Is it possible to factor #y=x^2/2 +10x + 22 #? If so, what are the factors?

1 Answer
Feb 9, 2016

Answer:

This can not be factored using integer values.

#" "x~= -2.57" or "x~=17.483" to 3 decimal places"#

Explanation:

given: #" " y=1/2 x^2+10x+22#

#y=1/2(x^2+5x+11)#

That did not work so back to the formula

Standard form#" "-> ax^2+bx+c=0#

where#" "x=(-b+-sqrt(b^2-4ac))/(2a)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#x=(-10+-sqrt(10^2-4(1/2)(22)))/(2(1/2) #

#x=(-10+-sqrt(100-44))/(1)#

#x=-10+-sqrt(56)#

#56=8xx7->sqrt(56)=2sqrt(14)#

So the factors are:

#[x-(10+2sqrt(14)color(white)(.))color(white)(.)] [x-(10-2sqrt(14)color(white)(.) )color(white)(.)]#

#x~= -2.57" or "x~=17.483" "# to 3 decimal places