is a quintic expression in x (NOT an equation).
x^5+2x^4+4x^2+2x-3
a(x5) + b(x5) + c(x5) + d(x5) + e(x5) = abcde(a+b+c+d+e) x5 = abcdeThis equation has at least 5 variables. To solve for all of them requires at least 4 more equations.
x*-4=8+(2-5*x) =-4x=8+2-x5 x*-4=8+(2-5*x) =-4x=8+2-x5 x*-4=8+(2-5*x) =-4x=8+2-x5
The additive inverse is x5 + 2x - 2.
1
5x + 5y = 5(x+y)
x2(x3 + 1) is the best you can do there.
x5+4x4-6x2+nx+2 when divided by x+2 has a remainder of 6 Using the remainder theorem: n = 2
x5 - 2x4 - 24x3 = 0 x3(x2 - 2x - 24) = 0 x3(x2 + 4x - 6x - 24) = 0 x3(x - 6)(x + 4) = 0 x = {-4, 0, 6}
7
x10 =========================== Another contributor says: If the question means [ x5 + x5 ] then the sum is [ 2 x5 ].
The x^5 at the beginning makes the degree of the polynomial 5.