Is it possible to factor #y=4x^2+19x - 5 #? If so, what are the factors?

1 Answer
Meave60
Mar 26, 2018

#y=(x+5)(4x-1)#

Explanation:

Factor:

#y=4x^2+19x-5#

You can factor the right-hand side by using the "splitting the middle term" method.

Multiply the coefficient of the first term by the constant.

#4xx-5=-20#

Find two numbers that when added equal #-19#, which is the coefficient of the middle term, and that multiply to #-20#. The numbers #20# and #-1# meet the requirements.

Split #19x# as the sum of #20x# and #-x#.

#y=4x^2+20x-x-5#

Factor out the common terms in the first two terms and the second two terms.

#y=4x(x+5)-(x+5)#

It is understood that #-(x+5)# is #-1(x+5)#.

Factor out the common term #x+5#.

#y=(x+5)(4x-1)#

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