Factor polynomial $$ -5x^{2}+14x-8 $$

$$ -5x^{2}+14x-8 = -( x-2 )( 5x-4 ) $$

Step 1 :

After factoring out $ -1 $ we have:

$$ -5x^{2}+14x-8 = - ~ ( 5x^{2}-14x+8 ) $$

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:

$$ a = 5 , b = -14 ~ \text{ and } ~ c = 8 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 5 }$ by the constant term $\color{blue}{c = 8} $.

$$ a \cdot c = 40 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = 40 $ and add to $ b = -14 $.

Step 5: All pairs of numbers with a product of $ 40 $ are:

PRODUCT = 40
1     40	-1     -40
2     20	-2     -20
4     10	-4     -10
5     8	-5     -8

Step 6: Find out which factor pair sums up to $\color{blue}{ b = -14 }$

PRODUCT = 40 and SUM = -14
1     40	-1     -40
2     20	-2     -20
4     10	-4     -10
5     8	-5     -8

Step 7: Replace middle term $ -14 x $ with $ -4x-10x $:

$$ 5x^{2}-14x+8 = 5x^{2}-4x-10x+8 $$

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

$$ 5x^{2}-4x-10x+8 = x\left(5x-4\right) -2\left(5x-4\right) = \left(x-2\right) \left(5x-4\right) $$

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