Re: factoring problem

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Posted by Soroban on October 03, 2002 at 17:50:31:

In Reply to: factoring problem posted by lisa on October 03, 2002 at 15:59:00:

: I am not sure what the easiest fastest way to do a problem like this. Do I expand first, it really gets messy when I do that.

: {[(x-3)^2][(2x-3)^4]}+{[12(x-3)^3][(2x-3)^2]}

Lisa,

I'm not sure WHAT we're supposed to "do" to the problem.
What are the directions? Multiply it out? Factor it? Solve for x?
(Yes, I KNOW that it's not an equation - duh! But how many non-equations
have we been asked to solve here?)

What Dennis said is absoluately correct.

I'll go a step further. Take a good look at the problem.

You have: (something)^2 (another)^4 + 12(something)^3 (another)^2

or: a^2 b^4 + 12 a^3 b^2

How hard is that to factor?

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Comments:
: : I am not sure what the easiest fastest way to do a problem like this. Do I expand first, it really gets messy when I do that. : : {[(x-3)^2][(2x-3)^4]}+{[12(x-3)^3][(2x-3)^2]} : Lisa, : I'm not sure WHAT we're supposed to "do" to the problem. : What are the directions? Multiply it out? Factor it? Solve for x? : (Yes, I KNOW that it's not an equation - duh! But how many non-equations : have we been asked to solve here?) : What Dennis said is absoluately correct. : I'll go a step further. Take a good look at the problem. : You have: (something)^2 (another)^4 + 12(something)^3 (another)^2 : or: a^2 b^4 + 12 a^3 b^2 : How hard is that to factor?

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