Find all real roots

j(x)=32x^10-33x^5+1
I understand this part of the working out:
32(x⁵)²-33(x⁵)+1
But I don't understand how they did this part of the working out: (32x⁵-1x⁵-1)=0
X⁵=1/32 or X=1/2
Can someone explain pls

Its a hidden quadratic to get the values for x^5 (youve missed a couple of brakets out in your factorization) so
(32x⁵-1)(x⁵-1)=0
Then its just looking at each factor and reasoning its zero as usual and then take the 5th root.

OK thank you but how did u get the -1 in both brackets

Thought you said youd understood it. You just factorise the quadratic
32z^2 - 33z + 1
where z=x^5

I get that part but I'm confused on how u actually factorise it

If all else fails, just use the quadratic formula, but why not expand the above and see how it matches.