How do you factor $18 + 8 r^{2} - 30 r$ $18+8r^2-30r$ ?

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Answer 1 · 2015-05-25T16:17:29

$18 + 8 r^{2} - 30 r$ $18+8r^2-30r$

$= 2 (4 r^{2} - 15 r + 9)$ $=2(4r^2-15r+9)$

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

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:

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

How do you factor 18+8r^2-30r18+8r2−30r18+8r^2-30r?