Polynomial question! (Enter if you dare.) (1 Viewer)

[inedible]

If P(x)=x^6+x^4+x^2+1 show that the solutions of P(x)=0 are among the solutions of the equation x^8-1=0. Hence factorize P(x) over R

and

The equation x^n +px - q = 0 has a double root show that (p/n)^n + (q/(n-1))^n-1 = 0

Thank you!

 

[Carrotsticks]

I won't bother doing all the working out because I'm just browsing BOS for a short break before resuming work. I'll tell you the method however

Question 1

Write it out as x^8-1 as (x^2-1)(x^6+x^4+x^2+1) = 0. Since we already know that 2 solutions of x^8-1=0 are 1 and -1, the other solutions of x^8-1 must be the solutions of x^6+x^4+x^2+1=0. From here, just use the complex roots of unity to factorise x^8-1=0 and state that all other solutions except +1 and -1 are the roots of x^6+x^4+x^2+1 =0.

Question 2

Differenciate p(x) and make x the subject. Sub it back into p(x) and play with algebra and factorising till u get what they told u to get.

 

[inedible]

Thanks! I just needed a method to start.