math help please?

Question

factor and find zeros...
f(r)=3r^4-108r^2

Answer 1

f(r) = 3r^2(r^2 - 36)

f(r) = 3r^2(r+6)(r-6).... now set = 0

and your roots are....

3r^2 = 0.... so r = 0 is one zero

r + 6 = 0 .... so r = -6 is the 2nd zero

and r - 6 = 0 .... so r = 6 is the 3rd zero

zeros are r = 0, -6, and 6

Answer 2

factor first to get
3r^2(r^2-36)
so either 3r^2=0 or r^2-36 =0
if 3r^2=0 then r=0
if r^2-36=0
(r+6)(r-6)=0
r= -6 or r= 6

so answer is 0, -6, 6

Answer 3

3r^4-108r^2
=3r^2(r^2-36)
=3r^2(r-6)(r+6)
r-6=0 or r+6 =0 or 3r^2=0
r=6 or r = -6 or r = 0

Answer 4

Use GCF to factor out the common terms. Then you'll be left with a difference of squares. Set all factors equal to zero and solve for x. The common factor of 3 and 108 is ___. The common factor of x^4 and x^2 is____ (always take the lower power when you have a common variable). You will be left with ___(x^2-36). Use difference of squares to factor out (x^2-36).... (x+__)(x-__). Set your GCF=0 and (x+__)=0, (x-__)=0...solve.

Answer 5

to be fully correct, you need the answer to be r = 0, 0, -6, 6

the 2 zeros are important since its r^4, not now, but in about 5 years, you'll be glad you learned it the right way

Answer 6

3r^2(r^2-36)=0

3r^2(r-6)(r+6)=0

3r^2=0 r=0
r-6=0 r+6=0
r=0,6,-6