Is it possible to factor #y=2x^2 + 9x - 5 #? If so, what are the factors?

1 Answer
Jan 1, 2016

Answer:

#y = (2x-1)(x+5)#

Explanation:

As it so happens, any quadratic can be factored into the form

#ax^2 + bx + c#
#= a(x - (-b+sqrt(b^2-4ac))/(2a))(x-(-b-sqrt(b^2-4ac))/(2a))#

That big block comes from the quadratic formula, along with the fact that given a polynomial #P(x)#, if #P(a)=0#, then #(x-a)# is a factor of #P(x)#.

However, sometimes (when #b^2-4ac < 0#) the factors involve complex numbers, and there is also a lot to calculate there. Let's see if we can find some real factors without resorting to that right away.

If we have the factorization
#(Ax + B)(Cx + D) = 2x^2 +9x - 5#
then expanding the left hand side and comparing coefficients gives us the system

#{(AC = 2), (AD+BC = 9), (BD = -5):}#

If we limit ourselves to integer solutions, then we quickly see from the first and third equations that #A# and #C# have to be #+-1# and #+-2# and that #B# and #D# have to be #+-1# and ∓#5#. A quick check will show that

#{(A = 2), (B = -1), (C = 1), (D = 5):}#

is a solution. Thus, substituting back, we have

#(2x-1)(x+5) = 2x^2+9x-5#