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

Answer 1

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

#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.

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Answer 2

To write y = 2x^2 - 10x + 5 in factored form, you can factor out the common factor, if possible, and then use the quadratic formula or completing the square method to factor the quadratic expression. However, since the given expression doesn't factor easily, you can leave it in standard form as y = 2x^2 - 10x + 5.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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