Factor polynomial $$ 3x^{2}-10x+8 $$

$$ 3x^{2}-10x+8 = ( x-2 )( 3x-4 ) $$

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:

$$ a = 3 , b = -10 ~ \text{ and } ~ c = 8 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 3 }$ by the constant term $\color{blue}{c = 8} $.

$$ a \cdot c = 24 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = 24 $ and add to $ b = -10 $.

Step 4: All pairs of numbers with a product of $ 24 $ are:

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

Step 5: Find out which factor pair sums up to $\color{blue}{ b = -10 }$

PRODUCT = 24 and SUM = -10
1     24	-1     -24
2     12	-2     -12
3     8	-3     -8
4     6	-4     -6

Step 6: Replace middle term $ -10 x $ with $ -4x-6x $:

$$ 3x^{2}-10x+8 = 3x^{2}-4x-6x+8 $$

Step 7: Apply factoring by grouping. Factor $ x $ out of the first two terms and $ -2 $ out of the last two terms.

$$ 3x^{2}-4x-6x+8 = x\left(3x-4\right) -2\left(3x-4\right) = \left(x-2\right) \left(3x-4\right) $$

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