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Factor: bca² + bcd² + adb² + adc²

sweatingbear

Member
May 3, 2013
91
The task is to factor

\(\displaystyle bca^2 + bcd^2 + adb^2 +adc^2\),

which undeniably is a non-trivial one. It turns out that the expression can be factored into \(\displaystyle (ab + cd)(ac + bd)\). I have absolutely no idea what strategy to employ in non-trivial cases such as these (although I did attempt factoring by grouping, however to no real avail). Forum, got any clues/suggestions/tips? Anything helps!
 

kaliprasad

Well-known member
Mar 31, 2013
1,309
The task is to factor

\(\displaystyle bca^2 + bcd^2 + adb^2 +adc^2\),

which undeniably is a non-trivial one. It turns out that the expression can be factored into \(\displaystyle (ab + cd)(ac + bd)\). I have absolutely no idea what strategy to employ in non-trivial cases such as these (although I did attempt factoring by grouping, however to no real avail). Forum, got any clues/suggestions/tips? Anything helps!
try to group it
bca^2 + bcd^2 + adb^2 +adc^2
= bc(a^2+d^2) + ad(b^2 + c^2)
no

bca^2 + adb^2 + bcd^2 + adc^2
= ab( ac+ bd) + bc(bd + ad) ( I got it)
= ( ac+bd)(ab+ cd)

it is done

did you miss something
 

SuperSonic4

Well-known member
MHB Math Helper
Mar 1, 2012
249
The task is to factor

\(\displaystyle bca^2 + bcd^2 + adb^2 +adc^2\),

which undeniably is a non-trivial one. It turns out that the expression can be factored into \(\displaystyle (ab + cd)(ac + bd)\). I have absolutely no idea what strategy to employ in non-trivial cases such as these (although I did attempt factoring by grouping, however to no real avail). Forum, got any clues/suggestions/tips? Anything helps!
Essentially it comes down to trial and error and practice.

I would seek to factor \(\displaystyle bc \) from the first two terms and \(\displaystyle ad\) from the second two simply because they stand out to me.

\(\displaystyle bc(a^2+d^2) + ad(b^2+c^2)\). This just leads to a dead end though.

[hr][/hr]

Let's try \(\displaystyle ab\) and \(\displaystyle cd\) respectively (some question setters aren't very imaginative (Wink))

\(\displaystyle ab(ac +db) + cd(bd + ac) = ab(ac+bd) + cd(ac+bd)\)

Now I have \(\displaystyle ac+bd\) in both terms I can factor again to get the answer:

\(\displaystyle (ac+bd)(ab+cd)\)
 

sweatingbear

Member
May 3, 2013
91
Thanks a lot, looks like trial-and-error will have to do.