Is it possible to factor #y=x^2+5x-3 #? If so, what are the factors?

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Answer 1 · 2016-04-09T12:22:33

Factors are #(x+5/2-sqrt37/2)(x+5/2+sqrt37/2)#

In #y=x^2+5x-3#, the discriminant is given by #5^2-4*1*(-3)=25+12=37#, which is positive but not a square of a rational number.

Hence, we can have irrational factors of #y=x^2+5x-3#.

As roots of #x^2+5x-3=0# are #x=(-5+-sqrt37)/2#

or #x=-5/2+sqrt37/2# and #x=-5/2-sqrt37/2#

Hence factors of #y=x^2+5x-3# are

#(x-(-5/2+sqrt37/2))(x-(-5/2-sqrt37/2))# or

#(x+5/2-sqrt37/2)(x+5/2+sqrt37/2)#