
factorising
hi, could someone help me with this please:
a polynomial is given by $\displaystyle p(x) = x^3  6x^2 + 9x  4$
a) factorise p(x) completely i did this question and came up with $\displaystyle (x  1)^2 (x  4)$ but then the next question i didn't get
b) hence solve the equation $\displaystyle y^6  6y^4 + 9y^2  4 = 0$
i'm guessing its something to do with the second equation being the same as the first one but with x squared, so $\displaystyle y = x^2$. but the answer in the book is $\displaystyle y = \pm 1$ and $\displaystyle y = \pm2$ could someone please explain this? thankyou, mark

$\displaystyle (y^2)^36(y^2)^2+9(y^2)4$ ... Do you see the similarity between the original polynomial? By taking out a common exponent, I've reproduced an equation extremely similar to $\displaystyle p(x)$, just with a different variable. I've created $\displaystyle p(y^2)$.
Let's just pretend y^2 is equivalent to x, for the sake of simplicity. You found the original polynomial, factored, to be http://www.mathhelpforum.com/mathhe...949518831.gif
By resubstituting (y^2) into the factored form of the polynomial we get:
$\displaystyle (y^21)^2(y^24)$
Do you understand my thought process? Are you able to finish it off?

if $\displaystyle y = x^2$ why are you making $\displaystyle x = y^2$?

x doesn't actually equal $\displaystyle y^2$... For the sake of simplicity, it's easier to imagine $\displaystyle y^2$ as x in order to understand that $\displaystyle y^2$ can be substituted into the factored form of the $\displaystyle p(x)$.

i can see now that by the answer given in the book that that your way of putting it actually does work, i'm just struggling to see why you can put y^2 in place of x in the brackets

I'll try to make this supremely clear!
http://www.mathhelpforum.com/mathhe...bff286851.gif= http://www.mathhelpforum.com/mathhe...949518831.gif
Then,
$\displaystyle p(1)=(11)^2(14)$ correct?
So,
$\displaystyle p(y^2)=(y^21)^2(y^24)$
http://www.mathhelpforum.com/mathhe...f5f636d71.gif
I derived y^2 from this equation by realising that I need to manipulate it into a form which somehow relates to http://www.mathhelpforum.com/mathhe...bff286851.gif... By using index laws I realised that for http://www.mathhelpforum.com/mathhe...a3250f341.gif to become http://www.mathhelpforum.com/mathhe...f5f636d71.gif the substituted value must be y^2.
http://www.mathhelpforum.com/mathhe...a3250f341.gif
Hence,
$\displaystyle p(y^2)=(y^2)^36(y^2)^2+9(y^2)4$
$\displaystyle p(y^2)=y^66y^4+9y^24$
I hope that helped :)

ok thanks, that has shed a bit of light on it

Hopefully somebody else will be able to articulate a response that best suits your frame of mind :)
These questions are somewhat intuitive and are definitely tricky if you've never dealt with them before :(

true, sometimes i'll read something in the book and not have a clue what it means, then someone will explain in a different way and i'll understand instantly. i do understand your explanation now though so thanks.

http://www.mathhelpforum.com/mathhe...f5f636d71.gif,
let y^2 = x
x^3  6x^2 + 9x  4 = 0,
Next step is to break down the middle terms, a la "trial and error";
x^3  4x^2  2x^2 + 8x + x  4 = 0, then group them,
(x^3  4x^2) + (2x^2 + 8x) + (x  4) = 0, factorize
x^2(x  4)  2x(x  4) + (x  4) = 0,
you will notice that we extracted the (x  4), factor it.
(x  4)(x^2  2x + 1) = 0, but notice that (x  1)^2 = x^2  2x + 1,
then we have,
(x  4)(x  1)^2 = 0. Now, it is easy to solve the roots
but x = y^2,
then (y^2  4)(y^2  1)^2 = 0.
by zerofactor theorem, roots can be obtained.
y^2  4 = (y + 2)(y  2) = 0,
y^2  1 = (y + 1)(y  1) = 0,
y = {2, 1, 1, 2}, solution set
Factoring a cubic polynomial is very hard, even a trained eye blinks. More practice helps . . . .