x3 + 4x2 + 6x + 24 = (x2 + 6)(x + 4)
x3 - 7x2 + 6x = x(x2 - 7x + 6) = x(x2 - x - 6x + 6) = x[x(x - 1) - 6(x - 1)] = x2(x - 1) - 6x(x - 1) = (x2 - 6x)(x - 1)
x3 - 7x2 + 6x =x (x2 - 7x + 6) =x (x - 6) (x - 1)
(x2 plus 40) (x minus 1)
x3-x2
x3 + 4x2 + 6x + 24 = (x2 + 6)(x + 4)
x3 - 7x2 + 6x = x(x2 - 7x + 6) = x(x2 - x - 6x + 6) = x[x(x - 1) - 6(x - 1)] = x2(x - 1) - 6x(x - 1) = (x2 - 6x)(x - 1)
x3 - 7x2 + 6x =x (x2 - 7x + 6) =x (x - 6) (x - 1)
(x2 plus 40) (x minus 1)
x3-x2
x3 + 3x2 - 6x - 8 = (x - 2)(x2 + 5x + 4) = (x - 2)(x + 1)(x + 4)
-x(x2 - 2)(x2 + 3)
(x - 4)(x - 2)
x4 - 1.We can not "solve" this as we have not been told the value of x. However, we can simplify this expression:We have an x and a minus x here which will cancel out. Likewise the x2 and x3 will cancel out with the -x2 and -x3 respectively. This therefore leaves us with just x4 - 1.
(x3-3x2+6x+5)/(x-2) = x2-x+4 remainder 13 or as x2-x+4+13/(x-2) It's difficult to show the workings out on this computer but division in algebra has a lot in common when doing long division with integers.
(3x + 5)(x2 - 2)
16 + 6x - x2 = 16 + 8x - 2x - x2 = 8*(2 + x) - x*(2 + x) = (8 - x)*(2 + x)