Factoring Problem

J

Jason76

Senior Member
Joined
Oct 19, 2012
Messages
1,180
This was from a calculus thread:

\(\displaystyle 1 - 9x^{2} + 24x^{4} - 16x^{6}\)

What's the first step? The different terms don't seem to have a common factor, and "factoring by grouping" doesn't seem to be the way.
 
Jason76 said:
This was from a calculus thread:

\(\displaystyle 1 - 9x^{2} + 24x^{4} - 16x^{6}\)

What's the first step? The different terms don't seem to have a common factor, and "factoring by grouping" doesn't seem to be the way.
Click to expand...

Jason I told you in your previous post:

The above is achieved through factorization and application of rational root theorem (http://www.purplemath.com/modules/rtnlroot.htm) - which would become obvious if you substitute k = x2

These are taught in algebra II - high school.
Click to expand...

Did you go to the website I had referred to and learn about rational root theorem?

If you did, please state the theorem for us.
 
\(\displaystyle 1 - 9x^{2} + 24x^{4} - 16x^{6}\)

\(\displaystyle 3x^{2}\) is the common factor of all the terms except the 1. That's all I see.
 
Last edited:
Jason76 said:
\(\displaystyle 3x^{2}\) is the common factor of all the terms except the 1. That's all I see.
Click to expand...

That vision is irrelevant for this problem.

Did you substitute k = x2 ?

Did you read and digest rational root theorem?
 
\(\displaystyle \pm \dfrac{1,3}{1,2,3,6} \)

Is this right? :confused: Leading coefficient is \(\displaystyle 6\), constant term is \(\displaystyle 1\) and factors are \(\displaystyle 1,3\)(numerator), and \(\displaystyle 1,2,3,6\)(denominator - leading coefficient)
 
Last edited:
Jason76 said:
\(\displaystyle \pm \dfrac{1,3}{1,2,3,6} \)

Is this right? :confused: .... Incorrect

Leading coefficient is \(\displaystyle 6\), ... Incorrect ...leading coefficient is -16


constant term is \(\displaystyle 1\) .... correct

and factors are \(\displaystyle 1,3\)(numerator), and \(\displaystyle 1,2,3,6\)(denominator - leading coefficient)
Click to expand...

Fix your original expression after substitution.
 
Jason76 said:
This was from a calculus thread:

\(\displaystyle 1 - 9x^{2} + 24x^{4} - 16x^{6}\)

What's the first step? The different terms don't seem to have a common factor, and "factoring by grouping" doesn't seem to be the way.
Click to expand...
The first thing I would do is to convert this to a cubic. (This strictly speaking is not necessary, but it simplifies things in my opinion.)

\(\displaystyle y = x^2 \implies 1 - 9x^2 + 24x^4 - 16x^6 = 1 - 9y + 24y^2 - 16y^3.\)

The advantage of this is that a cubic can always be factored into a linear term and a quadratic term by the fundamental theorem of algebra. Furthermore, there is a very ugly formula that will give you a root of a cubic. Sensible people avoid that formula if possible.

One possible way to avoid that formula, as Subhotosh Khan has told you, is the rational root theorem. It tells you what to look for if the polynomial has a rational root. It does not guarantee that a rational root exists.

But IF A RATIONAL ROOT EXISTS, it will be in the form \(\displaystyle \pm \dfrac{p}{q}\),

where p is an integer factor of the constant term and q is an integer factor of the coefficient of x^n.

In this case, that means a rational root \(\displaystyle = \pm\ any\ of\ \dfrac{1}{16}, \dfrac{1}{8}, \dfrac{1}{4}, \dfrac{1}{2}, 1.\)

For this problem, it turns out there is at least one rational root. What is it?

How do you use that information to factor the cubic?

What do you get?

Now use the quadratic formula to factor the quadratic term.

What do you get?

Now convert from y back to x.

What do you get?
 
You must log in or register to reply here.
Top