How do you factor #16 + 20y - 6y^2#?

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Answer 1 · 2015-05-16T13:50:00

#16+20y-6y^2 = -2(3y^2-10y-8)#

#3y^2-10y-8# is of the form #ay^2+by+c#, with #a=3#, #b=-10# and #c=-8#.

The general quadratic solution tells us that #ay^2+by+c = 0# when

#y=(-b+-sqrt(b^2-4ac))/(2a)#

So substituting our values of #a#, #b# and #c#, that gives

#y=(10+-sqrt(10^2-4*3*(-8)))/(2*3)#

#=(10+-sqrt(100+96))/6#

#=(10+-sqrt(196))/6#

#=(10+-14)/6#

That is for #y=4# and #y=-2/3#

So #(y-4)# and #(3y+2)# are both factors of #3y^2-10y-8#

In summary #16+20y-6y^2 = -2(y-4)(3y+2) = 2(4-y)(3y+2)#