answersLogoWhite

0

Factor x5 plus x2

Updated: 4/28/2022
User Avatar

Wiki User

11y ago

Best Answer

x2(x3 + 1) is the best you can do there.

User Avatar

Wiki User

11y ago
This answer is:
User Avatar

Add your answer:

Earn +20 pts
Q: Factor x5 plus x2
Write your answer...
Submit
Still have questions?
magnify glass
imp
Related questions

How do you factor -x5-x3 plus 6x?

-x(x2 - 2)(x2 + 3)


What is x to the fifth power plus x to the second power minus 4 divided by x plus 3?

7


How do you factor x to the fifth minus one?

x5-1 = (x - 1)(x4 + x3 + x2 + 1)


What is x5 divided by x3?

1.6667


What is the expntents of x5 - 3x4 plus x2 - 4 plus 5x4 plus 3x2 plus 2x plus 1?

x^5+2x^4+4x^2+2x-3


How do you factor when there's powers in the equation?

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.


What is the cubed root of x to the fifth?

Need to factor under radical cubic root[X5} cubic root[X2 * X3] now bring out the X3 X*cubic root[X2] -----------------------


What is the factor x to the fifth power - x to the third power plus x to the second power - 1?

x^5 - x^3 + x^2 - 1 = (x - 1)(x + 1)(x + 1)(x^2 - x + 1)


Who were the last 5 golden glove winners for the cubs?

Andre Dawson x2, Bob Derier, Ron Santo x5, Greg Maddux x5, Mark Grace x2


What is the area of a rectangle x2 by x3?

x5 square units.


What is the greatest common factor of x5 and x3?

Since x3 is a factor of x5, it is automatically the GCF.


What is x2 times x3?

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.