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)
To perform the multiplication of (x + 4)(x + 5), you need to use the distributive property. This means you multiply each term in the first parenthesis by each term in the second parenthesis. (x + 4)(x + 5) = xx + x5 + 4x + 45 Simplify this further to get x^2 + 5x + 4x + 20, which simplifies to x^2 + 9x + 20.
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
The x^5 at the beginning makes the degree of the polynomial 5.
3x3 + 6x2 + x + 2 3x2 (x + 2) + (x + 2) (x + 2) (3x2 + 1) OR 3x3 + x + 6x2 + 2x (3x2 + 1) + 2 (3x2 + 1) (3x2 + 1) (x + 2) Check: (x + 2) (3x2 + 1) = 3x3 + x + 6x2 + 2 = 3x3 + 6x2 + x + 2 x3 - 3x2 - 4x + 12 x2 (x - 3) - 4 (x - 3) (x - 3) (x2 - 4) (x - 3) (x + 2) (x - 2) Check: (x - 3) (x + 2) (x - 2) = (x - 3) (x2 - 4) = x3 - 4x - 3x2 + 12= x3 - 3x2 - 4x + 12=== === x5 - x4 + 8x3 - 8x2 + 16x - 16 x4 (x - 1) + 8x2 (x - 1) + 16 (x - 1) (x - 1) (x4 + 8x2 + 16) (x - 1) (x2 + 4)(x2 + 4) (x - 1) (x2 + 4)2 Real Solution (x - 1) (x + 2i) (x - 2i) (x + 2i) (x - 2i) (x - 1) (x + 2i)2 (x - 2i)2 Check: (x - 1) (x + 2i)2 (x - 2i)2 = (x - 1) (x + 2i) (x - 2i) (x + 2i) (x - 2i) = (x - 1) (x2 + 2xi - 2xi - 4i2) (x2 + 2xi - 2xi - 4i2) = (x - 1) (x2 - 4(-1)) (x2 - 4(-1)) = (x - 1) (x2 + 4) (x2 + 4) = (x - 1) (x4 + 4x2 + 4x2 +16) = (x - 1) (x4 + 8x2 + 16) = x5 + 8x3 +16x - x4 - 8x2 - 16 = x5 - x4 + 8x3 - 8x2 + 16x - 16