Factor polynomial $$ 40x^{2}-36x+8 $$

$$ 40x^{2}-36x+8 = 4( 2x-1 )( 5x-2 ) $$

Step 1 :

After factoring out $ 4 $ we have:

$$ 40x^{2}-36x+8 = 4 ( 10x^{2}-9x+2 ) $$

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:

$$ a = 10 , b = -9 ~ \text{ and } ~ c = 2 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 10 }$ by the constant term $\color{blue}{c = 2} $.

$$ a \cdot c = 20 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = 20 $ and add to $ b = -9 $.

Step 5: All pairs of numbers with a product of $ 20 $ are:

PRODUCT = 20
1     20	-1     -20
2     10	-2     -10
4     5	-4     -5

Step 6: Find out which factor pair sums up to $\color{blue}{ b = -9 }$

PRODUCT = 20 and SUM = -9
1     20	-1     -20
2     10	-2     -10
4     5	-4     -5

Step 7: Replace middle term $ -9 x $ with $ -4x-5x $:

$$ 10x^{2}-9x+2 = 10x^{2}-4x-5x+2 $$

Step 8: Apply factoring by grouping. Factor $ 2x $ out of the first two terms and $ -1 $ out of the last two terms.

$$ 10x^{2}-4x-5x+2 = 2x\left(5x-2\right) -1\left(5x-2\right) = \left(2x-1\right) \left(5x-2\right) $$

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