Factor polynomial $$ 3s^{2}-25s+50 $$

$$ 3s^{2}-25s+50 = ( s-5 )( 3s-10 ) $$

Step 1: 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:

$$ a = 3 , b = -25 ~ \text{ and } ~ c = 50 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 3 }$ by the constant term $\color{blue}{c = 50} $.

$$ a \cdot c = 150 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = 150 $ and add to $ b = -25 $.

Step 4: All pairs of numbers with a product of $ 150 $ are:

PRODUCT = 150
1     150	-1     -150
2     75	-2     -75
3     50	-3     -50
5     30	-5     -30
6     25	-6     -25
10     15	-10     -15

Step 5: Find out which factor pair sums up to $\color{blue}{ b = -25 }$

PRODUCT = 150 and SUM = -25
1     150	-1     -150
2     75	-2     -75
3     50	-3     -50
5     30	-5     -30
6     25	-6     -25
10     15	-10     -15

Step 6: Replace middle term $ -25 x $ with $ -10x-15x $:

$$ 3x^{2}-25x+50 = 3x^{2}-10x-15x+50 $$

Step 7: Apply factoring by grouping. Factor $ x $ out of the first two terms and $ -5 $ out of the last two terms.

$$ 3x^{2}-10x-15x+50 = x\left(3x-10\right) -5\left(3x-10\right) = \left(x-5\right) \left(3x-10\right) $$

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