Is it possible to factor #y=2x^2 -452x-68#? If so, what are the factors?
1 Answer
Jun 29, 2016
Explanation:
Step 1: Removing the obvious integer common factor:
From this point on there are no "pretty" factors but...
Step 2: Use the quadratic formula to find the roots
For the general quadratic:
#ax^2+bx+c# the roots are given by:
#color(white)("XXX")x=(-b+-sqrt(b^2-4ac))/(2a)# In this case
#color(white)("XXX")x=(226+-sqrt((-226)^2-4(1)(-35)))/(2(1))#
So we can continue the factoring as: