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Maths problem

lauzie2011

Hey guys ,this is the question ;
x^2-ax+9 is a factor of x^3+bx+c
-Express b in terms of a
-Express c in terms of a
If anybody could give any pointers on how to approach the question ,it would be much appreciated

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JonnyMcNamee

Its a long division question. Just consult your book and you should be able to get it out. I made a botched attempt at it and got something but I'm fairly sure it was wrong!

Sorry I couldnt be of more help!

Neodymium

x^2-ax+9 is a factor of x^3+bx+c

So start by dividing x^3+bx+c by x^2-ax+9

lauzie2011

Thank you so much !

Mwalimu

The important thing here is that, since (x^2 - ax + 9) is a factor, there is no remainder (i.e. any remainder you get must be zero).

So, when you do the long division [you get (x + a) as the other factor], what you have left over must be equal to zero.

By my calculations, that means (a^2 + b - 9) = 0 (coefficient of the remaining 'x' term) and (c - 9a) = 0 (remaining constant term).

Re-arrange these to get what you're asked for.:) [ b = 9 - a^2] [c = 9a]

Out of interest, go back to the original with some value of 'a' (say 2, which then gives b = 5 and c = 18).
The situation then is that x^2 - 2x + 9 is a factor of x^3 + 5x + 18. When you divide using long division, the other factor is (x + 2) and there is no remainder. This confirms that it is a factor and the (x + 2) confirms that 'a' = 2. Neat!;)

N.B. choosing a random value for 'a' at the start doesn't work; you need to work out the expressions for b and c. I only mentioned giving 'a' some value so that you would see how the result stacks up. A specific value doesn't prove the general case - an important point to keep in mind.

lauzie2011

Thank you! Much appreciated

lauzie2011

One more question

The parametric equations of a curve are ;
x=(t+2)^2(e^t) and y=t+ln(t+2)^2
-find dx/dt and dy/dt

tom_tom

pop it into www.wolframalpha.com and it will do it for you

brownacid

lauzie2011 wrote: »

One more question
The parametric equations of a curve are ;
x=(t+2)^2e^t and y=t+ln(t+2)^2
-find dx/dt and dy/dt

Differentiate x with respect to t and do the same for y. The rules for differentiating e and ln are in your log book or maths book.

Neodymium

x = (t+2)^(2e^t)
dx/dt = (2e^t)(t+2)^(2e^t - 1)

y = t + ln(t+2)^2
y = t + 2ln(t+2)
dy/dt = 1 + 2/(t+2) = (t+4)/(t+2)