Thread: Factorising cubic polynomials through short division

1. Factorising cubic polynomials through short division

Use short division to factorise x^3 - 5x^2 - 2x + 24.

I don't know how to use short division to factorise this cubic polynomial. I only know how to factorise it through long division. If anyone could show a very simple worked out solution to this it will be appreciated!

2. Hello, Joker37!

Use short division to factorise: .$\displaystyle x^3 - 5x^2 - 2x + 24$
I'm not familiar with a "short division method."

I would use the Factor Theorem and then use long division.

Given a polynomial $\displaystyle p(x)$, if $\displaystyle p(a) = 0$, then $\displaystyle (x-a)$ is a factor.

In this problem, the possible roots are factors of 24: .$\displaystyle \pm1, \pm2, \pm3, \pm4,\:\hdots$

We find that $\displaystyle p(3) = 0\!:\;\;3^3 - 5(3^2) - 2(3) + 24 \:=\:0$
. . Hence, $\displaystyle (x-3)$ is a factor.

Using long division: .$\displaystyle x^3 - 5x^2 - 2x + 24 \;=\;(x-3)(x^2-2x-8)$

. . which factors further: .$\displaystyle \boxed{(x-3)(x+2)(x-4)}$

3. Originally Posted by Joker37
Use short division to factorise x^3 - 5x^2 - 2x + 24.

I don't know how to use short division to factorise this cubic polynomial. I only know how to factorise it through long division. If anyone could show a very simple worked out solution to this it will be appreciated!
Hi Joker37,

$\displaystyle x^3 - 5x^2 - 2x + 24.$

The first thing I would do is try to find a rational root if f(x)=0. According to the Rational Root Theorem, the possibilites are +/-1, +/-2, +/-3, +/-4, +/-6, +/-8, +/-12, +/-24.

I found that f(3)=0, so I will use synthetic division to find the depressed quadratic which should be easier to factor.

Code:
3 | 1  -5  -2  24
3  -6 -24
--------------
1  -2  -8  0
The resulting quadratic is $\displaystyle x^2-2x-8$ which can easily be factored into

$\displaystyle (x-4)(x+2)$

So
$\displaystyle x^3 - 5x^2 - 2x + 24=(x-3)(x-4)(x+2).$

4. Originally Posted by Soroban
Hello, Joker37!

I'm not familiar with a "short division method."
Yes...it doesn't seem to be quite a popular method.

5. Originally Posted by Joker37
Yes...[short division] doesn't seem to be quite a popular method.
Could you tell us what it is, so we'll know in the future?

Thank you!

6. Originally Posted by stapel
Could you tell us what it is, so we'll know in the future?

Thank you!
I,too, am curious. I found Short division - Wikipedia, the free encyclopedia regarding whole number "short" division, but nothing on polynomial "short" division.

7. Originally Posted by Soroban
Hello, Joker37!

I'm not familiar with a "short division method."

I would use the Factor Theorem and then use long division.

Given a polynomial $\displaystyle p(x)$, if $\displaystyle p(a) = 0$, then $\displaystyle (x-a)$ is a factor.

In this problem, the possible roots are factors of 24: .$\displaystyle \pm1, \pm2, \pm3, \pm4,\:\hdots$
The possible rational roots, not the possible roots.

CB

8. Originally Posted by masters
I,too, am curious. I found Short division - Wikipedia, the free encyclopedia regarding whole number "short" division, but nothing on polynomial "short" division.
Maybe it can refer to Shorthand division of polynomials called Synthetic Division or Ruffin's rule