# Factoring a polynomial

I am trying to factor the following polynomial: $$4x^3 - 8x^2 -x + 2$$

I am trying to do the following: $4x^2(x - 2)-x+2$ but I am stuck.

edit: correction.

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It might be better to write it as $4x^2(x-2)- (x-2)$ to take advantage of what you have done so far. – Geoff Robinson Jan 24 '12 at 19:04
It may be of value to note that there is a general formula for finding roots of 3rd degree polynomials. The formula looks complex but may be of help if you can't guess a root. See this reference for more details: en.wikipedia.org/wiki/Cubic_function – NoChance Jan 25 '12 at 20:57

You can factor as

$$4x^2(x-2)-(x-2)=4x^2(x-2)-(1)(x-2)$$

Then factor out the $x-2$ to get

$$(x-2)(4x^2-1).$$

But, you may further factor $4x^2-1$ to get

$$(x-2)(2x-1)(2x+1).$$

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+1:nice explanation... – pedja Jan 24 '12 at 19:07
It wasn't "factored wrong"; it's just that it wasn't taken far enough. – Michael Hardy Jan 24 '12 at 19:19
I agree with Michael. Could you perhaps edit that out? I think it's a little bit impolite. Incomplete and wrong are two very different things. :) – 000 Jan 25 '12 at 1:08
If you look at the revisions, the original posting contained an error where $(x-2)$ was incorrectly $(x-2x)$. I was not being impolite. But, I'll edit it out since things have changed. – Joe Johnson 126 Jan 25 '12 at 20:02

I'll use your polynomial to illustrate a more general procedure for factoring polynomials with integer coefficients (and assuming it has at least one rational root):

First, guess a root of $4x^3-8x^2-x+2$. The so called "rational roots test" will be helpful here.

Eventually, you'll discover that $x=2$ is a root of $4x^3-8x^2-x+2$. This will imply that your polynomial has the form $$\tag{1}(x-2)(ax^2+bx+c),$$ for some constants $a, b, c$.

To find those constants, you could do one of two things (and maybe more)

1. perform the division $4x^3-8x^2-x+2\over x-2$.
2. expand (1) and set it equal to the original polynomial. Setting the coefficients of the two sides of this equation equal to each other will give you a system of equations that are solvable for $a$, $b$, and $c$.

Once you've figured out what $a,b$, and $c$ are, factor the quadratic.

Using method (2), we have:

$$4x^3-8x^2-x+2 = ax^3+(b-2a)x^2+(c-2b)x-2c$$

A moment's reflection reveals that $c=-1$; whence $b=0$; whence $a=4$. Thus \eqalign{ 4x^3-8x^2-x+2 &= (x-2)(4x^2-1)\cr &= (x-2)(2x+1)(2x-1).\cr }

Of course, the other answers are more suitable to your problem; but in the event that your polynomial doesn't factor nicely (such as for $x^3+6x^2+11x+6$), you might try using this approach.

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Franco, you said you are stuck. Did you have an approach?

In general, for a cubic equation to be factored means the person who is asking this question is indirectly giving you a hint that there is at least one real root.

Your approach should be as follows: Step 1: first to try to see if x=0, 1 -1, 2 or -2 is a root. Chances are in some cases that one of these is a root. In that case, let us say you figured that $x=2$ is a root of $4x^3 - 8x^2 -x + 2$, then if you know factoring polynomials a little bit then you could immediately figure $4x^3 - 8x^2 -x + 2 = (x-2)(4x^2-1)$ which means the other factors can be found by the fact that $(4x^2-1^2) = (2x-1)(2x+1)$. If in some cases you cannot find the obvious root as described in Step 1, then Step 2: Observe if there is a pattern like $4x^3-8x^2$ has two coefficients $4$ and $8$, which has the same pattern as $-x+2$, i.e. $4x^3-8x^2 = 4x^2(x-2)$ and $-x+2 = -(x-2)$, which means you can combine those as $$(x-2)4x^2-(x-2) = (x-2)(4x^2-1)$$

And in some cases as here you could also do $(4x^3-x) - (8x^2-2)$ which gives you

$$x(4x^2-1)-2(4x^2-1) = (x-2)(4x^2-1) = (x-2)(2x-1)(2x+1)$$

Luckily with this question, Step 1 and Step 2 worked. What if you have a cubic equation that does not have obvious ways to factor.

Step 3: For a general cubic equation $ax^3+bx^2+cx+d=0$, apply the substitution

$$x = y - \frac{b}{3a}$$

then we get

$$a\left(y-\frac{b}{3a} \right)^3 + b\left(y-\frac{b}{3a} \right)^2+c\left(y-\frac{b}{3a} \right)+d = 0$$

which simplifies to

$$ay^3 + \left( c-\frac{b^2}{3a}\right)y+ \left(d + \frac{2b^3}{27a^2} - \frac{bc}{3a} \right) = 0$$

This is called a depressed cubic equation, because the square term is eliminated. It is much easier to use this and then find the roots. (back substitute to get the roots in terms of $x$)

For example $$2x^3-18x^2+46x-30=0$$

Substitute $x=y+3$ and simplify this cubic equation to

$2y^3-8y=0 \Rightarrow y=0,2,-2$ which then gives the roots as $x=1,3,$ and $5$.

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$(4x^3-x)-(8x^2-2)=x(4x^2-1)-2(4x^2-1)=(x-2)(4x^2-1)=$

$=(x-2)(2x-1)(2x+1)$

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+1: Yours didn't appear until after I sent mine. – Joe Johnson 126 Jan 24 '12 at 19:04