How do you factor #y=x^2 + 35x + 36# ?

2 Answers
Dec 27, 2015

Answer:

You can not factor that trinomial, unfortunately.

Explanation:

You can always see if a quadratic function like the one above is factorable by using the discriminant of the quadratic formula:

#b^2# - 4ac
#35^2# - 4(1)(36)
1120 - 144
976

Since the final result, 976, is not a perfect square, this trinomial cannot be factored into even factors.

Hopefully you understand now.

Dec 27, 2015

Answer:

This quadratic can only be factored using irrational coefficients:

#x^2+35x+36#

#=(x+35/2-sqrt(1081)/2)(x+35/2+sqrt(1081)/2)#

Explanation:

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

So:

#x^2+35x+36#

#=x^2+35x+(35/2)^2-(35/2)^2+36#

#=(x+35/2)^2-1225/4+144/4#

#=(x+35/2)^2-1081/4#

#=(x+35/2)^2-(sqrt(1081)/2)^2#

#=((x+35/2)-sqrt(1081)/2)((x+35/2)+sqrt(1081)/2)#

#=(x+35/2-sqrt(1081)/2)(x+35/2+sqrt(1081)/2)#

We cannot simplify #sqrt(1081)# further since #1081=23*47# has no square factors.