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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: lqzeng on September 20, 2010, 05:36:55 pm

Title: How to expand cubic functions by hand
Post by: lqzeng on September 20, 2010, 05:36:55 pm: Hey guys, just need help on doing something I never actually learnt how to do by hand.

How do I expand
(z-1-2i) (z-1+2i) (z+3)

by hand? (show steps and method if possible)

Sorry, I don't know how to the mathematical font, but thanks in advance! :)
Title: Re: How to expand cubic functions by hand
Post by: Martoman on September 20, 2010, 06:14:07 pm: Decent question as this is overlooked and there is a shortcut that most don't recognise.

Usual process if it weren't as nice as you defined would be to expand one into another and then rinse and repeat. so say you had:

$(x+1)(x+2)(x+3)$
ignore the $(x+3)$ term momentarily.

$(x+1)(x+2) = x^2+3x+2$

Now $(x^2+3x+2)(x+3) = x^3 +3x^2+3x^2+9x+2x+6$

$= x^3 +6x^2 +11x+6$

Because of multiplication rules you could have done it $(x+2)(x+3)$ then $\times(x+1)$

The Yr6 reason for this is because A*B*C = B*A*C = C*A*B or any other arrangement you can think of. :smitten:

So now to your question you could do that method I described above or recognise that the two first brackets are complex conjugates and so you can apply difference of perfect squares.

$((z-1)-2i)((z-1)+2i)$

$(z-1)^2-(2i)^2 = (z-1)^2 +4$ (as $i^2 = -1$ )

so $z^2-2z+1+4 = z^2-2z+5$

Multiplying through $(z^2-2z+5)(z+3) = z^3 +3z^2-2z^2-6z+5z+15 = z^3 +z^2 -z+15$
Title: Re: How to expand cubic functions by hand
Post by: lqzeng on September 20, 2010, 06:21:49 pm: Thank you, good sir! You star.
Title: Re: How to expand cubic functions by hand
Post by: Martoman on September 20, 2010, 06:44:01 pm: Quote from: lqzeng on September 20, 2010, 06:21:49 pm
Thank you, good madame! You star.

:smitten: