Factorise fully this expression: dx + by + bx + dy

[Rob_Brown]

Could someone help me with this expression?

I can factorise most expressions fully. But having problems with this one. I have not seen one like this before. It has all common factors, so don't understand what to do.

Thanks.


dx + by + bx + dy

 

[Deleted member 4993]

Could someone help me with this expression?

I can factorise most expressions fully. But having problems with this one. I have not seen one like this before. It has all common factors, so don't understand what to do.

Thanks.


dx + by + bx + dy

dx + by + bx + dy = dx + bx + by + dy = x*(d+b) + y * (d+b) ....... continue.....

 

[stapel]

Could someone help me with this expression?

I can factorise most expressions fully. But having problems with this one. I have not seen one like this before. It has all common factors, so don't understand what to do.

dx + by + bx + dy

To expand upon the excellent reply provided earlier, let me point out that, any time you're faced with four terms like this (and assuming you've already taken out front the factors common to all four terms, if any), then you probably want to think "factoring in pairs". Then see if there's a way you can set up pairs of terms so that you can end up with a common factor, between the two sets of terms.

In your case, you've got:

. . . . .dx + by + bx + dy = (dx + by) + (bx + dy) (nothing factors)

. . . . .dx + bx + by + dy = (dx + bx) + (by + dy) = x(d + b) + y(b + d) = x(b + d) + y(b + d)

. . . . .dx + dy + by + bx = (dx + dy) + (by + bx) = d(x + y) + b(y + x) = d(x + y) + b(x + y)

Either way, you'll end up with the same factored form -- other than perhaps the order of the summands in one or another of your factors. But, since the order in addition doesn't matter, you're good!