Question

What are the restrictions for 4x^2-x+4x/5(3x-1)^2?? Please help!!!

Answer 1

@hartnn ?? Can you help me! I have been working on this problem for forever.

Answer 2

restrictions on values of 'x' right ?

Answer 3

Yes I already simplified it and multiplied it together, and this is the end product. But I don't know the restrictions bc it's a complex denominator.

Answer 4

complex denominator ?
denominator = 5 (3x-1)^2
right ?
restrictions : denominator = 0
so, ''x can't take values for which denominator = 5 (3x-1)^2 = 0
so,
3x-1 = 0
can you solve this for x ?

Answer 5

Oh, so it's 1/3, but what about the 5? And would you write the final answer-numerator like 4x^2-x+4x-1, or (x+1)(4x-1)?

Answer 6

about the 5 :
\(5 (3x-1)^2=0 \\ \)
divide both sides by 5 , 0/5 = 0
\((3x-1)^2=0\)
take square root,
3x-1 = 0
x=1/3 is the restriction.
numerator is a polynomial , so in this case it does not play any role.

Answer 7

THANK YOU!!!! :D

Answer 8

welcome ^_^