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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.

Thanks for your help.

edit: correction.

  • 0
    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: http://en.wikipedia.org/wiki/Cubic_function2012-01-25

4 Answers 4

4

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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    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.2012-01-25
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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.

3

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 $5$.

2

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

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