ok im havin a huge brain fart right now so can someone please factor this x^2 -1/4x +1/16?

Answer 1

It may be easier if you got rid of the fractions
x^2 - 1/4 x + 1/16
16x^2 - 4 x + 1
This is not factorable, are you sure you wrote it down correctly?

Answer 2

If you don't like the fractions multiply by 16:

16x^2 + 4x + 1

Better? If not, use the quadratic equation

x = (-4 +- sqrt(16 - 4*16*1))/2*16
= (-4 +- sqrt(-48))/32

So this does not factor over the reals. You can get the complex factors, if you want, from this equation.

Answer 3

x^2 -1/4x +1/16

1/16-4/16=-3/16<0 so there are no real factors. If there is a typo & the eqn shoud be
x^2 -1/2x +1/16 then it =(x-1/4)^2

Answer 4

16x^2+4x+1/16
factor the numerator using the Quadratic formula

Answer 5

x^2 - (1/4)x + (1/16)

Here is a trick, the quadratic formula

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-(-1/4) ± sqrt((-1/4)^2 - 4(1)(1/16)))/(2(1))
x = ((1/4) ± sqrt((1/16) - (4/16)))/2
x = ((1/4) ± sqrt(-3/16))/2
x = ((1/4) ± (i/4)sqrt(3))/2

x = (1/8)(1 ± isqrt(3))

therefore this problem cannot be factored.

if you tried to factor this, it would look something like this

(8x - (1 + isqrt(3))) * (8x - (1 - isqrt(3)))

Answer 6

It has imaginary numbers

(x- (1/8+ i sqrt(3)/8)

change the + to a -

Answer 7

think it was x^2-1/2x+1/16.

Answer 8

use the quadratic formula first to obtain the zeores and it will factor to (x - first zero) * (x - second zero). I dunno if it will be real #'s or imaginary #'s. Too lazy to figure out.

Answer 9

this is not factorable........