Factor polynomial $$ 25n^{3}+105n^{2}-270n $$

$$ 25n^{3}+105n^{2}-270n = 5n( 5n-9 )( n+6 ) $$

Step 1 :

After factoring out $ 5n $ we have:

$$ 25n^{3}+105n^{2}-270n = 5n ( 5n^{2}+21n-54 ) $$

Step 2 :

Step 2: Identify constants $ a $ , $ b $ and $ c $.
$ a $ is a number in front of the $ x^2 $ term $ b $ is a number in front of the $ x $ term and $ c $ is a constant. In this case:

$$ a = 5 , b = 21 ~ \text{ and } ~ c = -54 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 5 }$ by the constant term $\color{blue}{c = -54} $.

$$ a \cdot c = -270 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = -270 $ and add to $ b = 21 $.

Step 5: All pairs of numbers with a product of $ -270 $ are:

PRODUCT = -270
-1     270	1     -270
-2     135	2     -135
-3     90	3     -90
-5     54	5     -54
-6     45	6     -45
-9     30	9     -30
-10     27	10     -27
-15     18	15     -18

Step 6: Find out which factor pair sums up to $\color{blue}{ b = 21 }$

PRODUCT = -270 and SUM = 21
-1     270	1     -270
-2     135	2     -135
-3     90	3     -90
-5     54	5     -54
-6     45	6     -45
-9     30	9     -30
-10     27	10     -27
-15     18	15     -18

Step 7: Replace middle term $ 21 x $ with $ 30x-9x $:

$$ 5x^{2}+21x-54 = 5x^{2}+30x-9x-54 $$

Step 8: Apply factoring by grouping. Factor $ 5x $ out of the first two terms and $ -9 $ out of the last two terms.

$$ 5x^{2}+30x-9x-54 = 5x\left(x+6\right) -9\left(x+6\right) = \left(5x-9\right) \left(x+6\right) $$

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