If j(x) = x4 - 81, and k(x) = x + 3 then j(x) divide by k(x) =

(x-3)(x^2+9) or x^3 -3x^2 + 9x - 27 First, let's see about factoring x^4 - 81. Cursory examination indicates that it's the difference of two squares and so it initially factors into (x^2 - 9)(x^2 + 9) And the (x^2 - 9) term is also the difference of 2 squares so it too factors into: (x - 3)(x + 3) So a partial factorization of x^4 - 81 is: (x - 3)(x + 3)(x^2 + 9) The (x^2 + 9) term could be factored as well, but that's not needed for this problem, and so I won't do it. Now we can divide (x-3)(x+3)(x^2+9) by (x+3). The (x+3) terms will cancel and we get as the result (x-3)(x+3)(x^2+9) / (x+3) = (x-3)(x^2+9) We can leave the answer as (x-3)(x^2+9), or we can multiply it out, getting: x^3 -3x^2 + 9x - 27

He will need 2,898 pieces of paper