Question

Factorise each of the following polynomials.

Answer 1

\(x^{4}-x^{2}-72\)

Answer 2

The constant term for \(x^{4}-x^{2}-72\) is -72
The factors of -72 are \(\pm1,\pm2\pm3,\pm4,\pm6,\pm8\pm9,\pm12,\pm18,\pm24,\pm36,\pm72\)

Answer 3

Do I need to show all the factors of -72 using factor theorem?

Answer 4

Normally, with something scary like a 4th degree polynomial,
you would just need to do the hard work of finding `one root`.

So you'd plug each value in one by one,
until one of them gives you zero as a result.

But this problem is much easier.
It's actually a quadratic in disguise.

Answer 5

so,there is another method to solve it?

Answer 6

\(\large\rm x^{4}-x^{2}-72\quad=\quad (\color{orangered}{x^2})^2-\color{orangered}{x^2}-72\)

If we make this clever substitution, \(\large\rm \color{orangered}{u=x^2}\)

\(\large\rm \color{orangered}{u}^2-\color{orangered}{u}-72\)
Do you see how it will be more manageable from here?

Answer 7

If that substitution business was confusing, you can let me know.

Answer 8

I find it easy :)
I rather use this method than using factor theorem xD

Answer 9

Heh :)
So them umm... hmm it looks like it factors easily, doesn't it?

Answer 10

72 is 9 and 8

Answer 11

i tried finding x^2=8 and x^2=9
x^2=8 (got decimals) so,i will hv to use x^2=9

Answer 12

\(\large\rm u^2-u-72\quad=\quad (u-9)(u+8)\quad=\quad (x^2-9)(x^2+8)\)

I don't think you can set x^2 equal to 8, right?

Answer 13

But yes, the other bracket should factor down further (Hint: conjugates).

Answer 14

ah xD
Yes :)

Answer 15

i got x=3

Answer 16

Well they just want you to leave it factored,
not give the actual solutions :)

Answer 17

Conjugates, right?
\(\large\rm a^2-b^2=(a-b)(a+b)\)

So then,
\(\large\rm x^2-3^2=(x-3)(x+3)\)

Answer 18

yes

Answer 19

Yes, 8 is 2^3, but we're not going to be be able to do anything nice with that :) So probably best to leave it as an 8.

Answer 20

okay

Answer 21

So ya, there is your final answer after you expand out the conjugates,
\(\large\rm (x-3)(x+3)(x^2+8)\)

Answer 22

That is factored completely.

Answer 23

*Thumbs up*

Answer 24

This method is only use for \(x^{b}\)
b=2,4,6,8...

Answer 25

The conjugates thing?
Yes. Because you can write any even power of x as a square.

Example: \(\large\rm x^8-9\quad=\quad (x^4)^2-3^2\)
Here I was able to rewrite the first term as a square because the power is even.
Therefore,

\(\large\rm x^8-9\quad=\quad (x^4-3)(x^4+3)\)

Answer 26

okay :)
Thank you! @Zepdrix

Answer 27

np