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

Question

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

1 Answer

Impact of this question

1884 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-07T06:55:41

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

Explanation:

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

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

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

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

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

$6 and 8$ $" "6 and 8$
$↓ ↓$ $" "darr" "darr$
$21 \to 3 \times 1 = 3$ $" "2" "1" "rarr3xx1 = 3$
$" "3" "8" "rarr2xx8=ul16$
$color(white)(wwwwwwwwwwwwwwww)19$

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:

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

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you factor 16 = 38r - 12r^216=38r−12r216 = 38r - 12r^2?

1 Answer

Explanation:

Related questions

Impact of this question

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