1. ## factorising

hi, could someone help me with this please:

a polynomial is given by $p(x) = x^3 - 6x^2 + 9x - 4$

a) factorise p(x) completely- i did this question and came up with $(x - 1)^2 (x - 4)$ but then the next question i didn't get

b) hence solve the equation $y^6 - 6y^4 + 9y^2 - 4 = 0$

i'm guessing its something to do with the second equation being the same as the first one but with x squared, so $y = x^2$. but the answer in the book is $y = \pm 1$ and $y = \pm2$ could someone please explain this? thankyou, mark

2. $(y^2)^3-6(y^2)^2+9(y^2)-4$ ... Do you see the similarity between the original polynomial? By taking out a common exponent, I've reproduced an equation extremely similar to $p(x)$, just with a different variable. I've created $p(y^2)$.

Let's just pretend y^2 is equivalent to x, for the sake of simplicity. You found the original polynomial, factored, to be

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

Do you understand my thought process? Are you able to finish it off?

3. if $y = x^2$ why are you making $x = y^2$?

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

5. 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

6. I'll try to make this supremely clear!

=

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

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

I derived y^2 from this equation by realising that I need to manipulate it into a form which somehow relates to ... By using index laws I realised that for to become the substituted value must be y^2.

Hence,
$p(y^2)=(y^2)^3-6(y^2)^2+9(y^2)-4$
$p(y^2)=y^6-6y^4+9y^2-4$

I hope that helped

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

8. 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

9. 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.

10. ,

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 . . . .