Need to factor under radical cubic root[X5} cubic root[X2 * X3] now bring out the X3 X*cubic root[X2] -----------------------
1.6667
x^5 - x^3 + x^2 - 1 = (x - 1)(x + 1)(x + 1)(x^2 - x + 1)
(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
The additive inverse is x5 + 2x - 2.
-x(x2 - 2)(x2 + 3)
7
x5-1 = (x - 1)(x4 + x3 + x2 + 1)
1.6667
x^5+2x^4+4x^2+2x-3
A power can be factored in many different ways; for example: x5 = x times x4 = x2 times x3. When such a power appears as a term in combination with other expressions, you should look out for common factoring patterns. Here are two examples: x2 + x5 Here, you can use the pattern "common factor". In this case, both parts have the common factor x2, so you can factor it out. x4 - 1 Here, you can use the pattern "difference of squares". Note that x4 is the square of x2, and 1 is the square of 1.
Need to factor under radical cubic root[X5} cubic root[X2 * X3] now bring out the X3 X*cubic root[X2] -----------------------
x^5 - x^3 + x^2 - 1 = (x - 1)(x + 1)(x + 1)(x^2 - x + 1)
Andre Dawson x2, Bob Derier, Ron Santo x5, Greg Maddux x5, Mark Grace x2
x5 square units.
Since x3 is a factor of x5, it is automatically the GCF.
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.