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

mathmaniac

Active member
Mar 4, 2013
188
\(\displaystyle x^3-7x^2+14x-8=0\)

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

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
What does the rational roots theorem tell you about possible rational roots?
 

soroban

Well-known member
Feb 2, 2012
409
Hello, mathmaniac!

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

Note that [tex]x=1[/tex] is a root of the equation.

Hence, [tex](x-1)[/tex] is a factor of the polynomial.

Dividing, we have:

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

. . . . . . . . . . . . . . . . [tex]=\: (x-1)(x-2)(x-4)[/tex]
 

mathmaniac

Active member
Mar 4, 2013
188
I want it factorized without any roots actually found...
I want factorization to give the answer...
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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)
 

mathmaniac

Active member
Mar 4, 2013
188
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

Administrator
Staff member
Feb 24, 2012
13,775
$x^3-7x^2+14x-8=x^3-(6+1)x^2+(8+6)x-8=$

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

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

mathmaniac

Active member
Mar 4, 2013
188
You know the roots and I think thats 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

Administrator
Staff member
Feb 24, 2012
13,775
\(\displaystyle x^3-7x^2+14x-8=x^3-(6+1)x^2+(8+6)x-8=x(x^2-6x+8)-(x^2-6x+8)=\)

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

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

Administrator
Staff member
Feb 24, 2012
13,775
You know the roots and I think thats 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.
 

mathmaniac

Active member
Mar 4, 2013
188
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

Administrator
Staff member
Feb 24, 2012
13,775
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.
 

mathmaniac

Active member
Mar 4, 2013
188
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

Administrator
Staff member
Feb 24, 2012
13,775
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.
 

mathmaniac

Active member
Mar 4, 2013
188
Ok,so factorization is no skill,no procedure or anything,its just luck?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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

Active member
Dec 25, 2012
42
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
 

mathmaniac

Active member
Mar 4, 2013
188
Is there any (general) method to factorize CUBICS like quadratics (where you find p and q for pq=ac and p+q=b)?
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
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 :

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

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) \(\displaystyle \alpha, \, \beta \) such that

\(\displaystyle a = \alpha - \beta^2\) and \(\displaystyle b = \alpha \beta\)

Then \(\displaystyle \beta\) is a root of the cubic.

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

Balarka
.
 
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BestMethod

New member
Dec 11, 2019
9
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

Perseverance
Staff member
Feb 5, 2013
1,384
\(\displaystyle \begin{align*}x^3-7x^2+14x-8&=x^3-8-7x(x-2) \\
&=(x-2)(x^2+2x+4)-7x(x-2) \\
&=(x-2)(x^2+2x+4-7x) \\
&=(x-2)(x^2-5x+4) \\
&=(x-2)(x-1)(x-4)\end{align*}\)
 

Monoxdifly

Well-known member
Aug 6, 2015
299
Teach me master!
 

Greg

Perseverance
Staff member
Feb 5, 2013
1,384
\(\displaystyle (a-b)^3=a^3-3a^2b+3ab^2-b^3\)

\(\displaystyle a^3-b^3=3a^2b-3ab^2+(a-b)^3\)

\(\displaystyle a^3-b^3=3ab(a-b)+(a-b)^3\)

\(\displaystyle a^3-b^3=(a-b)(3ab+(a-b)^2)\)

\(\displaystyle a^3-b^3=(a-b)(3ab+a^2-2ab+b^2)\)

\(\displaystyle a^3-b^3=(a-b)(a^2+ab+b^2)\)