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!
6 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
(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
4 answers