# Is it possible to factor #y=-8x^2 +8x+32#? If so, what are the factors?

##### 2 Answers

Solutions to the quadratic requires use of the quadratic formula (or, equivalently, completion of the square)

#### Explanation:

It is possible to pull out a common factor of

I chose to pull out

The question then arises whether there are any further factorisations of the remaining quadratic in the brackets.

Comparing with the form

It might be seen that

These can be used to evaluate the "discriminant" part of the quadratic formula, which is the part:

which evaluates to

As 17 is not a perfect square, there are no further "neat" whole number factorisations of the quadratic. Non-integer (in fact non-rational but not non-real) roots may, of course, be found using the quadratic formula.

For completeness, these are

that is

#### Explanation:

Given:

#y = -8x^2+8x+32#

We can factor this by completing the square and using the difference of squares identity:

#A^2-B^2 = (A-B)(A+B)#

with

#y = -8x^2+8x+32#

#color(white)(y) = -2(4x^2-4x+1-17)#

#color(white)(y) = -2((2x-1)^2-(sqrt(17))^2)#

#color(white)(y) = -2((2x-1)-sqrt(17))((2x-1)+sqrt(17))#

#color(white)(y) = -2(2x-1-sqrt(17))(2x-1+sqrt(17))#