How do you factor $16 = 38r - 12r^2$ ?

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Answer 1 · 2017-08-07T06:55:41

$2(2r-1)(3r-8)=0$

Set it equal to $0$
Take out the common factor of $2$ first.

$12r^2-38r+16 =0$

$2(6r^2-19r+8)=0$

Find factors of $6 and 8$ whose products ADD to $-19$

As a hint, we see that $19$ is odd. Therefore the factors of $8$ cannot be $2 and 4$ , because their products will be even, so use $1 and 8$

$" "6 and 8$
$" "darr" "darr$
$" "2" "1" "rarr3xx1 = 3$
$" "3" "8" "rarr2xx8=ul16$
$color(white)(wwwwwwwwwwwwwwww)19$

The signs in the brackets will both be negative.

$2(2r-1)(3r-8)=0$

These are the factors which will lead to solutions of:

$r = 1/2 and r = 8/3$

How do you factor 16 = 38r - 12r^216 = 38r - 12r^2?