Is the term factorable in this Algebra problem?

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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.
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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...
 
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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.
 

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