Factorising Algebraic expression with no such factors [closed]

I am asked to factorise $12r^2 + 8r - 15$, but I do not get any two such factors. Do I leave the answer as $12r^2 + 8r - 15$?

closed as off-topic by Shailesh, R_D, Jack's wasted life, Alex Mathers, Behrouz MalekiSep 25 '16 at 20:33

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shailesh, R_D, Jack's wasted life, Alex Mathers, Behrouz Maleki
If this question can be reworded to fit the rules in the help center, please edit the question.

• Is it written correctly or did you mean $12 r^2$? – Moo Sep 25 '16 at 11:58
• Maybe $20r-15=5(4r-3)$? This depends on the context. – Bernard Sep 25 '16 at 11:59
• could it be that there is a typo? – Dr. Sonnhard Graubner Sep 25 '16 at 12:00
• yes sorry, i meant 12r^2 – Dan Khan Sep 25 '16 at 12:01
• is it $$12r^2+8r-15$$? – Dr. Sonnhard Graubner Sep 25 '16 at 12:01

solving the given equation after the solution formula of a quadratic equation we get $$r_1=-\frac{3}{2}$$ or $$r=\frac{5}{6}$$ thus we have the factorization $$12(r+\frac{3}{2})(r-\frac{5}{6})$$
One way might be $$12{ r }^{ 2 }+8r-15=12\left( { r }^{ 2 }+\frac { 2 }{ 3 } r-\frac { 5 }{ 4 } \right) =12\left( { r }^{ 2 }+\frac { 2 }{ 3 } r+\frac { 1 }{ 9 } -\frac { 1 }{ 9 } -\frac { 5 }{ 4 } \right) =12\left( { \left( r+\frac { 1 }{ 3 } \right) }^{ 2 }-\frac { 49 }{ 36 } \right) =12\left( r+\frac { 1 }{ 3 } -\frac { 7 }{ 6 } \right) \left( r+\frac { 1 }{ 3 } +\frac { 7 }{ 6 } \right) =\\ =12\left( r-\frac { 5 }{ 6 } \right) \left( r+\frac { 3 }{ 2 } \right)$$