x2(x3 + 1) is the best you can do there.
1.6667
Andre Dawson x2, Bob Derier, Ron Santo x5, Greg Maddux x5, Mark Grace x2
x5 square units.
If x2 and x3 are meant to represent x2 and x3, then x2 times x3 = x5 You find the product of exponent variables by adding the exponents.
3²x5³x2²
x2 - x10 + x5 - x12 - x3
X5/X3 5 - 3 = X2 =====That is the exponent part; subtract the bottom exponent from the top exponent. X3/X5 3 - 5 X - 2 =====or, 1/X2 ===
(x1+x2+...+x5)/5 = 34 (x1+x2+...+x5)= 5 * 34 = 170 [(x1+x2+...+x5) + x6]/6 = 35 170 + x6 = 6*35 = 210 x6 =210-170=40 Answer is 40
-x(x2 - 2)(x2 + 3)
x5 / x2 = x3.
x5 = x3 times x2. In this case x3 = 64 so x = cube root of 64 ie 4