How do you factor #16b^2+60b-100#?

1 Answer
May 19, 2015

I use the new AC Method (Google, Yahoo Search)
#f(x) = 4(4x^2 + 15x - 25) = #4(x - p)(x - q)

Converted function: f'(x) = 4(x^2 + 15x - 100) = 4(x - p')(x - q')
Factor pairs of (-100): (-2, 50)(-4, 25)(-5, 20). This last sum is 15 = b.
Then p' = -5 and q' = 20
Then,# p= (p')/a = -5/4 #and #q = (q')/a = 20/4 = 5#

Factored form of f(x) :
#f(x) = 4(x - 5/4)(x + 5) =4 (4x - 5)(x + 5)#

Check by developing:#f(x) = 4(4x^2 + 20x - 5x - 25).# OK