Step 1 :

After factoring out $ b $ we have:

$$ 3b^{3}-13b^{2}+10b = b ( 3b^{2}-13b+10 ) $$

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:

$$ a = 3 , b = -13 ~ \text{ and } ~ c = 10 $$

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

$$ a \cdot c = 30 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = 30 $ and add to $ b = -13 $.

Step 5: All pairs of numbers with a product of $ 30 $ are:

PRODUCT = 30
1     30	-1     -30
2     15	-2     -15
3     10	-3     -10
5     6	-5     -6

Step 6: Find out which factor pair sums up to $\color{blue}{ b = -13 }$

PRODUCT = 30 and SUM = -13
1     30	-1     -30
2     15	-2     -15
3     10	-3     -10
5     6	-5     -6

Step 7: Replace middle term $ -13 x $ with $ -3x-10x $:

$$ 3x^{2}-13x+10 = 3x^{2}-3x-10x+10 $$

Step 8: Apply factoring by grouping. Factor $ 3x $ out of the first two terms and $ -10 $ out of the last two terms.

$$ 3x^{2}-3x-10x+10 = 3x\left(x-1\right) -10\left(x-1\right) = \left(3x-10\right) \left(x-1\right) $$

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