MHB How Can I Factorize the Equation x^3-7x^2+14x-8=0?

[mathmaniac1]

$$x^3-7x^2+14x-8=0$$

Anyone please lead me to factorize this eq on my own...and if you don't want to waste your timr,just show me how you do it...

 

[MarkFL]

What does the rational roots theorem tell you about possible rational roots?

 

[soroban]

Hello, mathmaniac!

\text{Solve: }\:x^3 - 7x^2 + 14x - 8 \:=\:0

Note that x=1 is a root of the equation.

Hence, (x-1) is a factor of the polynomial.

Dividing, we have:

. . x^3 - 7x^2 + 14x - 8 \:=\: (x-1)(x^2 - 6x + 8)

. . . . . . . . . . . . . . . . =\: (x-1)(x-2)(x-4)

 

[mathmaniac1]

I want it factorized without any roots actually found...
I want factorization to give the answer...

 

[MarkFL]

Then you need to cleverly rewrite the function so that you may factor by grouping, but using the rational roots theorem and division is much easier and more straightforward. (Wink)

 

[mathmaniac1]

Then you need to cleverly rewrite

How to cleverly rewrite?Teach me how to cleverly rewrite polynomials so as to factorize.Thanks a lot

 

[MarkFL]

$x^3-7x^2+14x-8=x^3-(6+1)x^2+(8+6)x-8=$

$x(x^2-6x+8)-(x^2-6x+8)=(x-1)(x^2-6x+8)=(x-1)(x-2)(x-4)$

I don't know why the LaTeX isn't rendering...but you get the idea.

 

[mathmaniac1]

You know the roots and I think that's why you selected 1,6 and 8.
Or is there any reasoning for what you did?I was asking for a well reasoned way of factorizing this eq.
Do you have any reasons?

 

[MarkFL]

$$x^3-7x^2+14x-8=x^3-(6+1)x^2+(8+6)x-8=x(x^2-6x+8)-(x^2-6x+8)=$$

$$(x-1)(x^2-6x+8)=(x-1)(x-2)(x-4)$$

I know why my previous post was problematic...I quoted the OP and there was some funky formatting code in it. That was fun...(Swearing)

 

[MarkFL]

You know the roots and I think that's why you selected 1,6 and 8.
Or is there any reasoning for what you did?I was asking for a well reasoned way of factorizing this eq.
Do you have any reasons?

I essentially worked backwards when pressed to show how you could factor by grouping...I would not do this, I would find a rational root (if it exists), then divide and factor the remaining quadratic. Much easier.

 

[mathmaniac1]

Code?

- - - Updated - - -

So you say you factorizing is difficult?Ok...

So what if you are not able to find any zeros,what if they are rationals?

 

[MarkFL]

Factoring is not difficult if there are rational roots. Tedious perhaps, but not difficult.

If there are no rational roots, then I would look for ways to obtain higher order factors.

 

[mathmaniac1]

Anyway if the roots are rational or if atleast one is rational,you can't do putting values for x,so how to factorize in such cases?

Please show an example,if you can...

And what do you mean by "ways to obtain higher factors"?

 

[MarkFL]

An example would be a quartic that has no rational roots, but may be factored with two quadratic factors, both of which have irrational roots, but are easy to find via the quadratic formula.

 

[mathmaniac1]

Ok,so factorization is no skill,no procedure or anything,its just luck?

 

[MarkFL]

It takes some skill, there are case by case procedures, and sometimes luck may be involved. But there is no general "this is how you factor all polynomials" method.

 

[Farmtalk]

When I was in high school and took pre-calculus, my teacher preached how important factorization was. I was given a 98 problem worksheet, all which had quartics, cubics, and quadratics that needed factored.

What I learned was to try and get the problem factored down to a quadratic. Then once I got down to a quadratic, I could usually factor it down more if needed. Factorization is a very important skill that can be used in nearly all branches of mathematics.:D

 

[mathmaniac1]

Is there any (general) method to factorize CUBICS like quadratics (where you find p and q for pq=ac and p+q=b)?

 

[mathbalarka]

Fortunately, yes but unfortunately it's not popular cause it is made by me (I don't think it is very well-known since I haven't found any material on the net regarding it)

But it's applicable only to the Bring-Jerrard form of the cubic :

$$x^3 + a x - b = 0 \;\; a, \, b \in \mathbb{Z}^{+}$$

If you can find (and you will probably find one in polynomial time complexity if the cubic above has all the, or at least, one integer root) $$\alpha, \, \beta $$ such that

$$a = \alpha - \beta^2$$ and $$b = \alpha \beta$$

Then $$\beta$$ is a root of the cubic.

This can be arbitrarily generalized for any degree (Excersise : How?)

Balarka
.

 

[BestMethod]

In this kind of factoring,a good way is the following;

As a result of the remainder theorem,if x=b makes any polynomial vanish,then (x-b) is a factor of it.

Now,how could you find any number such b?

According to the rational root theorem,a possible integer root of such a polynomial should be a divisor of the constant term(here is -8) which are ;
1,2,4,8,and their negatives.

So,we should substitute them,one by one,starting from x=1 to see which ones make it vanish.

Here,x=1,2,and 4 make it vanish,then (x-1)(x-2)(x-4) is a factor of it.

Now,since the degree of the polynomial is 3,and the coefficient of x^3 is 1,then,

Those are all the factors.Substituting can be annoying.I mean,how would you evaluate the polynomial for x=4?!
Let’s calculate it by a fast substituting method;

x^3=x^2.x

Now plugging 4 into x,

x^3=x^2.x=x^2.4=4x^2

So,f(x)=x^3-7x^2+14x-8=4x^2-7x^2+14x-8=-3x^2+14x-8

Again,writing -x^2=-x.x,and setting x=4,

-3x^2=-3x.x=-3x.4=-12x

So,f(x)=-3x^2+14x-8=-12x+14x-8=2x-8

Setting x=4,f(x)=2(4)-8=8-8=0

Here,i had to explain it step by step.So,if you do it by yourself,you'll find it's fast.

 

[Greg]

$$\begin{align*}x^3-7x^2+14x-8&=x^3-8-7x(x-2) \\
&=(x-2)(x^2+2x+4)-7x(x-2) \\
&=(x-2)(x^2+2x+4-7x) \\
&=(x-2)(x^2-5x+4) \\
&=(x-2)(x-1)(x-4)\end{align*}$$

 

[Monoxdifly]

Teach me master!

 

[Greg]

$$(a-b)^3=a^3-3a^2b+3ab^2-b^3$$

$$a^3-b^3=3a^2b-3ab^2+(a-b)^3$$

$$a^3-b^3=3ab(a-b)+(a-b)^3$$

$$a^3-b^3=(a-b)(3ab+(a-b)^2)$$

$$a^3-b^3=(a-b)(3ab+a^2-2ab+b^2)$$

$$a^3-b^3=(a-b)(a^2+ab+b^2)$$