How do you factor #18+8r^2-30r#?

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Answer 1 · 2015-05-25T16:17:29

#18+8r^2-30r#

#=2(4r^2-15r+9)#

#4r^2-15r+9# is in the form #ar^2+br+c#, with #a=4#, #b=-15# and #c=9#.

This has discriminant given by the formula:

#Delta = b^2-4ac = (-15)^2-(4xx4xx9)#

#= 225 - 144 = 81 = 9^2#

Since this is a perfect square, the quadratic equation #4r^2-15r+9 = 0# has two distinct real rational roots, given by the formula:

#r = (-b+-sqrt(Delta))/(2a) = (15+-9)/8#

That is #r = 3/4# and #r = 3#

From this we can deduce:

#4r^2-15r+9 = (4r-3)(r-3)#

So:

#18+8r^2-30r = 2(4r-3)(r-3)#