Q: Factor x5 plus x2

Write your answer...

Submit

Still have questions?

Continue Learning about Algebra

1.6667

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)

(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.

Related questions

-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.