Factoring algebraic complex expression

AI Thread Summary
The discussion focuses on factoring the expression ac - bd + adi + bci. The initial attempt involved factoring out 'a' and 'b', leading to a form of a(c + di) - b(d - ci). Participants highlight the similarity between the factors (c + di) and (d - ci), discussing how to identify common factors. A key point raised is the importance of recognizing differences in the real and imaginary components of these expressions. The conversation concludes with a question about whether there is a systematic approach to follow when factoring complex algebraic expressions.
Maxo
Messages
160
Reaction score
1

Homework Statement


Factor the expression ac-bd+adi+bci

Homework Equations




The Attempt at a Solution


We can factor the variable 'a' which gives:
a(c+di)-bd+bci
The common factor in the remaining terms is b, and if we also factor out b we get
a(c+di)-b(d+ci)

But this is not the way I want it factorized. I want it to be factored completely. How can that be done step by step?
 
Physics news on Phys.org
Maxo said:
The common factor in the remaining terms is b, and if we also factor out b we get
a(c+di)-b(d-ci)
Do you realize the similarity between the factors (c+di) and (d-ci)? How can you make them the same?
 
Last edited:
Fightfish said:
Do you realize the similarity between the factors (c+di) and (d-ci)? How can you make them the same?
I made a misstake, it should be a(c+di)-b(d-ci). Anyway so you're saying (c+di) and (d-ci) can be made the same. I actually don't see how that could be done? I mean both the real and the imaginary parts of these expressions are different. How are they the same?
 
Maybe I wasn't very clear with what I meant by "make the same", but what I was trying to convey is identifying the common factor between these two terms. Notice that they differ by a factor of i
 
  • Like
Likes 1 person
For your second step, factor out bi instead of just b. Be more careful with signs and brackets.
 
  • Like
Likes 1 person
Now I see it :)

Is there some rule (algorithm) / procedure to follow when factoring like this?
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top