How do you solve #15x^2 + 26x + 8 = 0#?

1 Answer
Jan 4, 2016

Find a suitable split of the middle term, then factor by grouping to find:

#15x^2+26x+8 = (5x+2)(3x+4)#

Explanation:

Find two factors of #AC = 15*8 = 120# whose sum is #B=26#.

The pair #20#, #6# works:

#20 xx 6 = 120#

#20 + 6 = 26#

Then use these to split the middle term and factor by grouping:

#15x^2+26x+8#

#=(15x^2+20x)+(6x+8)#

#=5x(3x+4)+2(3x+4)#

#=(5x+2)(3x+4)#