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#1 2008-04-14 19:16:24

soha
Real Member
Registered: 2006-07-07
Posts: 2,530

co efficients

Find the co efficients of x and x² in the following products:-

(3x-1)(3x-5)(3x+4)


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#2 2008-04-15 04:08:35

Dragonshade
Member
Registered: 2008-01-16
Posts: 147

Re: co efficients

For 3x could only choose (-5 ,4).   -1 could choose (3x,4) and (-5,3x)
3x*(-5)*4+(-1)*3x*4+(-1)*(-5)*3x = -60x-12x+15x=-57x

I'll leave the second one to you

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#3 2008-04-15 04:51:07

LuisRodg
Real Member
Registered: 2007-10-23
Posts: 322

Re: co efficients

Expanding it is the fastest way in my opinion

So coefficient of x^3 is 27, coefficient of x^2 is -18 and coefficient of x is -57.

Last edited by LuisRodg (2008-04-15 04:53:10)

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#4 2008-04-15 06:01:16

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: co efficients

(3x-1)(3x-5)(3x+4)

Going for the coefficient of x^2:

3x*3x*4 = 36x^2
3x * -5 * 3x = -45x^2
-1 * 3x * 3x = -9x^2

x^2: -18

This won't have much of an advantage over expanding the entire thing for small multiplications, but as you get larger this method is probably prefered.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#5 2008-04-15 18:38:35

soha
Real Member
Registered: 2006-07-07
Posts: 2,530

Re: co efficients

thanks


"Let us realize that: the privilege to work is a gift, the power to work is a blessing, the love of work is success!" smile smile
- David O. McKay

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#6 2008-04-15 21:59:43

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

Re: co efficients

It is interesting to compare this type of problem
to finding the volume of a cuboid or rectangular "cube".
If each side is broken into 2 parts, then
you get:
volume = (a+b)(c+d)(e+f)
Note that this is different than
finding the area of the cuboid:
surface area = 2(a+b)(c+d) + 2(a+b)(e+f) + 2(c+d)(e+f)
The volume again broken up is:
volume = ace + ade + acf + adf + bce + bde + bcf + bdf


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