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

?

Question

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

?

Savannah Anderson · Accepted Answer

You can not factor that trinomial, unfortunately.

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.

Isaiah Anthony · 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)#

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.

Eli Assouline · Answer

undefined To factor the quadratic expression $y = x^2 + 35x + 36$, we need to find two numbers that multiply to 36 (the constant term) and add up to 35 (the coefficient of the linear term). The two numbers are 1 and 36. We can rewrite the middle term as $x^2 + 1x + 36x + 36$. Now, we can factor by grouping:

$x^2 + 1x + 36x + 36$

$= x(x + 1) + 36(x + 1)$

$= (x + 1)(x + 36)$

So, the factored form of $y = x^2 + 35x + 36$ is $y = (x + 1)(x + 36)$.