How do you factor #5y^2-26y+5#?

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Answer 1 · 2015-05-28T19:25:41

Use a version of the AC method...

#A=5#, #B=26#, #C=5#.

#AC=5xx5 = 25#

since the sign of the constant term is #+#, look for a pair of factors of #AC = 25# which add to give #26#. #1# and #25# work.

Use this pair to split the middle term into two, then factor by grouping...

#5y^2 - 26y + 5 = 5y^2 - 25y - y + 5#

#= (5y^2-25y)-(y-5)#

#= 5y(y-5) - (y-5)#

#= (5y-1)(y-5)#