### Welcome to our community

#### sweatingbear

##### Member

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

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!

##### Well-known member

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

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
= bc(a^2+d^2) + ad(b^2 + c^2)
no

= 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

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

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 )

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

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

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

#### sweatingbear

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