That doesn't factor neatly. Applying the quadratic formula, we find two real solutions: (1 plus or minus the square root of 13) divided by 2
x = 2.3027756377319946
x = -1.3027756377319946
x(x4+x3+x2+x+1)
x3 + 4x2 - 16x - 64 = (x + 4)(x + 4)(x - 4).
x cubed, or x^3 x(x2)= x3
Answer: x (x2-x)
x3 - 4x2 + x + 6 The sum of the odd coefficients equals the sum of the even coefficients, so (x + 1) is a factor. So x3 - 4x2 + x + 6 = x3 + x2 - 5x2 - 5x + 6x + 6 = x2(x + 1) - 5x(x + 1) + 6(x + 1) = (x + 1)(x2 - 5x + 6) = (x + 1)(x - 2)(x - 3)
X3 X(X2) X2(X) and, X * X * X
x(x4+x3+x2+x+1)
x3 + 4x2 - 16x - 64 = (x + 4)(x + 4)(x - 4).
Greatest common factor of x4 and x3 is x3.
x cubed, or x^3 x(x2)= x3
Answer: x (x2-x)
x3-x2 Both terms in this expression have x2 in them, so "divide" each term by it using the distributive property in reverse. x2(x-1) = x3-x2 If you "re-distribute" you should see that they are equal.
x(x3 - 24x - 22)
x times x times x
x3 - 4x2 + x + 6 The sum of the odd coefficients equals the sum of the even coefficients, so (x + 1) is a factor. So x3 - 4x2 + x + 6 = x3 + x2 - 5x2 - 5x + 6x + 6 = x2(x + 1) - 5x(x + 1) + 6(x + 1) = (x + 1)(x2 - 5x + 6) = (x + 1)(x - 2)(x - 3)
X squared plus 1 and x minus 1 ---- Actually it is the following: x3-x x(x2-1) x(x+1)(x-1)
x3 + 8 = x3 + 23 = (x + 2)[x2 - (x)(2) + 22] = (x + 2) (x2 - 2x + 4)