# How do you factor y= -12r^2+5r+3 ?

Jan 5, 2016

$\left(3 r + 1\right) \left(- 4 r + 3\right)$ or indeed $\left(3 r + 1\right) \left(3 - 4 r\right)$

#### Explanation:

to factor this trinomial $- 12 {r}^{2} + 5 r + 3$

compare to general form: $a {x}^{2} + b x + c$

first multiply ac in this case $- 12 \times 3 = - 36$

now consider the factors of - 36 which also add together to give b =5 in this question .

the pairs of factors of - 36 are ( - 2 , 18 ) reject as sum ≠ 5

( - 3 , 12 ) reject as sum ≠ 5

( - 4 , 9 ) accept as sum = 5

now rewrite the function as $- 12 {r}^{2} - 4 r + 9 r + 3$

(Note 5r has been replaced by $- 4 r + 9 r$)

factor the expression in pairs to give: $- 4 r \left(3 r + 1\right) + 3 \left(3 r + 1\right)$

$\left(3 r + 1\right)$ is a common factor here

$\left(3 r + 1\right) \left(- 4 r + 3\right)$3

$\Rightarrow - 12 {r}^{2} + 5 r + 3 = \left(3 r + 1\right) \left(- 4 r + 3\right)$