Is the term factorable in this Algebra problem?

  • Thread starter Thread starter Miike012
  • Start date Start date
  • Tags Tags
    Algebra
AI Thread Summary
The term 9a^2 - 6a + 1 - x^2 - 8dx - 16d^2 is factorable. By grouping the first three and the last three terms, it can be expressed as (3a - 1)^2 - (x + 4d)^2. This leads to the factored form (3a - 1 + x + 4d)(3a - 1 - x - 4d). The discussion clarifies that while the expression has factors, it does not represent an equation with roots unless set equal to zero. The final insight emphasizes that if treated as a function in x, roots can be expressed in terms of the constants a and d.
Miike012
Messages
1,009
Reaction score
0
Is this term factorable?

9a^2 - 6a + 1 -x^2 - 8dx - 16d^2

I don't see anything that I can group or any simple factors to factor out...
 
Physics news on Phys.org
Miike012 said:
Is this term factorable?

9a^2 - 6a + 1 -x^2 - 8dx - 16d^2

I don't see anything that I can group or any simple factors to factor out...

Group the first 3 and the second 3 with careful management of - signs
 
9a^2 - 6a + 1 - (x^2 + 8dx + 16d^2)

= (3a - 1)^2 - (x + 4d)^2

= (3a - 1 + x + 4d)(3a - 1 - x - 4d)

How do you know what the roots are?
 
Miike012 said:
9a^2 - 6a + 1 - (x^2 + 8dx + 16d^2)

= (3a - 1)^2 - (x + 4d)^2

= (3a - 1 + x + 4d)(3a - 1 - x - 4d)

How do you know what the roots are?

Not exactly sure what roots you are after? Expressions have factors - you have done that. Equations have roots - you don't have an equation?

EDIT: The following post spilled the beans about what I was trying to get you to think about.
 
Last edited:
If it were a function in x where a and d are unknown constants, you could set the factored expression equal to 0, solve for x, and express the roots in terms of a and d.
 
thank you for your guys help.
 

Thread 'Family of lines that are at a distance of 5 from the origin'

I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
View full post »

Thread 'Making the denominator of a certain fraction real'

Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
View full post »

Similar threads

Rational functions: combine and simplify terms
Replies
10
Views
2K
Factoring difference of squares not working?
Replies
3
Views
2K
Prove that terms cyclic in ##a,b,c##, are in ##\text{AP}##
Replies
7
Views
2K
Solve the simultaneous equations involving x, y, and their squares
Replies
2
Views
1K
Write each polynomial as the product of it's greatest common
Replies
16
Views
4K
Factorize a number in a different base
Replies
1
Views
1K
Prime factors of a unique form in the each term a sequence?
Replies
2
Views
2K
Use Remainder theorem to find factors of ##(a-b)^3+(b-c)^3+(c-a)^3##
Replies
4
Views
2K
Understanding Cubic Factorization: Solving for Roots with a and -2a
Replies
6
Views
1K
Doubt solving a polynomial inequality
Replies
4
Views
2K

Hot Threads

Recent Insights

Back
Top