How do you write #y=2x^2-10x+5# in factored form?

1 Answer
Sep 20, 2015

Answer:

#y=2(x-(5+sqrt15)/2)(x-(5-sqrt15)/2)#

Explanation:

#2x^2-10x+5=2(x^2-5x+5/2)=#

#=2(x^2-2*x*5/2+(5/2)^2-(5/2)^2+5/2)=#

#2((x-5/2)^2-25/4+10/4)=2((x-5/2)^2-15/4)=#

#=2((x-5/2)^2-(sqrt15/2)^2)=#

#=2(x-5/2-sqrt15/2)(x-5/2+sqrt15/2)=#

#=2(x-(5+sqrt15)/2)(x-(5-sqrt15)/2)#

General case:

#P_2(x)=ax^2+bx+c=a(x^2+b/ax+c/a)=#

#=a(x^2+2*x*b/(2a)+(b/(2a))^2-(b/(2a))^2+c/a)=#

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

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

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

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

You can see that #P_2(x)=0# for

#x-(-b+sqrt(b^2-4ac))/(2a)=0 => x=(-b+sqrt(b^2-4ac))/(2a)#

or

#x+(-b+sqrt(b^2-4ac))/(2a)=0 => x=(-b-sqrt(b^2-4ac))/(2a)#

which is famous formula for solutions of the quadratic equation.