can you do x^5-x^2?

Question

i was wondering if it is prime or you can solve it

Answer 1

You can certainly factor it:

x^2(x^3 - 1)

x^2(x - 1)(x^2 + x + 1)

So, its roots are 0 (repeat root), 1, and two imaginary roots

Answer 2

X^3

Answer 3

Well, whether it is prime depends on the value of x. For example, if x = 1, then the result is zero, which is not a prime number. And if x = 2, then the result is 28, which is also not a prime number.

But it can be reduced to x²(x³ - 1).

Answer 4

x^5-x^2=
x^2(x^3-1)
x^2 will never always be prime and never will be x^3-1 will be prime always

Answer 5

depends on the value of x:
x=2 then x^5-x^2 = 28
x=3 then x^5-x^2 = 234
x=4 then x^5-x^2 = 1008
x=5 then x^5-x^2 = 3100

don't see any primes yet.

Answer 6

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

Actually, you can get an insight by substituting x=10
10^5-10^2
=100000-100
=99900
=100*999
=(10^2)(9*111)
=10^2(10-1)(10^2+10+1)

It does not always work, but when it does, it helps.

Answer 7

You can get 3 real roots and 2 imaginary roots.

You can simplify it as:

x^2(x^3-1)

Further simplification:

x^2(x-1)(x^2+x+1)

From there you get:

X1=0
x2=0
x3=1`

applying quadratic formula to x4 and x5 you get:

x4= -0.5+ 0.87i
x5= -0.5 - 0.87i

Answer 8

It is solveable...factor out x^2 out of each term:
x^2(x^3-1)

set both factors equal to zero....and solve.

Answer 9

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

x^2+x+1 has no real factors.

Answer 10

wouldn't it be 2x^3