How do you factor the expression #-6x^2-25x-25#?

2 Answers
Leland Adriano Alejandro
Feb 28, 2016

the factors are #-(3x+5)(2x+5)#

Explanation:

Given #-6x^2-25x-25#
common monomial factor #=-1#

so #-6x^2-25x-25=-(6x^2+25x+25)#

factors of 6: could be (3 and 2)
or (6 and 1)

factors of 25: could be (5 and 5)
or (25 and 1)

by inspection and thru practice

best numbers are 3, 2, 5, 5 which could result to 15+10=25
because #3*5+2*5=25# giving us the middle number

so

#-6x^2-25x-25=-(6x^2+25x+25)=-(3x+5)(2x+5)#

God bless ....I hope the explanation is useful.

Nghi N.
Feb 28, 2016

y = -(3x + 5)(2x + 3)

Explanation:

I use the new systematic, non-guessing new AC Method to factor trinomials (Socratic Search)
y = -(6x^2 + 25x + 25) = -6(x + p)(x + q)
Converted trinomial: y' = - (x^2 + 25x + 150) = - (x + p')(x + q').
p' and q' have same sign since ac > 0.
Compose factor pairs of (ac = 150) --> ...(6, 25)(10, 15). This sum is 15 = b. Then p' = 10 and q' = 15.
Therefor: #p = (p')/a = 10/6 = 5/3# and #q = (q')/a = 15/6 = 5/2#.
Factored form: #y = -(x + 5/3)(x + 5/2) = -(3x + 5)(2x + 5)#

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