How do you factor #x^2 - 4x - 2#?

1 Answer
Feb 10, 2016

Answer:

Use the quadratic formula to get:
#color(white)("XXX")(x-2+sqrt(6))(x-2-sqrt(6))#

Explanation:

For a quadratic expression:
#ax^2+bx+c#
the zeros of the expression are given by the formula
#color(white)("XXX")x_z=(-b+-sqrt(b^2-4ac))/(2a)#

For the given expression
#color(white)("XXX")a=1#
#color(white)("XXX")b=-4#
#color(white)("XXX")c=-2#

So
#color(white)("XXX")x_z = (4+-sqrt((-4)^2+4(1)(-2)))/(2(1))#

#color(white)("XXX")=(4+sqrt(16+8))/(2)#

#color(white)("XXX")=(4+-2sqrt(6))/2#

#color(white)("XXX")=2+-sqrt(6)#

If #x_(z1) and x_(z2)# are zeros of a quadratic then the quadratic can be factored as:
#color(white)("XXX")(x-x_(z1))(x-x_(z2))#