Factor polynomial $$ 6x^{2}-19x+8 $$

$$ 6x^{2}-19x+8 = ( 3x-8 )( 2x-1 ) $$

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:

$$ a = 6 , b = -19 ~ \text{ and } ~ c = 8 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 6 }$ by the constant term $\color{blue}{c = 8} $.

$$ a \cdot c = 48 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = 48 $ and add to $ b = -19 $.

Step 4: All pairs of numbers with a product of $ 48 $ are:

PRODUCT = 48
1     48	-1     -48
2     24	-2     -24
3     16	-3     -16
4     12	-4     -12
6     8	-6     -8

Step 5: Find out which factor pair sums up to $\color{blue}{ b = -19 }$

PRODUCT = 48 and SUM = -19
1     48	-1     -48
2     24	-2     -24
3     16	-3     -16
4     12	-4     -12
6     8	-6     -8

Step 6: Replace middle term $ -19 x $ with $ -3x-16x $:

$$ 6x^{2}-19x+8 = 6x^{2}-3x-16x+8 $$

Step 7: Apply factoring by grouping. Factor $ 3x $ out of the first two terms and $ -8 $ out of the last two terms.

$$ 6x^{2}-3x-16x+8 = 3x\left(2x-1\right) -8\left(2x-1\right) = \left(3x-8\right) \left(2x-1\right) $$

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