Question

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

Answer 1

`-8` and `(x^2 - x -4)`
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 `8` (or `-8`) from the outset to yield:

`-8(x^2 - x -4)`

I chose to pull out `-8` to leave the coefficient of the term in `x^2` as positive for convenience.

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

Comparing with the form

`ax^2 + bx + c`

It might be seen that
`a` corresponds to `1`
`b` corresponds to `-1`
`c` corresponds to `-4`

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

`sqrt(b^2 - 4ac)`

which evaluates to

`sqrt((-1)^2 - 4 (1) (-4))`

`=sqrt(1 +16)`

`=sqrt(17)`

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

`(-b +- sqrt(b^2 - 4ac))/(2a)`

`= (-(-1) +- sqrt(17))/(2(-1))`

that is

`x = -1/2 +- sqrt(17)/2`

Answer 2

`-8x^2+8x+32 = -2(2x-1-sqrt(17))(2x-1+sqrt(17))`

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 `A=(2x-1)` and `B=sqrt(17)` as follows:

`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))`