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

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

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  • $\begingroup$ Is it written correctly or did you mean $12 r^2$? $\endgroup$ – Moo Sep 25 '16 at 11:58
  • $\begingroup$ Maybe $20r-15=5(4r-3)$? This depends on the context. $\endgroup$ – Bernard Sep 25 '16 at 11:59
  • $\begingroup$ could it be that there is a typo? $\endgroup$ – Dr. Sonnhard Graubner Sep 25 '16 at 12:00
  • $\begingroup$ yes sorry, i meant 12r^2 $\endgroup$ – Dan Khan Sep 25 '16 at 12:01
  • $\begingroup$ is it $$12r^2+8r-15$$? $\endgroup$ – Dr. Sonnhard Graubner Sep 25 '16 at 12:01
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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 $$12(r+\frac{3}{2})(r-\frac{5}{6})$$

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One way might be $$12{ r }^{ 2 }+8r-15=12\left( { r }^{ 2 }+\frac { 2 }{ 3 } r-\frac { 5 }{ 4 } \right) =12\left( { r }^{ 2 }+\frac { 2 }{ 3 } r+\frac { 1 }{ 9 } -\frac { 1 }{ 9 } -\frac { 5 }{ 4 } \right) =12\left( { \left( r+\frac { 1 }{ 3 } \right) }^{ 2 }-\frac { 49 }{ 36 } \right) =12\left( r+\frac { 1 }{ 3 } -\frac { 7 }{ 6 } \right) \left( r+\frac { 1 }{ 3 } +\frac { 7 }{ 6 } \right) =\\ =12\left( r-\frac { 5 }{ 6 } \right) \left( r+\frac { 3 }{ 2 } \right) $$

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