Factor polynomial $$ 24y^{3}+6y^{2}-20y $$

$$ 24y^{3}+6y^{2}-20y = 2y( 12y^{2}+3y-10 ) $$

Step 1 :

After factoring out $ 2y $ we have:

$$ 24y^{3}+6y^{2}-20y = 2y ( 12y^{2}+3y-10 ) $$

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:

$$ a = 12 , b = 3 ~ \text{ and } ~ c = -10 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 12 }$ by the constant term $\color{blue}{c = -10} $.

$$ a \cdot c = -120 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = -120 $ and add to $ b = 3 $.

Step 5: All pairs of numbers with a product of $ -120 $ are:

PRODUCT = -120
-1     120	1     -120
-2     60	2     -60
-3     40	3     -40
-4     30	4     -30
-5     24	5     -24
-6     20	6     -20
-8     15	8     -15
-10     12	10     -12

Step 6: Find out which factor pair sums up to $\color{blue}{ b = 3 }$

Step 7: Because none of these pairs will give us a sum of $ \color{blue}{ 3 }$, we conclude the polynomial cannot be factored.

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