MHB Factor: bca² + bcd² + adb² + adc²

[SweatingBear]

The task is to factor

$$bca^2 + bcd^2 + adb^2 +adc^2$$,

which undeniably is a non-trivial one. It turns out that the expression can be factored into $$(ab + cd)(ac + bd)$$. 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]

The task is to factor

$$bca^2 + bcd^2 + adb^2 +adc^2$$,

which undeniably is a non-trivial one. It turns out that the expression can be factored into $$(ab + cd)(ac + bd)$$. 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]

The task is to factor

$$bca^2 + bcd^2 + adb^2 +adc^2$$,

which undeniably is a non-trivial one. It turns out that the expression can be factored into $$(ab + cd)(ac + bd)$$. 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 $$bc $$ from the first two terms and $$ad$$ from the second two simply because they stand out to me.

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

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

$$ab(ac +db) + cd(bd + ac) = ab(ac+bd) + cd(ac+bd)$$

Now I have $$ac+bd$$ in both terms I can factor again to get the answer:

$$(ac+bd)(ab+cd)$$

 

[SweatingBear]

Thanks a lot, looks like trial-and-error will have to do.