1. ## factorising polynomial

i am given a question to show that $\displaystyle x^4-a^4= (x+a)(x^2+a^2)$

hence factorise $\displaystyle x^4-4=$ and find the roots of the equation..

i have used the binomial expansion theorem to prove the first part of the question but i am having difficulty in factorizing the second equation.... anyone can show me how this is done..

i write $\displaystyle x^4-4$ as $\displaystyle x^4-2^2$ but i still cant get it in a form to factorize it.

2. Originally Posted by sigma1
i am given a question to show that $\displaystyle x^4-a^4= (x+a)(x^2+a^2)$

hence factorise $\displaystyle x^4-4=$ and find the roots of the equation..

i have used the binomial expansion theorem to prove the first part of the question but i am having difficulty in factorizing the second equation.... anyone can show me how this is done..

i write $\displaystyle x^4-4$ as $\displaystyle x^4-2^2$ but i still cant get it in a form to factorize it.
Dear sigma1,

You can write, $\displaystyle x^4-4$ as $\displaystyle x^4-(\sqrt{2})^4$

Can you continue from here??

3. i dont think i can manage that one. can you show me how to factorize that to evaluate it for the roots.

4. It's repeated application of the Difference of Two Squares Rule.

$\displaystyle x^4 - 4 = (x^2)^2 - 2^2$

$\displaystyle = (x^2 - 2)(x^2 + 2)$.

$\displaystyle = [x^2 - (\sqrt{2})^2](x^2 + 2)$

$\displaystyle = (x - \sqrt{2})(x + \sqrt{2})(x^2 + 2)$.

So if this was equal to 0...

$\displaystyle (x - \sqrt{2})(x + \sqrt{2})(x^2 + 2) = 0$

$\displaystyle x - \sqrt{2} = 0, x + \sqrt{2} = 0, x^2 + 2 = 0$

$\displaystyle x = \sqrt{2}$ or $\displaystyle x = -\sqrt{2}$ or $\displaystyle x = i\sqrt{2}$ or $\displaystyle x = -i\sqrt{2}$.

5. Originally Posted by sigma1
i dont think i can manage that one. can you show me how to factorize that to evaluate it for the roots.
What he's suggesting is the differnce of squares.

Remember: X^2 - Y^2 = (x-y)(x+y)

In this particular case

We have: (x^2)^2 - (2)^2 = (x^2 - 2) ( x^2 + 2) = (x - sqrt(2))*(x + sqrt(2)) * (x^2 + 2)

Find the roots from there.

6. Originally Posted by Prove It
It's repeated application of the Difference of Two Squares Rule.

$\displaystyle x^4 - 4 = (x^2)^2 - 2^2$

$\displaystyle = (x^2 - 2)(x^2 + 2)$.

$\displaystyle = [x^2 - (\sqrt{2})^2](x^2 + 2)$

$\displaystyle = (x - \sqrt{2})(x + \sqrt{2})(x^2 + 2)$.

So if this was equal to 0...

$\displaystyle (x - \sqrt{2})(x + \sqrt{2})(x^2 + 2) = 0$

$\displaystyle x - \sqrt{2} = 0, x + \sqrt{2} = 0, x^2 + 2 = 0$

$\displaystyle x = \sqrt{2}$ or $\displaystyle x = -\sqrt{2}$ or $\displaystyle x = i\sqrt{2}$ or $\displaystyle x = -i\sqrt{2}$.

since i was given the initial condition
$\displaystyle x^4-a^4= (x+a)(x^2+a^2)$

cant i just substitute the value in to that form and evaluate the roots.

7. Originally Posted by sigma1
since i was given the initial condition
$\displaystyle x^4-a^4= (x+a)(x^2+a^2)$$\displaystyle$

cant i just substitute the value in to that form and evaluate the roots.
You can do anything. In fact, to prove left side = right side all you really have to to is expand the brackets on the right side and this will sufficiently prove the condition.

But getting back to the roots. If you sub in the values into that form where do you go from there? Instead of a difference of squares we now have (something + something) (something squared + something squared) = 0.

Of course we can find the root of the something + something, but how do we find the roots from the other bracket (which has the squared terms)? It's rather difficult yes.

So it is easiar to do it by difference of sqaures. At least as far as I can see.

8. Originally Posted by sigma1
i am given a question to show that $\displaystyle x^4-a^4= (x+a)(x^2+a^2)$

hence factorise $\displaystyle x^4-4=$ and find the roots of the equation..

i have used the binomial expansion theorem to prove the first part of the question but i am having difficulty in factorizing the second equation.... anyone can show me how this is done..

i write $\displaystyle x^4-4$ as $\displaystyle x^4-2^2$ but i still cant get it in a form to factorize it.
Hi everyone,

Hey maybe you guys have'nt noticed, but $\displaystyle x^4-a^4= (x+a)(x^2+a^2)$ is obviously wrong. I think the question must be, "show that $\displaystyle x^4-a^4= (x+a)(x-a)(x^2+a^2)$"

Now one can write $\displaystyle x^4-2^4= (x+2)(x-2)(x^2+2^2)$ and find the roots easily.

9. Originally Posted by Sudharaka
Dear sigma1,

Hey maybe you guys have'nt noticed, but $\displaystyle x^4-a^4= (x+a)(x^2+a^2)$ is obviously wrong. I think the question must be, "show that $\displaystyle x^4-a^4= (x+a)(x-a)(x^2+a^2)$"

Now one can write $\displaystyle x^4-2^4= (x+2)(x-2)(x^2+2^2)$ and find the roots easily.
Ha, yeah that's definately true. I just skimmed over that, never thought there that there might be a problem with a given question lol

$\displaystyle (x+a)(x^2+a^2) = x^3 + xa^2 + ax^2 + a^3.....!= x^4-a^4$

10. Originally Posted by Sudharaka
Hi everyone,

Hey maybe you guys have'nt noticed, but $\displaystyle x^4-a^4= (x+a)(x^2+a^2)$ is obviously wrong. I think the question must be, "show that $\displaystyle x^4-a^4= (x+a)(x-a)(x^2+a^2)$"

Now one can write $\displaystyle x^4-2^4= (x+2)(x-2)(x^2+2^2)$ and find the roots easily.

thanks alot i didnt think that the question in itself could be wrong. then again i should have expanded it.