Is it possible to factor y=−8x2+8x+32? If so, what are the factors?

2 Answers
David O.
Mar 13, 2018

−8 and (x2−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(x2−x−4)

I chose to pull out −8 to leave the coefficient of the term in x2 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

ax2+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:

√b2−4ac

which evaluates to

√(−1)2−4(1)(−4)

=√1+16

=√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±√b2−4ac2a

=−(−1)±√172(−1)

that is

x=−12±√172

George C.
Mar 13, 2018

−8x2+8x+32=−2(2x−1−√17)(2x−1+√17)

Explanation:

Given:

y=−8x2+8x+32

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

A2−B2=(A−B)(A+B)

with A=(2x−1) and B=√17 as follows:

y=−8x2+8x+32

y=−2(4x2−4x+1−17)

y=−2((2x−1)2−(√17)2)

y=−2((2x−1)−√17)((2x−1)+√17)

y=−2(2x−1−√17)(2x−1+√17)

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