Question

Factorise the polynomial.

Answer 1

\(x^{5}+x^{3}+x\)

Answer 2

First,factorise the x.
\(x(x^{4}+x^{2}+1)\)

Answer 3

How do we factorise \(x^{4}+x^{2}+1\) into \((x^{2}-x+1)(x^{2}+x+1)\) ?
If possible,can u pls show me the working.

Answer 4

can we use the method u taught me before to apply in this question?

Answer 5

@Zepdrix

Answer 6

Hmm

Answer 7

Must be some other weird trick...
Hmm trying to figure it out :d

Answer 8

okay

Answer 9

When you apply Rational Root Theorem for example,
it tells you about a specific root (which corresponds to a `linear factor` of the polyomial).

But there are no linear factors of this fourth degree polynomial.
It only breaks down into two quadratics.

So we can't use our usual methods. Hmm.

Answer 10

I c...Okay.

Answer 11

Hmm I dunno broski :c sorry
Maybe try OpenStudy or somewhere else?

Answer 12

It's okay...
I will try to ask this question in OS first.
Thanks for trying it.

Answer 13

@Zepdrix

Answer 14

I got the working on how to factorise \(x^{4}+x^{2}+1\)

Answer 15

http://prntscr.com/dl6mm7

Answer 16

http://prntscr.com/dl6qql

Answer 17

Ah sorry I'm here XD was busy... sec checking..

Answer 18

it's okay...take your time. ^

Answer 19

\(\large\rm p^2+p+1\)

So they found a clever way to apply our conjugates trick.

So what they're doing is, rewriting this as two squares.
So we start by grouping the terms like this,

\(\large\rm (p^2+1)+p\)

And we want to `complete the square` on the bracketed stuff.

Answer 20

So we're missing the middle term, which is 2p.

Answer 21

We're not allowed to do this,
\(\large\rm (p^2+2p+1)+p\)

We can't just introduce 2p into our expression, willy nilly.
We have to keep things balanced.
So we'll subtract 2p at the same time,

\(\large\rm (p^2+2p+1)+p-2p\)

Answer 22

Remember your old math trick, "add this to both sides".
This is the other trick that you want to be comfortable with.
Adding and subtracting the same quantity is the same as making no change at all. We're allowed to do that.

Answer 23

From there we rewrite the bracketed portion as a perfect square,

\(\large\rm (p+1)^2+p-2p\)

And combine like-terms,

\(\large\rm (p+1)^2-p\)

Answer 24

Oh they approached it a little bit differently than I explained.

They rewrote p as 2p-p.
\(\large\rm p^2+1+\color{indianred}{p}\)
\(\large\rm p^2+1+\color{indianred}{2p-p}\)
And then they used the 2p to complete the square,
\(\large\rm p^2+2p+1-p\)

not combining it with the -p.

Answer 25

Now,i get it... :)
Thank you! @Zepdrix

Answer 26

cool

Answer 27

Holy Jesus