How do you factor $x^2+1/2x+1/4$ ?

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Answer 1 · 2018-03-02T03:42:25

The quadratic has complex factors: $(x-1/4(-1+sqrt(3)i)) and (x-1/4(-1-sqrt(3)i))$

$f(x) = x^2 + 1/2x +1/4$

Now, consider $f(x) =0$

Then if $a and b$ are roots of $f(x) -> f(x) =(x-a)(x-b)$

$f(x)$ is a quadratic with a discriminant of $1/4-4xx1xx1/4 =-3/4$

Since the discriminant $<0 -> a and b$ will be complex.

$x= (-1/2 +-sqrt(-3/4))/2$

$= -1/4 +-1/4sqrt(3)i$

$= 1/4(-1+-sqrt(3)i)$

$:.$ with $a and b$ defined above:

$a = (x-1/4(-1+sqrt(3)i))$
and
$b= (x-1/4(-1-sqrt(3)i))$

Thus, $f(x)= (x-1/4(-1+sqrt(3)i))(x-1/4(-1-sqrt(3)i))$

How do you factor x^2+1/2x+1/4x^2+1/2x+1/4?