Is this polynomial prime or can it be factored further? 8x^2 + 6x – 5?

[26]
8x^2+6x-5
=8x^2+10x-4x-5
=2x(4x+5)-1(4x+5)
=(4x+5)(2x-1)

Yes it can (hopefully)...

8x^2 + 6x – 5
= 8x^2 + 10x - 4x - 5
= 2x(4x + 4) -1(4x + 5)
= (2x - 1)(4x + 5)

There you have it.

More information on factorization...

http://www.ltcconline.net/greenl/courses/152B/FactoringRatExpr/factor.htm

ax^2 + bx + c

check the discriminant:

b^2 - 4ac

if it is negative, there are no real roots (therefore, no factors in real numbers)
if it is zero, it contains a square of the form a(x-r)(x-r) where r is the unique root.
if it is positive, it has two real roots and can be factored into the form
a(x-m)(x-n) where m and n are the roots.

-----

8x^2 + 6x -5

roots = [ -b +/- SQRT( b^2 - 4ac ) ] / 2a
roots = [ -6 +/- SQRT( 36 + 160) ] / 16
roots = [ -6 +/- 14 ] / 16
root_m = [-6 + 14]/16 = 1/2
root_n = [-6 -14]/16 = -5/4

factorization =
8(x - 1/2)(x + 5/4)

you can "distribute" the 8 as 2*4, like this:
[2(x - 1/2)][4(x + 5/4)] =
(2x - 1)(4x + 5)

This whole process looks long, but it works every time, even if the numbers don't work out to nice fractions.

8x^2 + 6x - 5
= (4x + 5)(2x - 1)

8x^2+6x-5=8x^2-4x+10x-5
=4x(2x-1)+5(2x-1)
=(2x-1)(4x+5)

(2x-1)(4x+5)