x5+4x4-6x2+nx+2 when divided by x+2 has a remainder of 6 Using the remainder theorem: n = 2
5
x^5+2x^4+4x^2+2x-3
7
The x^5 at the beginning makes the degree of the polynomial 5.
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.
The additive inverse is x5 + 2x - 2.
5x + 5y = 5(x+y)
is a quintic expression in x (NOT an equation).
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
5
x^5+2x^4+4x^2+2x-3
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.
(9+8)x(7-6)x5+4(3+2)