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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: macostar on April 25, 2014, 09:07:24 am

Title: Polynomial equations
Post by: macostar on April 25, 2014, 09:07:24 am: Any help with this would be greatly appreciated. No matter what I do I cant get the real values for A and B.
Title: Re: Polynomial equations
Post by: Zealous on April 25, 2014, 09:57:53 am: Quote from: macostar on April 25, 2014, 09:07:24 am
Any help with this would be greatly appreciated. No matter what I do I cant get the real values for A and B.
$P(z)=z^3-az^2+bz-4 \hspace{4mm} P(2)=0 \ & \ P(1-i)=0$

if P(2)=0 and P(1-i)=0, then (z-2) and (z-1+i) are factors of P(z).

As all the coefficients of z in P(z) are real numbers, we can use the conjugate root theorem and hence (z-1-i) must also be a factor.

We can now express P(z) as:

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

Expand:

$P(z)=z^3-4z^2+6z-4$

Therefore a=4 and b=6.

The question asks for full working, so you'd need to show more working when expanding.

Another method (I like the first one better though):
Spoiler
Subbing in z=2
$P(2)=-4a+2b+4=0$

Subbing in z=1-i
$P(1-i)=b-6+2ai-bi-2i=0$

You can then use simultaneous equations to solve for a and b, equating real and imaginery components.
Title: Re: Polynomial equations
Post by: macostar on April 25, 2014, 11:15:24 am: Thanks for that Zealous big help.