How do you factor the expression #25y^2- 52y + 27#?

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2 Answers
Bhavin B.
May 21, 2018

#(25y-27)(y-1)#

Explanation:

We need to find the factors that when multiplied it gives #675# #(25 xx 27)# and when added it gives #-52#.

Multiplying:
#675# = #3xx3xx3xx5xx5# = #27 xx 25#

Adding:
#-52# = #-27-25#
Hence, #-27 xx -25 = 675#

So the factors are: #-27# and #-25#

#25y^2-52y+27#

#25y^2-25y-27y+27#

#25y(y-1)-27(y-1)#

#(25y-27)(y-1)#

Check the answer:
#(25y-27)(y-1)#
#(25yxxy)-25y-27y+27#
#25y^2-52y+27#

George C.
May 21, 2018

#25y^2-52y+27 = (y-1)(25y-27)#

Explanation:

Given:

#25y^2-52y+27#

Note that #25-52+27 = 0#

Hence #y=1# is a zero and #(y-1)# a factor.

The leading term of the other factor must be #25y# to get #25y^2# in the product and the trailing term must be #-27# in order to get #+27# in the product.

So we find:

#25y^2-52y+27 = (y-1)(25y-27)#

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