# factorising

• Sep 14th 2009, 03:18 AM
mark
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
• Sep 14th 2009, 03:38 AM
Finley
$\displaystyle (y^2)^3-6(y^2)^2+9(y^2)-4$ ... 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/math-he...94951883-1.gif

By re-substituting (y^2) into the factored form of the polynomial we get:
$\displaystyle (y^2-1)^2(y^2-4)$

Do you understand my thought process? Are you able to finish it off?
• Sep 14th 2009, 03:46 AM
mark
if $\displaystyle y = x^2$ why are you making $\displaystyle x = y^2$?
• Sep 14th 2009, 03:51 AM
Finley
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 $\displaystyle p(x)$.
• Sep 14th 2009, 03:56 AM
mark
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
• Sep 14th 2009, 04:00 AM
Finley
I'll try to make this supremely clear!

http://www.mathhelpforum.com/math-he...bff28685-1.gif= http://www.mathhelpforum.com/math-he...94951883-1.gif

Then,
$\displaystyle p(1)=(1-1)^2(1-4)$ correct?

So,
$\displaystyle p(y^2)=(y^2-1)^2(y^2-4)$

http://www.mathhelpforum.com/math-he...f5f636d7-1.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/math-he...bff28685-1.gif... By using index laws I realised that for http://www.mathhelpforum.com/math-he...a3250f34-1.gif to become http://www.mathhelpforum.com/math-he...f5f636d7-1.gif the substituted value must be y^2.

http://www.mathhelpforum.com/math-he...a3250f34-1.gif
Hence,
$\displaystyle p(y^2)=(y^2)^3-6(y^2)^2+9(y^2)-4$
$\displaystyle p(y^2)=y^6-6y^4+9y^2-4$

I hope that helped :)
• Sep 14th 2009, 04:03 AM
mark
ok thanks, that has shed a bit of light on it
• Sep 14th 2009, 04:08 AM
Finley
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 :(
• Sep 14th 2009, 04:10 AM
mark
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.
• Sep 14th 2009, 07:49 AM
pacman
http://www.mathhelpforum.com/math-he...f5f636d7-1.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 zero-factor 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 . . . .