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

2 Answers
Mar 13, 2018

Answer:

#-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

Answer:

#-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))#