1. ## Factoring Quartic equation

I really don't understand how they factorise this:

y^4+y^2+1

Can anyone lend me a hand?

2. Originally Posted by nabbiechan
I really don't understand how they solve this:

y^4+y^2+1

Can anyone lend me a hand?
Can you solve $y^2=z$. Then, this becomes $z^2+z+1=0$. Can you solve this?

3. Oh sorry, I put solve instead of factorise. I apologize! >_<

4. Originally Posted by nabbiechan
Oh sorry, I put solve instead of factorise. I apologize! >_<
Factor over the reals? Solve $z^2+z+1=0$ and then use this to solve $y^4+y^2+1=0$ and then use this to factor it into linear factors.

5. Originally Posted by Drexel28
$z^2+z+1=0$ and then use this to solve $y^4+y^2+1=0$
This is the part I'm really confused about.

6. Originally Posted by nabbiechan
This is the part I'm really confused about.
Which part?

7. $y^4 + y^2 + 1$.

Make the substitution $x = y^2$ so that the expression becomes

$x^2 + x + 1$

$= x^2 + x + \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 + 1$

$= \left(x + \frac{1}{2}\right)^2 + \frac{3}{4}$.

This is not factorisable over the Reals.

8. Originally Posted by Drexel28
Which part?
where you're going to use z^2+z+1 to solve the equation.

@prove it:

I only was able to solve it through the calculator and it's factorisable, but I wanted to know the method. The answer was [LaTeX ERROR: Convert failed] . A friend told me to use another equation similar to it which made it more confusing.

9. Originally Posted by nabbiechan
I really don't understand how they factorise this:

y^4+y^2+1

Can anyone lend me a hand?
hi

$y^4+y^2+1$

$=y^4+(2y^2-y^2)+1$

$=(y^4+2y^2+1)-y^2$

$=(y^2+1)^2-y^2$

$=(y^2+1-y)(y^2+1+y)$

using the fact that $a^2-b^2=(a+b)(a-b)$

ok ?

hi

$y^4+y^2+1$

$=y^4+(2y^2-y^2)+1$

$=(y^4+2y^2+1)-y^2$

$=(y^2+1)-y^2$

$=(y^2+1-y)(y^2+1+y)$

using the fact that $a^2-b^2=(a+b)(a-b)$

ok ?

ooh. okay, thanks for everyone's help!