How do you factor completely: #9x^2 + 25x -6#?

2 Answers
Gió
Jul 15, 2015

I found: #(x+3)(9x-2)#

Explanation:

What I did was to solve using the Quadratic Formula the second degree equation:
#9x^2+25x-6=0#
so: #x_(1,2)=(-25+-sqrt(841))/18# that gives you:
#x_1=-3#
#x_2=2/9#
So basically my equation can be written as:
#(x+3)(x-2/9)=0# or
#(x+3)((9x-2)/9)=0#
#(x+3)(9x-2)=0*9=0#

Nghi N.
Jul 16, 2015

Factor: #y = 9x^2 + 25x - 6#

y = (9x + 2)( x - 3)

Explanation:

#y = 9x^2 - 25x - 6 =# a(x - p)(x - q)
I use the new AC Method to factor trinomials. (Google , Yahoo Search)
Converted #y' = x^2 - 25x - 54 =# (x - p')(x - q'). p' and q' have opposite signs. Factor pairs of (-54) --> (-1, 54)(-2, 27). This sum is 25 = -b. Then p' = 2 and q' = -27
#p = 2/9# and #q = -27/9 = - 3#

Factor form: #y = 9(x + 2/9)(x - 3) = (9x + 2)(x - 3)#

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