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

2 Answers
Mar 13, 2018

-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

Mar 13, 2018

-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))