# simplifying long algebraic expressions

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• Sep 1st 2009, 01:43 PM
absvalue
simplifying long algebraic expressions
Hi,

I have an expression that I need to simplify:

\$\displaystyle (i - n)(i - m)(i - y)(i - s)\$

I know how to simplify this algebraically using the FOIL method. I did it and I have the answer, but I'm really hoping there might be a different way to do it. Is there an easier way to simplify long algebraic expressions like this, especially given that i is a factor in each part?
• Sep 1st 2009, 01:48 PM
Kasper
Quote:

Originally Posted by absvalue
Hi,

I have an expression that I need to simplify:

\$\displaystyle (i - n)(i - m)(i - y)(i - s)\$

I know how to simplify this algebraically using the FOIL method. I did it and I have the answer, but I'm really hoping there might be a different way to do it. Is there an easier way to simplify long algebraic expressions like this, especially given that i is a factor in each part?

Is this \$\displaystyle i\$ the imaginary unit? Or just a regular variable?

Keep in mind \$\displaystyle i\$ is NOT a factor in each term.

\$\displaystyle i(1-n)\neq(i-n)\$ We can't start pulling out \$\displaystyle i\$'s.
• Sep 1st 2009, 01:55 PM
absvalue
I can't believe that I've been staring at the problem for 2 hours and missed that. Thanks for your help! I'll be sure to remember that.
• Sep 1st 2009, 02:00 PM
absvalue
Wait... quick question.

\$\displaystyle i(1 - n)(1 - m)(1 - y)(1 - s)\$

Wouldn't this evaluate to be:

\$\displaystyle
(i - ni)(i - mi)(i - yi)(i - si)\$

And wouldn't that have a different value than the original expression? Please correct me if I'm missing something. Thanks.
• Sep 1st 2009, 02:02 PM
pickslides
Kasper has changed his/her post.
• Sep 1st 2009, 02:05 PM
Kasper
If the \$\displaystyle i\$ you are dealing with IS the imaginary unit, expand and you will end up with a \$\displaystyle i^4\$ and a \$\displaystyle i^2\$ term, which you can use to simplify into whole numbers.
• Sep 1st 2009, 02:05 PM
absvalue
Quote:

Originally Posted by Kasper
Is this \$\displaystyle i\$ the imaginary unit? Or just a regular variable?

Keep in mind \$\displaystyle i\$ is NOT a factor in each term.

\$\displaystyle i(1-n)\neq(i-n)\$ We can't start pulling out \$\displaystyle i\$'s.

Sorry, I didn't see that you had updated your post. i is just a regular variable.
• Sep 1st 2009, 02:14 PM
Kasper
Quote:

Originally Posted by absvalue
Sorry, I didn't see that you had updated your post. i is just a regular variable.

I'm not sure then, are you sure the question was to simplify or to expand? This expression is in its simplest form.
• Sep 1st 2009, 02:29 PM
pickslides
Quote:

Originally Posted by Kasper
This expression is in its simplest form.

(Rofl)
• Sep 1st 2009, 03:35 PM
absvalue
Quote:

Originally Posted by Kasper
I'm not sure then, are you sure the question was to simplify or to expand? This expression is in its simplest form.

Yes, just double checked. The question was to simplify.
• Sep 1st 2009, 04:20 PM
HallsofIvy
Quote:

Originally Posted by absvalue
Wait... quick question.

\$\displaystyle i(1 - n)(1 - m)(1 - y)(1 - s)\$

Wouldn't this evaluate to be:

\$\displaystyle
(i - ni)(i - mi)(i - yi)(i - si)\$

No, it wouldn't. Since you have only one i, it can be multiplied into only one term you could write it as
(i-ni)(1- m)(1- y)(1- x) or
(1- n)(i- mi)(1- y)(1- x) or
(1- n)(1- m)(i- yi)(1- x) or
(1- n)(1- m)(1- y)(i- xi).

Quote:

And wouldn't that have a different value than the original expression? Please correct me if I'm missing something. Thanks.