For the polynomial below list each real zero and its multiplicity.

f(x) = (x + 1/1)^4 (x^2 + x)^5
Really need some help with this! Have worked it twice and got two different answers!

I think you have a typo, why would you have 1/1 ?

I do it's 1/4 instead. sorry

the correct equation is F(x) = (x + 1/4)^4 (x^2 + 1)^5

for the zeros , set F(x) = 0

(x+ 1/4)^4 (x^2+1)^5 = 0
(x + 1/4)^4 = 0
means (x + 1/4)(x + 1/4)(x + 1/4)(x + 1/4)=0
so x= -1/4 or x = -1/4 ... do you get the idea?
x = -1/4, multiplicity of that zero is 4
or
x^2+1 = 0
x^2 = -1
no real solution , so the only real zero is -1/4

Ok, think I finally got this. Got to a certain point and was wondering what now? Thank you so much!

f(x)=3|x-1|, find the value of f(3)-f(-2)/3

To find the real zeros and their multiplicities of the polynomial f(x) = (x + 1/1)^4 (x^2 + x)^5, we need to set each factor equal to zero and solve for x.

1) Setting (x + 1/1)^4 = 0
To solve this, we can simply solve for x as follows:
(x + 1/1)^4 = 0
Taking the fourth root of both sides, we get:
x + 1/1 = 0
x = -1

So, the real zero for this factor is x = -1, and its multiplicity is 4.

2) Setting (x^2 + x)^5 = 0
To solve this, we can use factoring or the quadratic formula. Let's use factoring:
(x^2 + x)^5 = 0
x^2 + x = 0

This equation can be factored as:
x(x + 1) = 0

So the possible solutions are:
x = 0
x + 1 = 0
Which gives:
x = 0
x = -1

Since the factor (x^2 + x) is raised to the fifth power, both values of x = 0 and x = -1 have a multiplicity of 5.

So the list of real zeros and their multiplicities for the polynomial f(x) = (x + 1/1)^4 (x^2 + x)^5 is:

Real zero: x = -1, Multiplicity: 4
Real zero: x = 0, Multiplicity: 5