Factor polynomial $$ 16x^{2}-24x+8 $$

$$ 16x^{2}-24x+8 = 8( x-1 )( 2x-1 ) $$

Step 1 :

After factoring out $ 8 $ we have:

$$ 16x^{2}-24x+8 = 8 ( 2x^{2}-3x+1 ) $$

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:

$$ a = 2 , b = -3 ~ \text{ and } ~ c = 1 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 2 }$ by the constant term $\color{blue}{c = 1} $.

$$ a \cdot c = 2 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = 2 $ and add to $ b = -3 $.

Step 5: All pairs of numbers with a product of $ 2 $ are:

PRODUCT = 2
1     2	-1     -2

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

PRODUCT = 2 and SUM = -3
1     2	-1     -2

Step 7: Replace middle term $ -3 x $ with $ -x-2x $:

$$ 2x^{2}-3x+1 = 2x^{2}-x-2x+1 $$

Step 8: Apply factoring by grouping. Factor $ x $ out of the first two terms and $ -1 $ out of the last two terms.

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

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