Is it possible to factor #y=x^2+8x+14 #? If so, what are the factors?

1 Answer
Jan 2, 2016

Answer:

Yes, you have to solve the equation though.

#x^2+8x+14=(x+4-sqrt(2))(x+4+sqrt(2))#

Explanation:

By finding the roots #x_1# and #x_2# you can factor using the formula:

#ax^2+bx+c=a(x-x_1)(x-x_2)#

#x^2+8x+14=0#

#Δ=b^2-4*a*c=8^2-4*1*14=8#

#x=(-b+-sqrt(Δ))/(2*a)=(-8+-sqrt(8))/(2*1)=-8/2+-sqrt(4*2)/2=#
#=-4+-(sqrt4)*sqrt(2))/2=-4+-(2*sqrt(2))/2=-4+-sqrt(2)#

Now that the roots #x_1=-4+sqrt(2)# and #x_2=-4-sqrt(2)# are found the factoring can be done as follows:

#x^2+8x+14=1*(x-(-4+sqrt(2)))(x-(-4-sqrt(2)))=#
#=(x+4-sqrt(2))(x+4+sqrt(2))#

An easier example for a better understanding

Factor the following function:

#y=x^2+8x+12#

We solve the equation:

#x^2+8x+12=0#

#Δ=b^2-4*a*c=8^2-4*1*12=16#

#x=(-b+-sqrt(Δ))/(2*a)=(-8+-sqrt(16))/(2*1)=-8/2+-4/2=-4+-2#

#x_1=-4+2=-2#
#x_2=-4-2=-6#

Now we have:

#x^2+8x+12=1(x-(-2))(x-(-6))=(x+2)(x+6)#