How do you factor # 7y^2+19y+10 #?

2 Answers
Aug 8, 2015

Answer:

#7y^2+19y+10 = (7y+5)(y+2)#

Explanation:

If #7y^2+19y+10# can be factored into two binomials:
#color(white)("XXXX")##(ay+b)(cy+d)#
#color(white)("XXXX")##color(white)("XXXX")##=acy^2+(ad+bc)y+bd#
then we need to find factors
#color(white)("XXXX")##a, c# of #7#
and
#color(white)("XXXX")##b, d# of #10#
such that
#color(white)("XXXX")##ad+bc=19#

The only integer factors of #7# are
#color(white)("XXXX")##(a,c) = (7,1)#
The (non-negative) integer factors of #10# are
#color(white)("XXXX")##(b,d) in {(1,10), (10,1), (2,5),(5,2)}#

This gives us 4 possible combinations to check and we find:
#color(white)("XXXX")##ad + bc = (7xx2) + (5xx1)# gives the required #19#

So #(ay+b)(cy+d)# becomes #(7y+5)(1y+2)#

Aug 8, 2015

Answer:

Factor: #y = 7y^2 + 19y + 10#

Ans: (7x + 5)(x + 2)

Explanation:

I use the new AC Method.
Converted #y' = y^2 + 19y + 70 = #(y - p')(y - q').
p' and q' have same sign (Rule of signs)
Factor pairs of 70 --> (2, 35)(5, 14). This sum is 19 = b.
Then, p' = 5 and q' = 14.
Therefor, #p = (p')/a = 5/7,# and #q = 14/7 = 2#

Factored form: y = 7(y + 5/7)(y + 2) = (7y + 5)(y + 2).