polynomial question please help! (2 Viewers)

yorkstanham

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Can anyone please help with this question:

The equation x³ + x² - 2x - 3 = 0 has roots a, b, c. Find the equation with the roots:

a) a²bc, ab²c, abc²

b) 2a+b+c, a+2b+c, a+b+2c

Thanks in advance
 
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Nope. It only has 1 real root. The other 2 are complex.

See http://www.angelfire.com/ab7/fourunit/Image-5.jpg for the roots a,b,c

See http://www.angelfire.com/ab7/fourunit/Image-7.jpg for a solution to questions a) and b)

See http://www.angelfire.com/ab7/fourunit/Image-8.jpg for an alternative solution to a) and b) (a bit shorter)

Answers for a) and b):

a) x<sup>3</sup>+3x<sup>2</sup>-18x-81=0
b) x<sup>3</sup>+4x<sup>2</sup>+3x-3=0

BTW, this is Arnold & Arnold (4 Unit Cambridge Mathematics) Ex. 4.3 Q14 or Problem 26 in Vladimir Yastreboff's pdf: http://members.optusnet.com.au/hoahie/4unit/T2POLYL2.PDF
with &alpha;, &beta;, &gamma; replaced by a,b,c.
 
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yorkstanham's school motto says, De virtute in virtutem. How very apt!

So why concern yourself with such pedestrian rubbish?

See if you can do this one instead.

If two distinct points on E: y<sup>2</sup>=x<sup>3</sup>+x<sup>2</sup>-2x-3 are A(x<sub>1</sub>, y<sub>1</sub>), B(x<sub>2</sub>, y<sub>2</sub>) with x<sub>1</sub>&ne;x<sub>2</sub> show that if &lambda;=(y<sub>2</sub>-y<sub>1</sub>)/(x<sub>2</sub>-x<sub>1</sub>) then there is a third point C on E through AB given by
(&lambda;<sup>2</sup>-x<sub>1</sub>-x<sub>2</sub>-1, &lambda;(&lambda;<sup>2</sup>-2x<sub>1</sub>-x<sub>2</sub>-1)+y<sub>1</sub>).

And when you are finished, read

Silverman, J. H., The Arithmetic of Elliptic Curves, Springer, 1986.
 
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