It is the number (WITH its sign) that comes before the x3 and so it is -2.
use two phase method s.t x1+x2+x3>=3 -x1+2x2>=2 -x1+5x2+x3<=4 x1,x2,x3>=0
The variable term, X^3, is a third order polynomial term and will render three solutions, though one of those may be zero.
y2=x3+3x2
x3 + 8 = x3 + 23 = (x + 2)[x2 - (x)(2) + 22] = (x + 2) (x2 - 2x + 4)
4X2+X3+3X+7Y+2X3 can only be simplified to 3X3+4X2+3X+7Y, because X3 and 2X3 are the only like terms.
x2 • (5x2 + x + 8)
(3x + 5)(x2 - 2)
Answer this question…A. x4 + 2x3 + 9x2 + 4 B. x4 + 4x3 + 9x2 + 4 C. x4 + 2x3 + 9x2 + 4x + 4 D. x4 + 2x3 + 9x2 - 4x + 4
(x3 - 3x2 + 4x - 7) - (2x3 + x2 - 3x - 5)
Six.
6
(x4 - 2x3 + 2x2 + x + 4) / (x2 + x + 1)You can work this out using long division:x2 - 3x + 4___________________________x2 + x + 1 ) x4 - 2x3 + 2x2 + x + 4x4 + x3 + x2-3x3 + x2 + x-3x3 - 3x2 - 3x4x2 + 4x + 44x2 + 4x + 40R∴ x4 - 2x3 + 2x2 + x + 4 = (x2 + x + 1)(x2 - 3x + 4)
x3 + 5x2 - x - 5 = (x2 - 1)(x + 5) = (x + 1)(x - 1)(x + 5)
3 - 3x + x2 - x3 = (1 - x)(x2 + 3)
x3 + x2 - 3x - 3 x(x2 + x - 3) - 3
They are -1, 1 and 5.