to get:

${x}^{3} - 6 {x}^{2} + 11 x - 6 \text{ "=" } \left(x - 1\right) \left(x - 2\right) \left(x - 3\right) .$

## 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 $D = \text{-} 6$? 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.)